COMPLEX YARIABLE

LrcruRE Norrs FoR A[ln 95n

P, G. SNFFMAN

-1-

l . Let a and b berealnurnbersr -€(a(o and _6(b<o I f i isan algebraic qtrantity satisfying the relation i2 = - I, then a = a * ib is acornplex nurnber. a is the real part of a (written a = Ra) and b is tfieimagina,ry part of a (b =Jot. rr F = (c + id), then by definition, o *g,-,'ft(a+ c)+ i (b+ d) and ap= (ac - bd)+ i (ad+ bc). The cornptexnumber6"$Khr*the commutative, associative and d,ietributive raws of algebra, and

""r, ot%''t*"

manipulated exactly as (a + xb) is manipulated, with the additional propertythat x2 = - I.

(a + i0) behaves in every way like a, hence complex numbers containthe real numbers as a subset. we rnake no distinction between (a + i0) anda.

r f ' x and y arerealvar iablesrthen z=xt iy isacomprexvar iable,

a=F isdef inedtomeanthatboth a=c and b=d hold, i .e. , everyequality of complex numberg involves two distinct equa,tions in real variables.

one motivation for the introduction of complex numbers iE that they allowthe polynomiel equation zz +'L = 0, to be sorved (its roots are ri). fuforegenerally, there is tlre following important result:

: Every polynornial p*(z) =

root ! satisfying p*((,) = 0.

If a = a * ib is a cornplex number, then & = a - ib is its @.

Two irnportant properties of the cornplex conjugate are easily verifiedfrom the definition: (&) = e, and a& = az + bz.

The modulug or absolute varue of a (written l"ll is defined as l*l =,r,/EE =k

va" + bz ' (The sguare root is always to be understood as a positive realnurnber when its argument is positive and real, rrnress otherwise specified,,i. e. take the positive root. )

NI"- rn has a7n

-Z-

Sorne properties of the absolute value are:

l "p l = l " l lp l ,l l " l - lp l l< la+ BlI f a is real, then l* I = a sgn a where sgn x =

€ and n.

0

0

fr I ir(

t - l i f

x

x

coril

scbct

z=;

mofi

th.t

r= |

mnffi

f t i r

-nf l |

l "q

intc;

intcf

rootr

I fo

are I

4.!

ceil

is er

2..

w = E + in is cal!.ed the multiplicative inverse of z = x * iy if

wz = I

(€ + in)(x + iy) = I

(€x - ny, + i(€y+ r1x) = 1

€x.. ny =1, €y+ W=0

and, is imrnediately possible to solve for

Alternativelyt wz = 1

1rZ\ l [ / |=c=-S

zr zz

a

;

l " l '*=f f i ,€=*F rr t=?t-+

The square root of i t w = oE, is defined by i= u . Then Ez - nz = * aad

2€r? = y. Solving these relatioasr w€ find that € = * /x +\/{ + va and

oiqo" l f f y)0, then € andIZ

n must both be chosen of the sarne sign, either positive or negative. If y ( 0,

then € and n rnust be chosen of opposite sign. In either case, there are two

possibilities for the square root. By convention, ff = lyl

A conaplex nunb€r, n = x + iy, can be

represented as a point in an rr-y plane,

where the real part is the x-coord,inate and

the irnaginary part is the y-coordinate"

The plane is called the Argand plane or the

Pr ftry)

0

0

-3-

complex plane, and the whole graph is called an Argand diagram. Ilt such

schernes, the cornplex nurnber z is a directed line segrnent or vector,

" = f,p. The di.fference of two cornplex nurnbers is zz - zr = pr*p, The

rnodulusis the lengthof the l ine segrnent: f " |=Op. 1", - zr | = PrPz.

It is natural to consider polar coordinates in the cornplex plane such

that x=rcos0 and y=rsin0. Then z=r(cose+is in0)where

, = l" l and o = arg z = tan-L *. Note thatarg z is undetermined to a

multiple of 2r because of the periodicity of the trigonornetric firnctions.

It is convenient to define -7t < atg z 4 1r as the PRINCIPAL VALUE of arg

and to write this principal value as A "g

z o

Clear ly ztzz = 11 rz I cos(Or + 0z ) + i s in(Or +

lrrrr l = l r , | . l r r l , and arg(qzz)=arg 21 +arg z

integer.

By induction it is for:nd that ,t = ,t(.os n0 + i sin n0), n being

integral. This result is cal led De Moivre's Theorem. Using this result,

roots rnay be written in a convenient forrn. U

un = o = p(cos O + i s in $)

!,= pn1"o"t**&r, I * i s in[**#, , , k= 0,L,2, . . . ,n-1.

rf a = 1r these are the n roots of unity. They 1ie on a circle l" l = I and

are separated from one another by an angular separation of zr/n.

4. Regions and Curves

The relation l, - ,o I = ",

describes a circle of rad.ius a having its

center at zo. I t is one example of a CURVE.

The relation lr - ,o I . ",

is the set of points insid.e t*re circle. It

is an example of a REGION.

Re zz = az irnplies that *z - yz = 42 . It is a rectangular hyperbola.

Re z2 >- a2 is the shaded region plus the borrndary in figure I.

ee )l and. l = f1"o" e -zr

z + Zkn, k being an

i s in 0),

x and

l

and

:y

rre two

rbe

E€,

eate and

ate.

e or the

-4-

The region defined by such relations

rnay be bor:nded or infinite. If the region is

bo.unded, there is an M such that Ir l . Vf

for all z in the region.

The region is connected if any two

points of the region can be joined, by brokea

straight lines lying in the region.

The region is open if for any zo in R, there

for each z sat isfy ing lu - uol a r , u beloags in R.

region is called a DOIvIAIN.

5. Mappings, functions, irnages+

727-+Ly z

FlG URE I .

isan e>0 suchthat

An open and conne

w:,r+!v l l

suppose that the poinls in two distinct comprex planes are associated

by a rule which maps each point of the first into sorne point of the second,

i .e. P*Q ot 2,+.w. 'We saythat w= f(z) where (u+ iv) = f (z) , u= u(x,and v = v(xry)' Then f is a fi:nction of the complex variable z. A firnctionrnaps regions of the one plane into image regions,curves or points in the otherplane. The rnapping need not necessari ly be one-one. For example, f(zl =

constant,maps the complex plane into one point. f,(z) = lzl rnaps the planeonto the non-negative rear axis, u ) 0. Except for the point z = L and w = -fi:nction w = f(z) =# is (r - 1). The point z = L is said to map into'

i l;.o*'io-"r:ot"'at infinitv) since l" ' rf - o implies rtrat ltt"l l + €, and

6. Bilinear Traqsformation

az+b\ t r t=-' - cz + d,

dw- b-cw* a ,4d-bc*0

that

uected

rted

t ,

= u(x, Y

ltion

other

z)=

1e

w = -1,

and z

-5-

This is a (1 - l) mapping completely deterrnined by any three pairs of

corresponding points. I t rnaps circles onto circles, straight l ines being

special cases of c i rc les, i .e. c i rc les passing through the point at m. The

inside of the circle tnay map onto either the inside or the outside of the

irnage circle. The equation of any circle or line in the cornplex plane can

be wri t ten as (1) Azl+E.z+ BZ + c = 0, where A and c are real and

Ac < lg l ' . r f A = 0, the c i rc le degenerates to a l ine, i .e. , the c i rc le

through the point at - .

In order to see this, set z =+. Then5

A*E+9+ c = oLEL E

A+EE+Bg+C[(=s

and z = 6 lies on the curve if 6 = 0 satisfies this equation.

In order to show that (I) is the equation of a line or circle, rernernber

thata c i rc le ofradius pt center zo, has equat ion lu _ ,o l = p,

i . e. (z - zo)(2 - Zo) = p, . This relat ion gives (1) wi th A = l , B = - "0,

Q = I"ol t - pz < le l t . For the case of a straight r ine, which is the per-

pendicular bisector of the l ine jo in ing points 21 and zz, l , - , r l=1, - uzl ,

whichimpl iesthat z(Zt -2"1 +Z(21 - zzl+ l " r l , - l r , f , = O.

References. It would be useful to read Chapter. l of Complex Variables and

Applications (Third Edit ion) by Ruel v. churchil l , Jarnes w. Brown, and

Roger F. Verhey. (L974, McGraw-Hil l) . Er future this reference is cal led

sirnply churchil l . Alternatively, read. chapter 1 of complex variablesby N. Levinson and R. Redheffer (Holden-Day I97O).

-

-o-

II. Elementarv Funstions of a Cornplex Variab-le

The definitions already given rnake clear the rules for rnanipulating a

large class of funct ions, e. g. :

(a) Polynornials, P(z) = ^nun

* arr-rzn- l * . . . * ao.

(b) Rational f irnctions, P(z)/Q(zl, where O is also a polynornial.

!(c) Fractional powers I P(z)] n of polynornials.

Now we consider other elernentary firnctions, the exponential, e*,

tr igonornetric f i :nctions (sin x,cos x, etc.) hyperbolic f i :nctions (sinh xrcosh x

etc. ) and the natural logarithm (1og x).

Def ine " '="*po="*("o" i* is iny)

where z=xf iy and y is

f ireasured in radians (1 radian = L}O/r degrees). (Radian measure is always

used in order to preserve the differentiat ion relations * "os y = - sin y;oy

d* s in y = cos y) .(ty

Then """u"

= "(* t+xz)(cos

yr cos yz - s in y1 s in yz]

+ cos y2 sin yr ] ) = "(zr tzz)

aoortroN FoRMULA.

Notethat z=x*iy=r(cose+is in0)="" i0. f f n is

then (" t ) t = "nu.

Thus (cosy* i s iny)n= (cos ny* i s inny)

rnole, De Moivre's Theorern is ver i f ied.

lzHowever, ("u)t rt

"t, for the left hand side is rnany-valued.

!xrc"F = .frl.o"t{333) + i sin d+.Lk")f .

0-<k-(n-I

The nth power of each rnember of the right hand side is "u.

Negat ive powers eze-z = ! . Hence e-u =Lr= (""1-t . " '

is per iod. ic.e

eu, for a l l z. Then the per iod is Zr i ' . luul = "*

> O for al l

and ^tg.o=y*Zl tk.

* i [ cos ys sin

an integer,

and, once

z*2r ie=

finite z,

- I -

xxu=e cosy, v=e slny, l * l =.* , argw=y.

rting a

dal.

xI

cosh x,

is

Irays

Yl

yr sin 1

! ! t

e

The mapping. * = " ':

Restr ict y to range - 7t < Y -{ f i . Then Arg w = y.

FIGURE 2.The horizontal str ip is mapped onto the whole w plane, and corres-

ponding points rnap as indicated in figure 2. On BF, x = 0, - ?r -(

y -(

r,

u = cos y and v = sin y. The w plane is not mapped continuously. There

is a jurnp across v = 0, u ( 0, which rnay be el irninated byttcutt ingrr the

w plane, i .e. by agreeing never to cross the negat ive real axis. Then the

strip is mapped continuously into the cut w plane. Sirnilarly, the strip

- e6 (x (oo, T 1y -<

3a' is rnapped onto the cut w-plane. A whole ser ies

of w planes is forrned, corresponding to the various str ips of the z-plane.

It is convenient to think of these w planes stacked together like playing

cards. They are called Riemann surfaces or Riernann sheets. (See

Levinson and Redheffer, Cornplex Variables, Holden-D.y, Cln. Z, section 2.

Henceforth th is book is referred to as LH).

Trigonornetric functions and hrnr€rbolic functions.-- .

CIearIy, . i " =.-Ylcosx* i s inx) and " iu

= "y1"o"

x - i s inx).ix -ix ix -ix

put I = 0, cos x =Lp. and sin x = u ii" . We d,efine:

^ iz -- iz - iz - - iz z -zcosz=" -=t

, s in"=" =r" Sirni1ar ly, cosh"=#,4'ZL

z-ze -esinhz = - ; - . I t is nowa sirnplernatterto ver i fythe var ious addi t ion

forrnulae.

iz . - ize =cosz+1slnz; e =cosz-L s: 'nz

-8-

Then

i (2, . lz ' )e " -a '= cos (z l *zzr +is in (zt+zz)= (cosz1 * is inzl)(cosz2 * is in

= " izt " iz2

"- i (ut+zz) = cos (zv * zz)- is in (zt I z2)=(cosz1- i s inzl)(cos z2 - i s in

= .- iz1 " ' i " ,

Multiplying out these expressions and, alternately adding and subtracting theglves: cos(21 * uzl = cos 21 cos z2 - a in z1 s in z2

sin(21 * zzl = sin 21 cos zz + cos zt s in z2

Note: Frorn the definitions, cos iz = cogh z and sin iz = * i sinh z. Hencecos iy= coshy and siniy= i s in l ry{and cos z = cos(x+ iy) = cos xcoshy

- is inxsinhy. Further, cos}z+ sinzz=Li coshzz- s inhzz=L. Al l theusual trigonornetric identities are satisfied if x or o is repraced by z.

f cos zf = (cos2 x coshzy + s inzx s inhz";*= lcosr* + " i r rhry;*

firrrctions tatt z, sec z, cosec z, tanh z, etc., are defined in the usuar

tanz=l i tu- . 1 tcos Z, sec z =

#?, cosec "

= #, tanh z = sin? z .

Zeroes-

sinx = 0 impl ies that x = nz, , n= 0, * I , t 2r . . .

sin z = 0 irnplies that sin x cosh y = O and, that cos x sinh y = g

Now cosh y >- r. Hence gin x = 0, but then cos x * 0. Hencesinh y = g. Thus sin z = 0 implies that 2 = r,Lrt n = 0, t l , t 2r.. . and al lroots are real .

Simi lar ly, cos z=0 impl iesthat z= (n+i , )o, n= 0r *1, * 2r ." . . .Alternative definition

For x real, the functions "x,

sin x, cos x are known to have convergentpower series (Taylor series) representations.

The

way:

I sLn zz

I SlZa Zz

rg thern

ffence,

rsh y

I the

z"

lsual

-9-

.*2 x3=I+*+TT+Tf*

.x3x5slnx=x-Tf.5i ' -

.x2x4cosx=t-7i 'TT

for x in these series provides alternative definitionsSubstitution of z

of a ' , s in z, and cos z.

s in z. Consider the str ip - 5.* . !Zr - 6 (y (o.

u= sinxcoshY ; v=cosx sinhY

x=0 impl iesthat u=0 and v=sinhy

7l* =; irnpl ies that u = cosh Y and v = 0

y=O irnpl iesthat u=sinxand v=0'

The l ine y = yo rnaps into u = sin x cosh Yo, v = cos x sinh y6. Then

==:ir= + -#

= r, an ellipse with foci at E and. Ef . Thus the stripcosh'yo s lnn-yo

shown in figure 3 rnaps into the whole w-plane. However, to ensure that

the rnapping is continuous, the w-plane is cut along the real axis frorn

l to o and from - l to - o (as shown in figure 3)'

z - p lone w-plane

gent

References: Churchil l , chaPter 3.

FIG U RE

- t0-

13. The logarithm is a many valued function, rike ,4, \,rle give

a meaningtoLogz, so thatwhen z= x) O, Logz= ! .nx is apossible value.

Define w so that w = log z means that "*

= ,. The function is defined for at z. since v/ =

-o covers the whore w-prane,

the equation z = ew always has a solution (notice the interchange.

of symbols). " ln

x = x , so the defiait ion is consistent. Letz =

"" i0 . Then "*

= "o " i t

= "" t0

. Hence l"o " i r1

= "o= l"" io1 = r .

t- Flence Log z = ln r + i(0+2rk) is a !n4Lvatuea runction. For real posit ive quantit ies, rn x1 + !.n x2 = .0n x1The same relationship holds for complex variabres but with

modif ication.

Special valges

Iog z1 * Log z, = log (2, zzl + hik.

log 1 = Zrik; log (- l) = ni + Ztr ik, for (- l) = .oL

log , =T + Zr ik.

Many valued functiong cannot be handled with ease.

geous to make them single valued.

Single valued branch

The principal branch of w = log z is chosen

-zr ( Im w qi fir and it is written Log z

Logz=lg- l" l + iArgz

(see section 3 for definit ion of Arg z)

The function is not continuous everywhgrs. It has a

arg z = v" In order to rnake it continuous, cut the

It is advanta-

so that

JumP acroas

z -plane along

e give

isa

he fgnc

'-plans,

lange

at

I

l?rk.

gL

= ln x1

lnta-

7tL

.11-

the negat ive real axis (see f ig. 4) . Then Log z is cont inuous and

single valued in the cut z-plane. The function Logk z = Log z+ZrkL

is also cont inuous and single valued in the cut z-plane. I t is an-

other branch. Not ice that the pr incipal branch and the next branch

Logr z are cont inuous with one another across the cut, i . e. the

l imit of Log z as Atg z - I t - 0 agrees with the l imit of

Logr z as Atg z - ' i r + 0 Not ice that the relat ion

Log z, z, = Log z, + Log z, is not general ly true. The funct ion

can be made single valued everywhere by introducing Riemann surfaces.

togk z ls the value on the kth Riemann sheet. Further explana-

t ion may be found in LR, p. 70.

14. Branch points. Moving along a circ le centered at z = 0

one complete revolution ( incrementing the angle continuously) leads

from one branch of the logarithm to the next. A single valued

branch can be def ined by any cut f rom 0 to oo because any cut pre-

vents arg z from increasing by 2r , so that arg z is uniquely

def ined in the cut plane. ,Figure 4 shows some possible cuts. The

point common to al l possible cuts is a branch point.

Theor ig inz= 0 is

consider the function

a branch point of Log z.I

* = zd which has two

As a further exampls,

branches,as

ng

E = lnr +rfi

LogS: Inr - t f

FIGURE 4.

r = rk(.o" I * r sin $) and rt = reos ($ + n) + i sin t$ + trlr

lxee=-;2 tcos.}+is i t i l . Anycutwhateverfron z=Q to z=@ makes

the function single valued. Then the branch points are z = 0 and z =

Alternatlvely, if z descrtbes a circle of unit radius, center z = 0t

beginning at g = 0 and ending at 0 = Ztt, the function varies con

from w = *1 to w = -1. Thus, going around a branch polnt leads to a

neru branch of the function. In thls erample, moving from 0 = 2n to

0 = 4n returns the function to w = 1, which is the value on the orlg

branch. Ttre two branches of thls function may be rePresented on Lwo

Riemann surfaces. SLnce the tnro branches are contlnuous with one another

across the cut, O < O < 2T" 0 ( f (or thesurfaces arecross jo ined

across the cut. The function Ir is continuous on the resulting space

-L2-

(Churchi l l p" 305).

Mul-tivaLued functions need not have branch Points.

Log ez = z + 2tr Lk has neither branch points not cuts.

J

Ttre funcElon

15. Complex powers.

It is natural to

- ^cl los l r l ^L a atg z=ee

define poerers by the rule "o

= .o Log z

. In the case of rational exponents' o = t

,

thls definitlon agtees wlth pre'vious ldeas. It follows that the values

of o, log z ate sorne of the values of Log za. It also presenres the

usual rules for manipuratJng exPonents. rn particular, ("o)B = .9 1og

= .o B Log z = ,o3. Notlce, however, that alL these expressions are nolt

many-valued, and equality means a branch of the l.h.s. is the same as a

branch of the r .h.s.

16. Inverse trigonometric functions.

since w = silr z maPs a strlp in the z plane into the whoLe

w-plane, w = sin z has a solut ion for al l w (sect ion 12).

ctz

Define w=sin- l t

-13-

to mean that sin w = z .

The function ig nlany_JelEg because sin (w * Zrl = sin w .lv/ -l\I'

s inw=.e-e,T,L-_Ziw_-r2_iwT=u.

' I ' t ren e-- -Zize'" - l=0

Solving the quadratic, eiw = i z *. , / f f i i f . 'z is real and legs

thanuni ty. Forcornplex z, . i *= !z+ (L-zzl t , whereI

(1 - ,z l2 ig two-valued. Thus .r / = - i tog { i , + (1 - " , )4 + Ztr k.

I

Now l i " + (1 - ' r , )z l + 0 or 6 for any f in i te z, and the log

function has no branch point. To make the inverse sine single valued,I

we must make i z + (1 - zz )z single valued. Now z ie single

valued, but (1 - ," l i is not. arg (1 - , , )* = arg (1 - z t* *11

arg (1 + ?)" . (1 - z lz has a branch point at z = L whi leI

(1 + z)z has a branch point at z = - t

L-z = l r -"1" i (n+0)

(1+z) = l t+"1. i+

(L-2ft= lr-z ,*. i l !z*31*tno, b=o,l.o. . .

( t+r1i = l1+zl*" t7*t*o

I( l 'z)z is made single valued by cut AB. Then o < + 1 zr The branch

withk=0 gives ( l -zy*-- , at z=0(where g= t i . Thebranch

with k=l g ives (1-z;*- l . at z=O (where S=zr) . The

function (1 + 4* is made single varued by cut cD. Then -r1,{1n , At z

k=0 gives (1 + r ' )*= I whi le k=l g ives ( I+ r l+= -1. TheseI

cuts make (1 - ,2 )Z single valued. Note well that only two branchesI

(not four) resul t . arg ( l - zzlT = 4+ 4* 9f kr where k = 0,1,' z 'z z

FIGURE

in plane cut as in fig

An alternative method ofas in f ig (4.2),

-14'

(4.1).

introducing cut does not work. If plane is cut

plane. But now the argument of the logand sin-1 z is not single valued,.

i tg, l= 0 is a branchpoint . However,

Strict ly

1

and ((z: l tz

\

r (z)

f . (z l = (ax-by) + i (bx+ay)

{IIIl

sPea

have the same mult ivalued behavior. In this case, the branch point

at oo can be forgotten. The function is inf inite at | = O. This

type of singularity is cal led a pole.

I8. By def ini t ion,

i v(x, y)

i v(x, y)

i v(x, -y)

i v(x, -y) =

+ib then

I r i{*$+ikzr+*(L-zz)z= l t -z} lze

z " , - r1g1n, - t1 l t1r

has continuous argument. in cutbehaves l ike Tiz for large z

f . (z) = u(x, y) +

-t(z) = u(x, y)

f .171 = u(x, -y) +

f (Z) = u(x, -y)

f . lz)=qzi a=a

L7. poi*-=!----eg The poiat at € is investigated by making the

transformation I"

= t

and examining the behavior at I z, O. Thus

w = log z = logf = - Iog E. There is a branch point at (= o

Hence, z = 6 ; said to be a branch point of w .

Consider w = (1 - ,z) i = (1 - l1 l . - ( r 'z-r l*

' (Lz )z

F16URE +.2

E:rample.

- 15-

6

t(zl

T@l

=ez=(a- ib)(x- i

=ez=(a+ib)(x- i

= i , = (a- ib) (x+i

v)

v)

v)

(ax-by)

(ax+by)

(ax+by)

(bx+ay)

(bx-ay)

-1

+i

Log

+ i ( -bx+ay)

Further exarnple of branch points and cuts.

Consider

w=[z(z-1)(z-+11]

l l lars w =i^rez+|ztg(z - I l+|arg(z -4\ . l * l = Vt l ' f l " - t l l " - + l j .

int

t

C denotes a sirnple closed contour. Iarg w]a denotes change

describing C, when arg w varies continuously along C. If

z=0,2= I or z=4r [" tg*]C=0. I f C containsonerand

1rcints, [ "tg *] C

=

The points z = 0r l r4 are branch

points. w can be made single valued by

introducing cuts so that [ "tg w]

"

= Zkr

for all allowable C. The cuts canu.be

introduced in an infinity of ways. They

in alg w after

C does not contain

only one, of these

?branch poi:otsaeed not be straight lines but usually are for convenience.

are points corrunon to all cuts. Exarnples of possible cuts.

(i)

However the cuts are

'w=

is single valued [.rg

points such as 0r l in

I

The

introduced

\/rqrT;;r, I ei(e+++rr) / z+it<r k = 0, I

*JC = 0 or 2n, the lat ter i f C encloses

(i) or Qr4 in ( i i) . For ( i), we can take

two branch

r191T,

-15a-

- t <S ( n, O < (r < Zr. If the cuts are curved as in (ii), there may be no

sirnple expressions for the bounds on the angles.

3uppo"" f(21 = li, then f(-1) = -i\6d' for case (i) aBd f(-I) = i,gg

for case (ii), as can be 6een by evalrrating changes in 0, S and (/ in mo

from 2 to -1 without croasing a cut.

i

- 16-

UI Continuitv. Convergence. and Differentiabi l i tv

Gonsider a funct ion w = f (z) = u(x,y) + i v(x,y) def ined as

a single-valued function in some domain. (Remernber, a domain is

an open, connected region. )

19. Lirrr i ts. u$of.@l = q rneans that for every e ) 0, there is

a 6(e) >0suchthat l f (u)-al (e wheaever l r -"ol< 6 I f

zo is the point at s, the limit means that lt("| - a | < € when-

ever l " l > N , for some N which, in general , depends on € .

f(z) is continuous at z = zo if l im f.(zl = f(zo )z*zo

These l imi ts obey the usual propert ies: I f f and g are

continuous,

lim (f+g) = lim f + lim g ; lim fg = (lim f) (lirn g)

Continuity of f implies continuity of Re f, of Im f, of T, of

l f l , etc. , but the converse is not necessar i ly t rue.

ZO. Clggggerc.l Tf. on i" I ""quence

of complex numbef,e,

a +aasn+oo meansthat lo--ol<e whenever n>N(e)nn

{"J isa@if la--a- l *0 as n,m+6 A- l l l l ! f .

sequence converges if and only i f i t is a Cauchy sequence (see any

calculus text. )

ZI. OifferentiaUiUtv A function f(zl is differentiable at zo if

r \lirn lf'(zL

- f'tz"\l exists. The limit is called the derivative and ie

z-zol z-zg I

writ ten f ' (zo) or df /az with z = zo understood. Alternatively,

wr i t ing z=zq}: ,zo-2, t im{4:- I l3.) . =! tzt=4

t-A h -L\2 ' -dz

-L7 -

txamole: Let f . (z l=v. Since %rr=t=t , ; | t=tconsider f '(z) = zn . By the binorniar theorem, (z + h)n = uo+n,

- G*l) ,n'2 yz + + hn . Therefore

# = ouo-l .

consider ez ""*on- "o = .u "h-- I

Let h = a * i b. Doing

the ser ies expansions, .h =."("o" b + i s in b) = (r+a)(r+ib) + o(az,h=1+h+o1trz1. Thus +"=1+o(b)*1as [*0, . Thedzz

E e = e , and the usual differentiation relation holds.

The erernentary propeg:ties of differentiation can be verifiedd

f ; ( f + B) = f ' + B' , $ trs l = f ,g * s, f , * ( l = + - rs,

"!u - 4z ^Lz , -rY

(? = t

;? '; etc'

Since sin z = g-2f , cos

" =

t ' . , and, d: | = i "Lu,

i t follows that # sin z = cos u .oa

$ cos z = - ein z

22. Chain rule: Suppose f.(zl and, g(z) are differentiable, thenF(z) = f le@)l has der ivar ive F'(z) = f , I e@)l g,@1 , or , i fw=g(z), then S=4!-Sg,

dz dw dz

23.

Most comprex functions are Eg! differentiabre. Letf '(zl = u(x, y) + iv(x, y) I f we write down a generar u and, v, assrnooth as we pleass, then f is not necessari ly differentiable. Foexarnple consider f.(zl = y - iy = Z. f (z+h) - f (z l = a- ib

This is g$_cont inuous at a = b = O . , I f

-1. Thusb = 0, the ratio is +l; i f a = 0, the ratio is

t-

- l im f, does not exist.h*0 -

(C auchy_Riernann conditions

rhat lirnh*0

o* = ty and oy

ls written above

Let

* a.v *

8u0x

Notice

bi l i ty. For

f , (z+h)- f (z). - i -_*- u= v+a,

1y

+

u(

a

- 18-

ib. Then, since f =

+ i {v f a.v + b)-va+

u(x' y) * iv{x, y), t '(z + hL ', f '(z) =

. Taking a to be zer.o,

= v" - iu". Taking b to be zexo,

o* * ir*. If a derivative exists, then o* * ir* = v" _ iu*,

= - t* must sirnultaneously hold. Clearly, all the

must exist if the ftrnction is to be di.fferentiabre.

Example. " '

="*"o"y+iexsiny

"* "o" y = # r # = -ex siny=- 3:

that this is not yet a sufficient condition for differentia-

example, f (z) -#(1 t - - i ) , - - f (1 - i ) , f (0) =o Atxo lYo

E=0, o*=t=tr , oy= -1= -vx. St i l l f r (0) doesnotexist .

Sufficient conditions

The Cauchy-Riemann condit ions, o* = v, and *y = _vx , aresufficient for differentiability if all the partial derivatives involvedare continuous. For, i f f irst part ials are continuous at a point{=,y) , then u(xia, y+b) = u(x,y) + aux + bu" + e1ff i andv{x*a, Y+b) = v(x, y) + avx + or, * ,r, /W where €1 and 6, * oas h* 0. Then f(z+h) _f(z l =au** bur+ i (av*+bvr)+ (e1+e2)fhf

= u*(a+ib) + iv*(a+ib) + (e1+e2)f f r fusing Cauchy Riernann relations. Thus f, (r) = u*+iv* - ,o,

=o=- iur=ty* i t* .

- 19-

24. Analvtic functipns. A function is ANALyTIC in a

if it is differentiable at every point of D, A function

throughout a region is also said to be REGULAR or Ho

One very important constraint upon a regular function is that

and imaginary parts (u and v) satisfy the cauchy Riemann (c -R)relations, u* = v, and oy = -vx , throughout the region.

that al l second order part ials of u and v e=ist. Then

o** = ty* = t*y = -oyy, i'e' o** * oyy = 0 and t** * uyy = 0

In that case, u and v """ "lrr"d

HARMONIG functions, and each

said to be the harmonic coniuqate of the other. This relationship

conveniently writ ten using the Laplacian operator;

Vz u = o, Va v = o, where Vz = # - #

To investigats the analyticity of a rnultivalued function, like

log z, i t is necessary to speci fy a branch, e. g. Log z (see sec-

t ion 13). Then, real abd imaginary parts may be identif ied, in this

case u=ln (*r+yr l t r , v=Argu=tarr- l : . Final ly,

xu-- = # = v,- anci u-- = + = -v-- , hence ihe c-R relat ionx x- f} ' Y -y x.*yo 'x '

hold and the f irst part ials involved are continuous. Then

f' (zl = ff i ,

= , This is true for each branch, and ws can

therefore write in this case # Log z = , . The derivative is not

def inedat z= 0

25" Complex ExDonents

aV-- a -1=a;=dz

domain D

differenti

its r

dotdz

=# (eologur=!u.eLogz

20-

Although the forrnal manipulations of differentiation are

same for al l branches of the funct ion, one part icurar branch

used consistenf ly, throughout the calculat ion. with that bi t

and use of the chain rule, al l d i f ferent iat ions are rout ine.

Example. f (z) = s in-1 z

often the

must be

of care

its re

(c -R)

rPPose

I

ach is

hip is

ke

sec-

this

t ions

oot

. -1 , ? lsin - z = - i log { i z + (L-"211

d-1; |s in- 'z-

- i d t '

i z + (r-zzF i? t i ' * (L-" . ) l =

f - i r r) - l , r -a.)=-r[ iz+(r-zz1i JL $_zz1i J e_zz) i

once again, remember that the same branch of each functionis used throughout a calculation.

26. t""""". f"""t i* I f w .= f(z) is analyt ic and if f , (z) + O,then z = g(w) is also analyt ic, g being the inverse of f . whilethis result is intuit ively reasonable, i t does reg*ire proof. Notethat Aw * 0 as A,z * O , and assuming (as is reasonable) that6z- 0 as Aw*0, then umhf l =I i rnr#r, i .e. f , {z l=+

(ffi) s'(w)'consider *=sir- l , , z=sinw.

*=cosw=(l-s inzw;*= $-zzyiIt cannot be overemphasized that the branches of each function mustbe used consistently throughout one calculation.

zL-

tV tta"+"nic functl

Recall that a function Q(x,y,z) is harmonic i f i t satisf ies

Laplacers equat ion, Vz 0 = O, where Vz = 0 ' Az Az' 6F"WrW

In two dimensions, there is no z dependence, and Laplacef s

equat ion becorn, az 6 az aes # r ff

= 0 This class of functions is of

immense physical i rnportance because many potent ials, electrostat ic,

gravi tat ional, magnet ic, etc. are harmonic.

I f f . (z\ = 0 + i + is analyt ic, then Q and r! are harmonic.

The converse is more di f f icul t to prove. I f V2 Q = O, there is a

analyt ic function, f, whose real part is 0, i f i t is possible to

conetruct a functioa, +, satisfying the relations {* =-0, and Sy =

Then, bydef in i t ion, f = O+iq isanalyt ic. sornerestr ict ions

uPon the domain of definit ion are needed if this converse is to hold,

but we shall assurne any such condit ions are satisf ied, as they are

in norrnal cases. The function + may be constructed by integra-

t ions on a horizontal and a vert ical l ins, viz. Q(x,y) = r] l(xo,y) -x^y

lo+rt* t ,y) dxr andg(xo,!r)= S(xo,yo)+ {r . O=t*,yr)dyr. By construc-

t ion, r ! r i ,y) = o*(xo,y) - , [ j * r r ,* ' , r , dxt = Q*(*o,y) * / - ] *--

0*(x, Y) - 0*(xo , y) = S*(x, y) ; and

_Fr6uBES_21. other coordinate svsterns: For polar coordinates, X = r cos

y = r s in 0 Consider e(r ,O) = Q(x,y) and e(r ,O) = e(x,y) . The

C -R relations for ig and e are found from

6 =8Q..8x*90fu . \y, - -&.8=,O_U.Oy*r - ox at ' ay F ; i [e =5; ;e r ,T f f i

o" = (cos e) ff + (sin e) $f, i vo = -(r sin r) 3* + (r cos ol

l-(",t)

-zz-

Then Q, = i *, , and sirnilarly

i oo = -ifr . Since

v,o = *r", # (" 3f) - l BrF = Q, i .e. #- i3*- f ,#=This is the two dimensional Laplace equat ion in polar coordinates.

28. Equipotent ial l ines. The equat ions {(x,y) = constant and

*(x, y) = constant define two famil ies of curves in

the x-y plans. A part icular member of the family

of curves is defined by the part icular value of the

constant terms chosen. Furthe", ,O"aslopes of

these curves are, respect ively f i =;J! and

rL 'y

f i = t ' . However, from the Cauchy-Riemann'v

relat ions, r , ,=F=-5=-. ,1 - I - - , ,6=-q=7q=-t

andthe

\ur ltwo famil ies of curves are everywhere orthogonal. The curves Q =

constant are often called equipotentials, l ing of force, streamlines, etc.References: Churchi l l , Ghaptet Z, Chapter g .

29. General orthogonal coordinate svstems.

f = u * i' is a cornplex firnction. suppose u and v are given

not as firnctions of xry but as fi:nctions of general coord,inates Er\ii .e. u=u(€,r ' l ) , v=v(t , r ' l ) where €=€(xry), q=r l (xry) . Howdowe

find if f is analytic?

suppose the coordinate systern is orthogonal, i.e. the curves

€ = const. intersect t ihe curves I = const. at r ight angles. Then thermetricr is given by

0

of

at ic,

ic.

is an

;o

lr='vs

r ld,

re

a-

FUc -

t(x

5_s0,

le

dsz = dxz + dy, = hr rdt, + 1.zz dqz

-ZZa_

where hr , hz can be fo'nd in terms of the partial derivatives of €

n w.r.t . x and y. The Cauchy-Riernann relations are

10u10vE; En',= - E; 5'9.

These can be verified by direct substitution (laborious) or by taking a

local cartesian coordinate systern 0'x'y, with origin at (€rrr) and,

paral lel to 11 = const. and € = const. Then 0/0'd = (L/ht)o/al, etc.,

and the result follows.

If 6 = € + il is itself an analytic firnction of z, and dl/dz +we have the important special case of a TRANST'ORT4A

Now h1 = h2 = ldg/dzl-r , andthe G_R relations are invariant.

10u 10vE-, E'g = E; '56;

*0,

AT

'23'

V Compt.x lnteerati .

. An integral in the complex plane is specif ied both by the form

of an integrand and by a path or contour of integration. Define a

curve, C, parametr ical ly using the relat ions x = g(t) , y '= " t ( t )

r -< t -( b where I and q are piecewise continuously differentiable.

The slope of such a curve is *5 = I . I f . S(x, y) is a function,ox: | .

tben by definit ion I C dx = / O[g(t), q(t)]

*t dt andcc

b

/"+ av = I: + [g(t), q(t)] *l dt. rf + is piecewise continuous,

these integrals are well-defined. Thus 1"t1"1 d* = .f u dx * i

/ vdx is well defined. Further, I"nF't dx + i t.t lz)dy = I"t@l

(dx+idy) = I . r@)dz=/" toa*-*,dy)+ i / " tvax+ udy) = f te l# dt

Many well known propert ies of the Riemann ,._-1.

integral are t rue for contour integrals; Simple, not c losed

[laf.(z) + ps(z)] dz = " J t1"ld,z + gf s1laz: n

)/

[ -" tp1au = - !"e{4 au where -G denotes A

integration along a contour in the direction simple' otosed

opposi te to that for +C.

A contour G is simple i f i t doe " g_

not intersect i tself (except possibly at the ./

end points, t = a and t = b), i .e. Not s impLe

Fig. 6z(t l = z( t r ) impl ies that t = t r

Figure 5 i l lustrates some curves.

30. Integral of the derivative. Suppose that f(zl is ananalytic func-

t ion in a domain D, and that F'(zl = f(z) where f is continuous

in D. For any contour C joining the points zr an.d z",

-24-

J"t(")dz = F(zz) - F(zr) . For, i f F = U+iV, f = U** iV* = Vy-

Then !"t1"1a" = I"{ {u*+iv*)dx + i{vr-iur)dr} = /"{u*ax+u,rdy) +iJ"(v*a* + vrdv) = [u] Z1 * rt" lZ - F@) -F(zr ) . rn part icurar, irC is a c losed cbntour,

{ t@)U, = 0. This i " %,

proved here for functions f which are derivatives of analyt ic funct ions. cauchyrs theorem is actual ly t rue in generar, as 10ng asf is analyt ic in D, which means only that ft exists at every Fand provided certain connectedness propert ies of the contour aresatisf ied. As a coro*ary, notice that [: ' , f .(z)dz is independent rthe contour C jo in ing zt and 2".

cauchy proved the theorem with the assumption that f, (z) icontinuous. Goursat removed this restr ict ion, requir ing onlythatf '( ,) exists. I t wil soon be demonstrated that the derivative of aanalyt ic function is i tself analyt ic and it is therefore redundant toassume f ' (z) to be cont inuous. Three proofs wi l l be out l ined herFor fur ther detai ls, see' Ghurchi l l , pp. r15-rzr or LR ch. 3.Lemma n) lf rlttat | -. f I r1t1 | at if F is comptex.Proof: Define a real posit ive number J such thatrroor: uerlne a real posit ive number J such that

., l"bfat = J"iO.Then r = .["b F"-igdt = .,["bn"{r"-ig)at + i /"rtr*

""lir)or" (since)dt" (Since

J is real , I :

b l t t1F.- i0;dt = o). Hence r - . . [ f ne re-rr lur .

1l: : : " , lR" F.- io l - . lFe- ie l = l r l ; thererore r = l fbrary_.

/"ol" lur .

Lemma(2) l !"r@1a"1 < rur- where lr l _< M and Llength of the contour of integration.

Proof: | ["rp16"1 - l/"b{o+i,r)(i+ii16s1 -- / ro+i*,r. ri*i i

is the

M/ l i+ in lat .' I g+ittl dt -.

t/ -iuv

+

if

! ,

irnc-

I

rPo

!

Proof of Cauchvts Theorem (1' | . Consider a

sider a domain D containing at least one

every straight line connecting zo with any

bas all points between z and zo in D.

star domain, i . e. con-

point zo such tlE t

other p6int z ia D

Here are some examples

-25-

However, ds = .,/(?m = (Lz+n, )i u, = li*ii 1at, and then

.f"o l i* i i lat = / ds = L. rhus l! r1"1dzl -< ML

AStar

o

Take, without lose of general i ty, zo = O (i f zo is not zero a change

of var iables 21 2 z-ze wi l l make i t 2936. ) Congider F(tr) = I f f ( f ,u)a,

where I is real. I f f hag a continuous derivativerdiffersntiat ion

and integration commute , then * .,,0 f.Qtz)dz =

"6 * f.Qrzldz

= f zt'0"2)dz = ,Ft^L * {"i l \rt qrJdz = - I .,,0 r.()tz)dz. rhus,

#= f i lx4au-f t1r ldz= e ThenF=constant. However, F(o)-0.

Hence F(1) = Q, i .e. , Sf . , : . ldz = O.

rt of

1S

t

Ian

to

tere.

A convex domainIt must be star.

Not star

Proof 2: Greents Theorem states that

are continuous, then for the area A

contour G,

Not star

if the part ial derivatives

bounded by a sirnple closed

ce

Consider

/"/ €l - fi)dx dv = [{oar + vdx)

f t1"7a" =15 {u+iv)(dx+idy) =16 toa* - vdy) r i f (udy+ vdx) - -//€*+$sr)dxdr

r i l l€:- f f r*ur.

-26-

since f is analyt ic, u and v satisfy the cauchy-Riemann rela

tions. Then both integrands in the double integrals above are iden

tical ly zero. Thus f tp1a" = Q

Proof 3 (Goursat) f t is not assumed to be cont inuous here. T

theorem is provgd f irst for a tr iangle. Bisecting the three sides

triangle ABF and connecting the respective rnidpoints, f igure 7 i

obtained. {"t?)dz -- ! nOt@)ar, where C indicates tr iangle AB

FIGURE 7while 4L indicates the integrals taken separatgly over tr iangles I

through 4, and summed. Then I t"tl"ya" | _. 4 | J "rr@)a"1

whe re

Ct is the triangle, i = I through 4, giving the rnaximum value of

integral. Repeat the process of bisecting the sides and conneeting

midpoints for tr iangles 1 through 4 and for their ,progeny'r down to

th" nth Eeneration. Then lf "t lryaul

-. 4o l[.ot{4ar1 . As the

number of b isect ions, n, goes to oq the tr iangles coverge to a

point, zo

Since f(z) is differentiable at zo, for each 6 X 0,

isa 6>o suchrhat lW -f , (zo) l (e whenever

Now take n so large that Co is contained in lr_r, | < 6. '

f ' (z l=f{ro)+ (z-zo)fr(zs)+ (z-zsla, I r l (e . Now I"dz=n

there

I u ' ro

The

rdJc d"

nA

J._(A I dz = o , and I" zd,z =nn

( , zz) dz = O Thus

1 rela-

e iden-

, The

t ides of

eTis

I ABF

les I

9re

of the

ting

rD to

I

toa

here

z-zo I

Then

i l " f.(z\dzl < en

longest s ide, ie.

_27 _

( rnax l ' " -uo l t . l I az | -< 3 sn

s," "o=7.

Thus,

sl where "o

is the

l !" 16 dz | -< E . Hence

- -n 4t

Therefore l^ f , (z)dz =JC

= l - ' t@)a, whereJZo

the integral is to be taken along the straight l ine segment fromz+h

zo to z" By the resurt just proved, F(z+h) - F(z) = . f , f (E)dg

Hence t@fl@ - r(z) l = l l , f t*nt tw - r@)ld6l -< max

l r tEl - t@)l . However, f is cont inuous since i t is d i f ferent iable.

Then as h + O, l t tg l - f (z) l - O Hence F'(z) exists and ig

equal to f(zl. Then ftp1a" =

,Frt (zldz = 0 , for al l contours

inside the star dornain in which t(z) is analyt ic.

The restr ict ion to star domains is unimportant. The result

holds in any sirnplv connected domain. A domain is simply connected

if the interior of every sirnple closed curve is in the domain. It is

'obvioustthat a c losed curve in a s imply connected domain can be

decomposed into a sum of c losed curves, each of which is in a star

domain. Rigorous proofs of these robvioust geometr ical resul ts are

hard. For fur ther detai ls, see LR Ch. 4.

To surnmarizs, Cauchyrs theorern states that i f f (z l is ana_

lyt ic evervwhere inside a simplv connected domain, then f "r@)dz

= g

for every closed contour C in the domain. An immediate corol lary

is that F(z) = [ '^tp1a, is defined in the dornain independenily of theuao

path f rorn zo to z and F,(z) = f . (z)

I J " t1")dz

| -< 3e . However, t is arbi t rary.

Now consider a star dornain. Def ine f(z)

0.

-28-

Examples Evaluate '= f r i -1". Parameter ize using s = aeio,

lz l=a0

-< g * hr . Then dz = i2" i069. r= [_ 'o {""- io1 i . " igd,e = iaz fzt'ro -- Jo

= Zti a2. Because ; ig not an analyt ic function of z,

Cauchyrs theorem does not apply. tvaluate I = .f * U"

l " l=^'= Io'"

!i."i 'do =

'-ri. since i is an anlyric function (its

derivative being '$1, i t may appear that this integrar shourd bezetot by Gauchyts theorem. However,

, is not analyt ic at onepoint, u ' 0, within the dbntour, and Gauchyrs theorem does notaPPly.

Evaluate r = j uior. Ghoose the branch so that ut="i.i*lu l=" '

where -f i < g -< r. Then

= t '3/ ' [ ' , - i ] = - + aIit"i"t8l i""igdo = i.3/, I!r"r*,

I=

E/z

because of the branch point at

I

zz is not analyt ic in l " l _( aI

z = 0. zz ig not €vea analyt icthe pr:nctured domain obtained by

"e*oving the point z = O becaueezz is not single valued, unless the plane is cug, in which case thecontour ie not croged. (rf the function is made singre

'arued using

Riemann gheets, then the eontour consia.""a is no longer a closedonei but it would be closed if the contour went twice around the ori-gin and it is easily verified that the integral taken between the limi

The indefinite integral can

Consider the expreegion f " &-JrC

be used to define analytic functions.

= F(z). This defines an analyt ic

0 to 4r vanishes. )

function if z is inside a simply connected domain @

)

Id0

29-

coatain z = 0 where the integral is siagula,r. Directly from *5

dcfi.ition of f dF I ', E =

; and F(1) = 0. This is an alternative

dcfinition for the function F(z) = Log z. If the simply connected

do"nain is forrned frorn the entire plane with a cut along the negative

rcal axis, F(z) = Lcg z. By allowing the contour to go around the

origia, other branches are obtained, for F' + = 2zr i N, i f the- l I

coGtout goes N t imes around the origin in the anticlockwise sense.

@f is called the winding nurnber. )

3a. Deformation of contours

If f(zl is analyt ic in the region included between two open

contours, C1 and C", jo in ing zr and z, then by Cauchyrs

t tcorem, l_=, f . (z)dz + L", f . (z)dz = o ,'JztJzz

- " 'z nzzi . - e. J

" , f . {z ldz =

Jr , f . (z)dz.cr cz

f f f(zl is analyt ic between two simple closed

C.2 , where Ct is inside Gz i. then

"fi,"ll

6 r.@)d,z= 6 f lz)dz"cl tz

For a simple closed contour may be

f(z) is analyt ic inside this contour.

forrned from

Thus

contours, C1 and

czrAB, - Cl , BA.

Not€

33.

that f(zl need not be analytic inside

h, - fB r I -crr f f t@ldz = o

ct.

Cauchv's Inteqral Forrnula (C. I. tr. . )

Let f(z) be an analytic function inside and on a closed

contour C. I f

z is outs ide C,

€l | \

# is analytic

.

I

-30_

z is in c, then f(z)-#tVy,

I tF=0. For i f z isnot in e,

and

then

inside C, and I W ug = 0 by Gauchyts

I f z is inside C, then Cauchyrs theorern al lows the contour todeformed into a circle of radius 6 around z. Thenn €l r \

{ ffi a6 = \"t I # - I W d6 = zri t(z,t n I€#@ u,lg-Zl=o

E- " lg-?l :u

by the above integral evaluation (section 30). Irowever,

the circle. Then

may be arbitrary small

on

E

l lW d6l - . / |W lasl . Given r>o, rherelE- ' l=5

a 6 > 0, such thar l l lg;- f1zyl < e

{ FWI ag -. f zoo = ztrt. sincelE '" | -= 6

the conclusion fol lows.

34. Consequences of G.I .E.-

Gonsider the possibility of differentiating under the integral

t ' *+.") =#,6f { t rh6 #} as =# f f f idz+# f CffiLTj . The second inregral is bounded as h + o,

d{-_ I r lLt)d"

= d I ffif

ae. It now follows by "epe"ted differentiation

d.nf n! r €(rrAr

al l orders inside D-

Further, i f f(9,)

n! Mp- wnere M is the

is analyr ic in lg-"1 ( R,

rnaxirnum value assumed by

then

f in

l r (o) tu) |

D.T

dif

€n

_ 31-

is cal led Cauchyrs inequali ty.

Liouvil lers theorgm. If f(z) is bound.ed and analyt ic in thewhole z plane, i t is constant.

By hypothesis, l t@) | < vr. By cauchy,s inequali ty, i t fol lowsthat l t ' (" ' l t a

* where R is an arbitrary (very large) radius.

Heuce f ' (z) - O. Therefore t(z) is a constant.

corol larv: I f l r t " t l < M lu lo, then f(z) is a polynorniat of de-Bree at most n. For 1t(n+1)1211 .

* .

Thus, ENTIRE functions, which is the name given to functionsaaalytic in the whole complex plane r &!e identically constant or un-bouaded at €. txamples of such functions are polynomials,z

C , S1n Zr etc.

Morerats Theorem (Inverse of Gauchyrs theorem).

If t(zl is continuous in a domain D and. I"tk)a, = O forsvery closed contour C in D , then t(z) is analyt ic in D.

Take a poinr zo of D. .fefine

F(z) = Lirtflug. It is singlevz' lued' Also t ' (z) = f,(z). Then f is the derivative of an anary-t ic function and is, therefore, analyt ic by C.I.F.

A weaker version of Cauchyrs theorem.

For Cauchy's theorem to be true, i .e. / , t lzyaz = O, i t isqrfficient that C is a sirnple Jordan curve (i.e. thetrous and does not interseet itself, and that f(zl isdo'n'in bounded by C and continuous onto C.

contour is contin-

analytic inside the

heo

to

3la-

A general proof is hard. rt is not hard i f we suppose c is a

star contour, i .e. there exists 0 inside c such that al l l ines from

0 intersect c just once. Let z be on c, and take L < l . Then

f qxrya, =* j f r f tz)d(rz) =+f r@,)dz,= o,-c ^t n"chr '

for the contour on which zt = lyz is inside c and is therefore a c

Ieven though zz

z = 0"

on which f is analytic.

If t(z) is continuous inside and on C, it i" gg!{gggg}yco3glg

, ' i .e. g iven € ) 0, there exists 6 such that l t . . ) - f (z ' ) l < e

whenever l r - u, | . O, wherever z maybe on C. I t fo l lows (see

Ch. VIU on trniform convetgence) that

-t irn- S i lx"ya"= S q4du,)r. * I -C "c

as difference can be rnade less than e L. Hence right hand side is ze

Example (i)

r !I zzdz = 0,

fz- t f=1is not analytic at z = 0, because it is continuous at

Exarnple (ii)

r = d @z - . r t |a,-c

The integrand is not analytic on the

sides of the cut, - 3 -( x

-( ELr y = 0+

or )r = 0-, but it is continuous onto the

cuts. Hence

+1

I=.1 [ (*- i0)z--1

+1=-zr iVA-z

-1

-1_1

azlzdx+ I [ ("++-1

Ii0)z - . t ]zd*

1Sa

from

hen

-3 lb-

c, enaluating the appropriate branches of the rnultivalued fi:nction1

1* - azp which is equal to \ffi? for z = x. ) a.

Note that for l " l large,

(zz - az )l = "$ - Z:)i =

"G - * 2i * oQo)),

tr the binornial theorern. (This will shortly be proved for cornplex

re.r iables). Hence

t= { uau-} ' { *-" ,#)=- iaztr .fu l=n l , l=n

Tberefore

+1('

- = 't:22

J \a ' - xo dx =T .. I

This is not the easiest way to evaluate this integral but illustrates how

complex fi:nction theory can be used to calculate real integrals. Many

Gtle examples will be studied later in what is known as trhe Calculus

of Residues.

_32-

35" Tavlor ser ies (Power ser ies. )

' The expression f . (z) = 7o * ar k-a) + a2(z-a)z + . . .+ an(z.Sr+ = ) , a- , " is cal led a power ser ies in z about z = e.Tn

Without loss of general i ty, take some new variable s = z_et

consider power ser ies about z = O, i . €. r set a = O

A serieg is abgolutelv convergent for specif ied z i fS I nrL la- z- l0

Let "r ,

= .o f ar z t . . . + ao zn and let ,o = l "o l + l " r r l +

l.o "o | . Then | "r, -

"* | -: ' Iro - rrrrl Frowever, by the hypo-

thesis of absolute convergence, there is an N(e) such that Iro-,whenever n, m ) N(e) . Hence

lo is a Gauchy sequence and the

series has a f inite sum. Notice that convergence does not imply

abeolute convergence. I.or sxa,rnplg, IJiS converges whilesl I

L ; is not convergent" Non-convergeace means the series diverg

If a series diverges, the sum of the abesrute values of the terms

also diverges.

36.convergesfor l " f <nconvergence. Oir the

and

converge or diverge.

Example. I # rt

I r" is

$znL; ts

convergent for lu I -. I and, divergent for l, | >

convargsnt for lu I a I and divergent for I ul ,-

convergent for l= l - . L, z* l , and divergent for

l " I r r

(X

-#

is convergenr except when g = 0, I ft!9is convergent. )

'gent.

zl +,

hypo-

f to-t*

d the

rPly

,le

iverge

'm9

eseeofitFr

lu l > r

" l>r

I for

" l> L

I

,33-

Pu'oof: Suppose the series converges for , = R.io. ?hen there is

r l ( such that l "o rol < r , s ince th" oth term of the ser ies ap-

grurches o. Suppose zr = Rr . iot . Then l^o , , o l _. K ,Hr" "

Morcver, i f gl ( R, f, t$t"a * . Hence the serieg converges ab-

*lf fcry i f l" l < R by cornparison test. convsrssly, suppose the

rcries diverges for some z such that lr l = ft . I f i t also co1_

E:rged for any z such that l" l = Rr ) R, then by the previous

lumlt, the series must converge for l"l = ft. By contradiction we

oooclude that i f a series diverges for some l" l - R, i t must then

d,vcrge for al l z having l" l > R .

The radius of convergence may be R = o as i t is forq31 = f nt ,o, which converges only for z = O. The radius of

ffiyergeace may have any finite value, or it may be €, as it is

fur fht= t znu

" t (which converges everywhere).

Ir ' Tests. Except for delicate guestions of behavior on the circle

d convergence, i t is enough to check for absolute convergence.

[}Alesrbert 's test.

u ) N, where N is a f ixed number. The ser ies diverges i f

.au+l "o*I

,;-

--

I > I for aII large n. If lim =o-i

-B z- ' rvr d"!r rarEc

n*6 % exists ' then

n =J-urr, (#l

This is the most useful test. I t fol lows from corn-

n

gnrieon with the geometric e.eries.

A rnore general test is Cauchv's test . A ser ies converges

"" a -ro*LA ser ies converges i f l * I -" la f for a l lazn

-34-

Labsolutely i f l i . t l "o rolo ( l , for then l " - ,n l < ( I -e)n for

tsuff ic ient ly large n. The ser ies diverges i f l i * l " r , u* loo r ,

for then ^o

,o doeg not approach 0 for al l large n. Hence

L, |F

R l im la- l^ '= l ,I l'1. I

r lil

l "ol . ( fr) ror

The ser ies I

whenever l " l < R

(1 n - l= L Era,-z" - . The

rnust be the sarrle as

_1 _ Iand n = E l .o l

n = t i rn inf l "o l o,

i .e.

n > N(e)

n^o "

= tlz) represents an analytic function

Consider the formal derivative of f dt'dz

radS,us of convgrgence of this derived ser ies ,

the radius of the or iginal , for l im lo a I - i

, n|I-L

t tnl " l t l srnce However, i t remains to establish

I t ' l

-RI

lim

that

slJ0

| ("+

.r, Rt for some R'l < R whenever lt l < R _ lz,

ch a forrnal differentiation is valid. Gonsider I = +J - 4f.

nr l lnn =In-€

a-(z+h)n - a nz

h)o- ro- nh

n(n-1)z

Convsrsely, every

sion (a Taylor ser ieg).

that l " -"1 ( R, i . e. ,

d ius R centered on a.

uo-Ll * g l l ln l , t l " l + l r , l l " -2. rhus l r l - .

n

Az dz

Using the binomial theorel

exPan

z

of ra

lu 'o

su

J.L - o

"o uo- l }

's test

ash

$lr0

hy

0

Cauc

I*

quarantees the convergence of this f inal series, thus

+ o" Hence rim $JIaI = X r E- ,o-L .t 'n

analytic function has a power series .

Suppose that t(zl is analytic for all

there are no singularit ies in a circle

Then t(z l - $ r( t ) tot t - - -# trt F-o)o ror

-J5-

By Cauchyts integral formula, f . (z) =

c{o}(,)=#j, f f i

*"* f t =ff i=*{r*frr(frt" *the rernainder is

J-2ni f ts

lE-" l= pp<R

* (ffit" *

and

ool where

l rn

A - (r-ot1+1

--n (g-o)o+1(6-")

Thus

r (z)=frW*#I , f1$-drt" ' l r-"ol=o

(6-a)n*L (( 'u) - t

The rernainder terrn

r n*li lRol . { ,T-f

-#*} .*o as o+eor s ince pl= lz-al<p. g=minlL-"1,

M = srax l r fst l . Hence, the ," t t "" is convergent.

radlus of c

a-

r f

t'aritv' the radius could' not be larger because the series gives an analyticfrnction within the radius of convergence (provided, the singularity is isolated).R-'nples; The function log(z+1) has a branch point at z =.-! .

rt t)trt - (-t)k+1Gt-ttt

. Then for the principal branch, rog (r+z) =(z+1)

t - t - .L-24- , . T - ? + , which converges for l " l<

For al l z, " '

= L + "

, " 23

, o,

" =r+o*T +3T+ . For

l ; = I - z * z2 - z3 + z4 - zs +. . .

1.

l " l (1,

r (z l

k*n

and

38.

e@) is analyt ic

fore there exists

_36-

that

o@-aln = "k+l(r-o)k+I +

= (r-o)k* l [ "n*t

= ("-o)k* l sn

^k+z7-o)k+Z +

* "k* r?-a1 + a*r(z-alz

Notice

k- Ia

t

a convergent .ser ies for lu-ol < R. Direct ly, l "n*r , l -*

-+ r+E -r r^ | n+kr. 'u"k+ol

- '^ l * Thus i f f i l "n*rr l "=i i f i r - ' - iTE11 *6 o-* lok+n I

o. Hence t(z l - { O,r-a)n = 1z-<z;k+l g1z1 where e@l

e@) is analyt ic for I " -"1 < R .

%i"

={l

t0

is

39.

Let f (z) = [ ^o

,o for l r l < R . Then the ser ies F(z)

a "n*1n

lT- must also have a radius of convergence R,

anal f i ic for lu l < n. Since Fr(z) = f (z) , F(4 = foz

and F(z

f (E)dE .Zeros of analvt ic functions.

An analytic fl-nction t(z) is said to have a zero at z = e,if f(a) = o. At such a point, the function may be written asf '(z) = (z-a)m g(zl where m is a posit ive integer g@) is anal

in the domain of definition, and g(a) + 0. One immediate conse_9uence of Taylorrs theorem applied to f is tbat

f(*)(ol .g(o) = Lm !

ofThe function

D. There-

le(z)-s(a) | <

and

a

therefore continuous

6 > 0 such that for

in a domain

each t ) 0,

_37_

ff ienever l ( r -o) l <6 I f .6< , then gV) cannot vanish

z)=

{O. Uniqueness of power ser ies.

suppose that f(zl and g7l are anarytic in the domain D.(ffvcn a sequence { ro} with a limit, e, in D such that

f i l tol = g(zn), then f = g everywhere in D. For, consider

Ffr ' ! = f-g. F is analyt ic in D. and has zeros at , = ,n SincefrhG { tJ have a l irnit point, two zeros that are arbitrari ly close

oory be found. Continuity guarantees that F(a) = 0, but a is noten isolated zero because there is no neighborhood of a in which

s(r) + 0. Then F(z) = 0 in D and f = g, This resurt guaran-

tcee that if two functions are equal in any region, then they rnust be

cq*l wherever they are analyt ic. This provides the basia for the

tfraory of analytic continuation, which will be discussed later.

Era'srple: f ' (z)= s inzz+cosz zi E@r=r. when z isreal ,

l l l- c. , - xt f(x) = g(x). . Notice

that it was crucial to the der:ronstration that

ectuUy contained in D. Thus, sin z = O

"n = olr. However, i t may not be asserted

nhcre because { "J has no finite point of

Corollarv. I f f = I ao ,n = I br, "o

for

e linit point, a, in D, then .o = bo =

{1.

suppose that a function f is analytic in a domain D and con-t iauous in D. Either f is a constant fuaction or lr l in D isrlrays less than its maximum, M, which occurs on the boundary

the l imit point, e, was

at points of the sequence

that s inz=O every-

accumulation.

, = ,n, ana {zJ has

r(o)(orn!

38-

of D. suppose that l t l = ff i at z = e insid,e D. f(a) = J_2ni

I # dz=* I r 'o r1o*0. i0)ae rhen r=*1zor(** .opiorl r - " l lO

- t t -o - ZrJo t(al

-1," id= t Jo pe'Ydo where lpl < I becauee the maximal modulus

attained at 7 = e. Taking moduli, , _. * Ioro ,u,

can only hold i f p = I since p is continuous. Then

, -2t, =

* 1-" . iOde implies rhat , = * fo'o "o" o de, which can

only i f cos S = l , or 0 A O. Thue f,(zl = f(al on the circle

l" ' "1 = 6 Hence t(z) = 11q1 everywhere.

Let Re f = Re g on a c losed contour, C, the boundarysome domain D. Consider h(z) =

" f -8. l f r l = "Re(f-g;_.

I

However,

the re

everywhsre in D because

sider#="8-r lo l=

the value of lt I on C is l. Gon_

"Re(g-f) - . t . Then l t r l = ] and h =

I f ( r ld( 1Jr -€ I , where tr

O and of radius R in

or f-g = i b. Thus f(zr is uniquery determined to an addit ive cstant in D by its real or imaginary parts on the boundary. Theactual deterrnination of f is known as a bound,ary value problem.Example. Let the function f = u + iv be analytic for Im zand suppose that f + O as l " l - 6 in yn z

suPpone that either u or v is given for y = 0 and for al l x"

rhen r1z) =dirt {+ r * f(()d( * J_.q \ zTtL ,_R l-z Zf i

indicates the semicircle centered at z =

uPPer half plane. Since f * 0 as l" l * oo in yn z

r im + f I IEI{E=o, and r(z l=_t i rn{u*, f"rFLqtR*o LnL ,r ( -z " '

4 ' g**. zzi J_* E

-39-

+ ro f l()d( This is an exarnpre of an irnproper integral de-nd ,_- (-z

ref r

:lation

rn ho

e

rry of

Con-

h=e

e con

The

em.

r,

x.

I

in tbe usual vay, as a limit. since z Ls not inside the con-

of integration, f f((Ld( = [* &)j,!. = o Hence, raking

'c E-; '-* l-; €1rr l4.-conjugatgs, * t ?F=o rhen r(z)=#i f f

**- I W = Ln

{oog similarly, (subtracting 0 from r)

rF) = * {_H

raking the reat parrs,

r fo .nr(()d( - I ro (( -x lv(()d(u(x,y) =; J-*dlejryz =; /*ffip

Srhs-a.rz's Lernma

The following result is occasionally usefur. suppose lrt"l l -< M

o l t l = 1, f (0) = 6, and t(z) is analyt ic for lz f -< t . Then

' let" t l 'Mlr l for l " l . t

i : f@) = Nrueio. The resutr fo l tows frorn the

G.ct that f/" is anatytic and lt/zl < t 'A on l"l = t.

-40-

42" Let t(zl be analytic (and therefore single valued) in thetured diec, 0 < I"-"1 < R. That is, f , (z l exigts everywhereexcept at e, where it is not defined, like the functions i andz

e at z = a = 0 . There are three ways in which f(z) can behave.

(i) Removable sinqularitv: A singularity is removable if) ry [ @-a) f , (z) l = o , or i f l1" l l is bounded by some conM, as z * e . Consider, for example, the function f.(zl =

Although its behavior i" onl.tecmined. at

rto be 1, then the function f,(4 =l L . if

I Slnzl -\z

For, *= r-**

Consider, f (z l=# Lvl

+F *{ f l l ls , wherecl -z ?f i J" . b a

z = 0, i f f (0)

\z= 0 t) rft

otherwise fJ

Cz has radius pz and Cr has radius

Pt, Pz

Lett ing pz * g, the second integral vanishes

since the integrand is bounded in any neigh_

borhood of d and the contour length vanishes.Then t1z)=f i ; f (C)d( i| ,", F

for z * a. However,

analytic.

FICURE 8.

this integral

form defineg an anarytic function for alr z, including 7 = e. ,r6ning f(e)=! f f ( ( ld(r-, Zri J l i , the singularity is el iminated. ..g. (( i i ) Afunct ionof tbeform f , (z)=f f i , g(e)+ 0 , g@r analis said to'have a pole of order m at z = e. I f rn = t,

4t-

i r cal led simple. A necessary condit ion for the presence of a pole

1g thrt , l ! Ia l t t r l l = 6. That is, for each N > O, there must

- - ista 6>o suchthat l f t r l l >N whenever o<1"-"1

h order to show that this is also a sufficient condition, let F(z)

I= tr1 . F is analytic for O < l"-"1

1[t1z; | )N, | r (") | .* , i .e. , | r (") | i "boundedin0<|u-o|<

Then F(z) is analytic in 0 -{ l"-"1

Tberefore F(z) = (r-o)^ G(z), G(a) + O Then #,

= e(zl is

r',arytic at z = e. Finally, f.(zl = j l4.* . By this, the sufficiency1z-a)

is dernonstrated.

(ii i) lf the singularity at z = e is not removable and not a Ooti,

i t is said to be an essent ial s ingular i ty. One example is ' f . (z) = "v

LVrX

#1. "* **o i f x+O+ and'

is bounded for arbitrari ly small y.

lcos; f7- i sin

Iiy

rtant,

sin zz

lefin

ytic "

De-

(sia

nal

=e

Lx

C

I

+ 0 i f x+ 0 . e

1

t- "*

* oo for any m. This wild betravior is characteristic of.

essential singularit ies as fol lows frorn the result:

Picard's theorem: In any neighborhood of an essential singularity, a

fuaction takes on every possible valus, except perhaps or1e, an inf inite

nurnber of t irnes.

While the proof of Picardts theorern is diff icult, i t is easy to

show that the function comes arbitrarily close to a given value an

infiaitg number of t imes. Consider a circular neighborhood of radius

p about an essent ia l s ingular i ty at z = e, i . e. , consider l " -"1 ( p .

IFe show that for any constant, c, and any E ) 0, there is sornepole

42-

z in the neighborhood for which lt@) - c | < t . Suppose the

0< I"-"1 (p ' Then f ' (z)= !n "

couldhavearmostapoleof

assert ion is fa lse. Then there is

for al l z in l " -"1 ( p Then

an r suchrhat l tk, l -cf >

0(z) = =+r tz)-c is analYt ic in

is cla

rq)

+ has a zero of order m in the neighborhood.have 3rr essential singularity, the result is establis

$ These are also cal led essent ia lsingularit ies' For exampls, cosec (L/zr, has poles at o =

*, aa non- isolated essent ia l s ingular i ty at z = O.45. Point at o. The bbhavior of a function f(zl at €f ied by making the transformation

" =

| an. considering

E+0.

FIample. ( i) polynomials, pN(z) = "o

,o + * ao "*(l)"N* "N- l € + . . . +

"o EN

EN

oN(6)=

l- " Then

"*(gL) has a pole

of order N at ( = O. Then it is said thatorder N at oo .

PN(z) has a pole

L( i i )

" t =

"E has an essent ial s ingular i ty at , = 6t as doe

sin z.

r f l ra l +

at € and f .(z) =

i l r '1 - g '(z l

trn ' ' (z lz

constant * 0).

Meromorphic funct ions: A furiction whose only singularities a

order m i f

f is known to

6 as l " l + oo ( for a l l atg z) , f (4 has a

'^ g(r) where s@) is bounded at oo. If

has a zeto of order m at 6 (again, g(eo) =

46.

Pole s rs a meromorphic. function. If t(z) is meromorphic €ver

4"3 -

including z = 6t i t is rational, i . e. f .(zl = f i{ f ,+

where P and O

are polynornials. rf t(zl had an inf inite nurnber of poles, the poles

would have a l imit point (possibly at oo) which would be an essen-

t ial singularity. Hence, f(z) has only a f inite number of po1es.

g@l ._**"-rr , *1(z-zr)+. . . +a

-r(z-r , ) - - l+ (z-zt)* f r ( r )At a pole, f .(z) = a. =

(z-rr)^ (u-rr l^

rf

s

she

rnd

)=

oIe

aa-rn m+I

rT

( r - r r )^ (z-2, ) - -1

._l is cal led the residue of

t ion F(z)= f (z l - I t -3 * , - . . .* " t t - l lpole s (z-z

l " ' 1z -z r)

The coeff ic ient

f inite plane, i . e. , i t ism

o( lz l *) where **

is a polynomial. Thus

47. Laurent expansion.

annulus Rl

vergent for

l " l ) Rr"

an entire function. At €,

is the order of the pole at

f . (z l = P(zl /a@1.

Consider a function f(zl

determined by the integrals

a- l+ h(z)

z'%l

at zr . The func-

is analytic in the

F(z) is

oo. Then F. (z)

con-

convergent for

annulus Rt0*

eqser ies ) l a ,o

Tnbn; is absolutelyz

analytic in

$bn/ , - ln th

LZ

s absolutely

an

e

l " l<ar, and tI

The coeff icients are

1 f (( td(

,n+1

_1n ?ri f

cfc

and b Et-I r (6)aE, and c may

44-

be any closed contour which encirc les the or ig in once. By cauch

theorem f(z)=5 6 +l ! l A, - I f l ( r )h, i t rF oE-fr

€, !?-dE. For 6 on

and lu lcnr, *=tQ*t*F*." . ) . For 6 on cr,

l " l )Rr,* = - i ( I+*r** . . . ) . Resurt fo4owsasfor

Taylor series. I f z = 0 is an isolated singularity, then t(zl i

analyt ic for 0 < ' l r l ( R, where R = the distance to the near

singular i ty. Then for O < l " l ( R, r(21 = | ,o ,o , where-oO

^ = I AI IEIto =

"ri f ;fi-

dE This is the Laurent expansion of f, andI

is unique. If t "o ut = f bo l,l

z- = f,(zl holds at a set of poi-oO -6

having a l imit point in O < l" l ( R, then "o

= bo. If "o

= 0

all n ( 0, the singularity is removable, and the function rnay bedefined to be analytic. If.

"o = 0 for n ( -m, the function has

pole of order m. If there is an inf inite number of "o

having

then z = 0 is an esgential singularity.

The coeff icient of , in the Laurent expansion,l^

"-1 =

f f i f f (g) dE, is cal led the residue. For exarnple,Iz, I I

e =L*; + ** .* , hasresidue l . Forasimple

pole, the residue can be fouad easily by sett ing "_l

= l irn [ (z_alf.(az+Q

If t(z) = ff i

and h has a simple zero at e, this relation be-

comes "-l

= ff i

. If the singularity is a pole of order *, ' ' t

_45_

4g. Bample of a Lauts4t txpansion: Expand the function t(z) - #,

in a Laurent series about z = 0. The simple pole at z = 0 has

a residue of -1, while the simple pole at z = L has a residue of

l . The function is analyt ic in the two dist inct regions O < l" l <.1

and I < l " l

ponding to the two regions. Consider f i rst the region 0 < l " l < 1.

€qh. l rdzf . (z l= I " . ro

where "o=*

P + . However, by-oo

n 'T*I r lar ,n* "1r-r1

the Cauchy Integral Formula, # I ft, = # #

L r o^d' '( ; t ) l r=o= -1 i f n*1)o For rn7 o, f f f= o,

since the integrand is analyt ic inside the contour. Thus

If . (z l = - i - t - z - zz - zs . .o, for 0 < l " l < I

Now consider the region 1<l" l < oo. + Fhri I" lrt un+z1r-Ly

L r dz L r dz (+i tn i l2of . ,

zt i J,=o ,n+z1,-L1 Lt t t t =L ,ot"( , - t ) \ )

Hence a_-=0 for i> -1 anda-= I for n(-1, and f . (z l=nn

.1 I+7 +.. .+;+. . . , l " l>1z

In practice, we would use part ial fractions, writ ing

I7

1z(z-L)

l ' l < I

Iz- l

for

1I=-;* f i

I, f r = l+

11= .T'=;

"(L-71 '

l " l > I .

z+zz+

=, (r

I=z

For

Fpr l " l ) I ,

which converges

t\(-)r - -

2

*L*4-*zz- . . . ) ,

;Ii

49.

as shown in f igure 9,

inside the "qo."",

fN(z)

-46-

nitc number of

ffi , which has simple poles with

= rwt. Let f, ,(z) = cot z - fl IN' -ft z'wr

Let CU be a square of s ide (ZN+l)r

is analyt ic inside and 611 CN. For

Consider cot z =

residue of +l at points z

= cot , -L - $ "' -LNm

fN(6)dEOn the cont6ur CN, lcot z IL(?z)

FIGU.RE 9

=* tanh- l tAt . Atsozz 1N+j;

' N+* '

'

(K tanh =Klog

- r .e I l . rhus lr*(z) |

f f i -ocotz=lnT;#

l " l < r /z (N+i l , . Hence lL -JtLt zZ_

2N when z is on CN. (If x (( 1, tanh -l 1t_

( K, tog N on C*. Then l f* t " ) - fN(o) | <

as N*o. Since f*(0)=0, fN(z)*0,

Integrating, I,

" - I l dz = 169

" '##?=ro' e-#l,s lnz. - |t r ) for lz lcwr

rN(z) - rN(o) = f UN(E)d€(# - f ,,

=z F-cN

( M, for I cot z1 = t :o:1.T + sintrz y . I i' ' (coSh"y -coszx)f ,

which is l ike I on the horizontal sides and 1iketanh y on the vert ical sides. Further,

ror zonc*, l I ; { : ; r l . * IL- . + f* 9"7t- u (N+*)z -ot -- ,ra J o (N+j;z -oz

for lz l < n* and Io' ,"o,

.ith

;

l+L)tr

orz

[1-x)

-47-

Hence, los (ff1 = F

log $- #) = roe il (1 - 7ft1 " rhe

val idi ty of interchanging operat ions in this way is assumed"

rhus Y =T (1 - f i1 . Boththelert

the r ight hand side of this equation are entire functions,

the aquali ty must hold everywhere. Sett ing u = t

=T (1 -#,=fr %, whichshows

I = ' - . T (2n)

1J. c-6. q-gL5 T7 B . . . , a resul t g iven by Wal l is 300 years

hand side and

consequent ly,

shows that Z

that 5

ago.

fi

=#

Inf ini te Products

6

P = I I (1 + a_) converges i f1n'

NPN = I I . (1 +n=r

oO

Then I ro*1

The product

convergence

for absolute

arr)*P as N*o, where pre O or oo.

11 + arr) converges to a

converges absolutely i f

impl ies convergence.

convergence is that XI

l imit log p.

00

f t (1 + l " - I ) converges. AbsoluteI * '

The necessary and suff icient condit ionl l

I " r , I converge.

-48-

50' Let c be a s imple c losed contour, and suppose that f (zis analytic within c except for a finite number of singurarities at

'= oj" (Note that a l l these singurar i t ies must be isolated because

there are only a f inite number of singularit ies.) The R.ESIDUETHE.REM states that Ia t1"1 dz = zri L (residue at e,). Drawij J 'smal l c losed contours

" j , each enclosing only ong

Cauchyts theorem guarantees that

L t@)dz = l , [

" . r(z) d,z, but by

JJ

s ingularity, a.q

i

Laurent expansion

f,(z)dz = zf i ( "_t) j ,

[ . t1"ya, = Zi l

Example: co =

circ le of radius

t1 zz 24r r - ;e 'F - . . . ) -

Having expanded about

I (a-r ) .j 'J

Lro' tTi f .@ az,

R > l" l and centered

The simple pole at z

FIGURE IO

where the contour c is

at the origin.

= + ai has a residue of

- iat

tf r+

residue at z =

the

1..J

formulas,

and

iat

fu. The sirnpls pole at z = - ai

double pole at the origin has residue

has residue ft I ; ' . The

##, l ,=o=*.Then

'=* Alternatively, doing the Laurent series

expansion, , ""z-( \ = =# Q+zt

. . . ) -#rfound that the

nffn

II

-a 'z ' +zt+

it is

(1

0,

(fi c;v6

. q, :

t

6

ta- , as before.

Y, and

As

-49-

beforg,

t= -;'d

the residues at the sirnple poles

sin ata3

rrnation

al on the

1o=t @+

=Zri I

is ,

a

a

;&frI

illlh-fr

,ffi,OEffi

, = " i0,

de = dz/ iz

uni t c i rc le, l " l - l ,

l l when l" l = ! .

n2,, =

J o f (cos g, s in O) dO. The

gives the equivalent contour

since cos o = ip + l ) ,

Hence t = , f r . ( !@++f ,* pr lnl r l=r

( resi

7r=rr0

Par a1- affi ) - e^r+ 1+ afar;;r)hcf

FIF

-1!Bt.t

dues).

dz=i"6

7*

= f-" adoro zaz+r-coi [ - =?

t

-

v r z(zaz+r_ j1u+!t t

6r=dzt @

. The integral has sirnple poles at

2zz + I t ZJffiF . Supposing that a is real, it is immediatefor the posit ive root, l" | > t, and only ths singulari ty atZzz + 1 - bFTir lies within C. The resid.ue is

=-I .fifff- Thus , -

(Zfil(Zail

-Na4 +az

ecos 0

I,

ff i, =rlo'o;*ffi-zf az -t p-llz)

as i t stands, a very messy calculat ion. The denominatorfourth degree polynomial, and the result ing calculation ofis labor ious. Al ternat ivsly, I = f

o ,^dQ == =ro ae+ff

Zrcos (n0 sin 0) d0. C onsider

-50-

2*! t l

r - f "CoS0l= | eu0

integral . I = F

?*- !-n!

- i (n0-sin 0)

fu- tu. rn* l - n!

real part is the

cos 6e - cos(nO

d0,

Then

whose

Ztr

Io

reguir

sin 0)d

52" ,soJ-* f{*;a*" This is an imprope

integral , formal ly def inedR

be gl1!g J-* r{*)a*, whenever that lirni

exists. A contour as shown in f igure 1l

is used. We require that f("1 have

only isolated singularit ies in one of the

half planes lrn z >z 0 or Irn z < 0 and

that f(z) have no branch points in that

a1

to

half plane. Suppose it is the upper half plane,

residue theorem gives that Ia f,(z)dz = J* f(x)dx

t t l : R

FICURE I Iy > 0. Then the

* /* * r,(z)dz

lu l=nThe usual circurnstance is that l r tot l =0(

= Z'r i L (residue)y>0

as lu | * * in the half

IrR

R-z*0 as

= kr i | ( residues)y>0

Examoler f* - ' *, -* 1f t ' Consider

poles at points where 24

r of interest . Then l /*" t (z)dzl

^6RHence j*r(*)a* = #[r*.In f (*)a*

f (z) = #, which has simple

= ei(tr+ Zkr) , i. e. , where

plane

i (#Zkr lz=e,

these poles are

i7r'4z=e

I

;fo- '4e- 4

=- l

i37r-4r€ residue s

respect i

- ; lL - , 3r

r € -4

, e 4

The

I I

, and

4e '4.3r-4

" ' t t '4e-LT 4e

- 51-

f i ,* l tgglsf fng over the contour shown in f igure 12, the

tbe large arc vanishes since

r{z)=0t*) as l r l ** .

contribution over

FIGURE I2

Th"o I" # dx = zri. I-a

( residuefor y>0)=+

- i+ - ;z( r + +e-4,) =+ (-z is in i l=*

Wccice that whenever f(x) is an even funct ion, i . e. , wheneverf i i r ) = f ( -x) , that /

* r1=;a* = + I : f (x)dx. consider, for exampls ,

trh€ Fourier i ^'ntegral I _ f f i dx, where raf is a real number.-oo

Tbe integrand is not bounded for cornplex arguments of large modulussruce there is an essential singularity at oq Then the obvious ana_fi 'ysis using residues wil l not suff ice. Notice instead thatI

"qry' dx = ps/a# o- = {: # dx , since /-ff i dx='-0O -oO

-€

i l ln js odd integrand is integrated over a symmetr ic inter 'al) . Nowiaz , -. r"? I = e - t , which is bounded i f ay > O. For

.a ) 0, chooser ' contour l ike the one shown in f igure 12. Since .?ra1 = O , t ,L+za-Jt f f iz) ,frf t 'c large arc gives no contribution, and r*

" iax - -a'

J-- F dx= ?r i+

= tre-" = ,[ : f f dx . For a ( 0, consider the contour of

ffigu're 12 reflected in the real axis (in order that ay > o). Then,,8 ^iax +a

.-- i;;? dx = -zri { = re' = ,[: ff dx

ltrhire this method of contour integration is often useful, it

-52-

is not always applicable. Consider a continuous, integrable real

funct ion, u(x) . suppose, there. is a funct ion f (z l , analyt ic for

^oo#iey ) 0, such that Re f.(z) - u(x) as y * 0 . Then Ji3rb

R"J J(z)d- oo,f ig

00f,= J-* u(x)dx . Howev6r, i f t (Q = Q ,th, as lyl * € r as is

often the case, the integral over the large arc wil l not vanish, and

method wil l not work. For example, we cannot use sirnple contotrr inte

gration to evaluate I* "-* ' dx .

-6

Example: Consider I: dx The ess ential singularity at

oLzf e dz.

. f -6Z

sin xx

oo prevents closing the path. Consider instead Im

This al lows closure of

leaves a singularity on

avoided by indenting a

contour of f igure 13 is

the path by a large semi circle at 6, but

the path of integration, at z = 0. This is

small semicircle around the origin and. theLZ

used. By Cauchyrs theorem, O = I"T dz

Rix- | e dx+uE x

' { " " ,=*

iz -E ix iz= dz+ f-

e_-- dz+ | e-- dzZ J n )( Jt I Zlz l=e

=Ziiz

feJtr-zlz l=x

R

I6

\ , t=,

sln x ctx +x az+[, tlz l=e

dz. l imt*0

. i0. 'nt dg

ize

z

Rr s lnx

utx

Rr s lnx=

Jo : dx' lirnt*0

LZe d,zz

=I imi foE*0

ur l = -in

figure 13

ea1

for' ie

f . (z)dzFit

as is

and

r inte-

'a t

z.

but

sie

;he

.dz

I,

\<

-53-

The integral around the large semicirc le

Jordanrs lernma.

52. Jordanrs lernrna: I l " t ' l la ' l .l , l=R, y>o

1t

oae

, 2R0,(- ; ,

< -tr--zR

to the

f t* o

€axConsider I= f

e' ' f -6

ex+I

gral exists since the integrand is

the contour of f igure 14. The pole

l ta t r "^ '

, ax-e"- . l , =-dzl-<zre

-Bc eo - ! "*_1

vanishes by virtue of

7r. l l " i " l laul =nl , l=n

e-R sin o de = zRfoz . 'R sin

zg

1tale

Hence

- zRg

. -R sin 0-<

exp

- .-*J

lemrzra

lau I

7t do

For 0-<0-(

,||

. Then fZ

Q. E. D.

ny sin o

e-R sin 0 U,

16 sin xJ--

---

clx = 77

There : t re a few standard methods for doing

grals" When these fai l , there ar" some useful t r icks

appl icable. These are now i l lustrated by exarnple.

LZpresent exampls, i f T d" l

| , l=R, y>oas f t+oo. Hence

contour inte-

les s general ly

dx, whe re 0 <

o,. -x( l -a) , as x

at ,=ni g ives

+0asx**o.

a ( 1. The inte-

+ -€. Use

a residue of

azl f * -a, 1-DA.o-1

axfeJo"

"E crx +

ax\< zr* - 0 asx,

e-r

{" #-, u" =t€

I-6

X4-e.

axF-^2t ia1-. t ia? dxl [ l - e I = -zlr lee - t

00 axThen f

e dx

, I X,-00 e - I

'54-

='rZtLr- f i

e-1trta-e7r'l 'a sin zra

f igure 14

contour integrat ion can be us& to show that I* " ' * ' dx = r [n,

-6

the *ethod is t r icky (see courant, carculus, vor. z p. 56r) .

Given this00

=J cos-€

the contour

R"io de +

resul t , i t is possible to calculate the Fresnel

x2 dx (it is not obvious that the integral converge

R"iode+oasR*€

(apply Jordanrs lemma). Then

FI6 URE I5-'.z dx + fo

(t+l!y'2

'Rz

tegralr I

C onside r

r4 'zz,e1

'tl

r4 -22,e1

of f igure 15. 0 =

ro t t - i ra

e dr.

.f "'"' dz =

{* "-*' dx +

z

o= [ ".rO

R(cos rz - i s in rz)dr.

{ ' t { "o, * - i s in xz) dx = ( l - i } , /z

J"R

^66Thus, J cosxz *={ s in* *=uft

53' Dir iehletrs Test: f t ,*, g(x) dx existg i f f is continuously

differentiable and monotone decreasing and l/*t Ut*) U*;x1

0

< Ad for

55-

constant M.

one important application of Dir ichletf s test is to integrals

of the form .,f* r1*1 "o"

kx dx, f ,,*, sin kx dx, "oa f

f(*) "ik*d*.

These integrals converge if f(x) is proport ional to any of the func-t ions *-" , ( log x) *- t , or (1og x)-s for large x. To ver i fy thetest , def ine G(x). = ! cf i t dg. , f , , r , c(€) a6 = [ f (g) c(g)I*

a \r , _\ : /J

a

xx- I r ' (g) c(g) d€. However, l l^* r , (g) c(g) dgl < M l r ( . ) l * oaa

as 3 * oo and x * oo . Thus the original integral converges.54. Cauchv principal Value: Suppose that f(x) is continuouspoints except * - :rr where f = o . , [o tt*) dx e:cists as an

a

at all

impro-

per integral if -ta i -f(x)

dx exists. For examplg, consid,ert*0 -a+t

'11t=* L5=z[* i1

'= z-E+-zasr-0. r f thediscon-

t inuity occurs at a point intcrior to the range of integration, the re-

sult is similar ' Let f(c) = 5 where a ( c ( b. Then .,fb r1*1 a*a

exists if 1im I I_t-"' f(x)dx * I^o](x)dx] exists. The improper in_t*0 -a c+6r+1

k r^ ^ 'Ez r-- Itegral , { ;*

does not exist , €!s I f + f + ='-L x

'r, *

II rog *]

e, + [tog (-*)] -i '

= rog (f), for which no rirn* exists as

61 and 5, approach zero independent ly. However, i f the l imit .existsfor t1 = Ez , this part icular value is cal led the Cauchy pr incipal

Value of the improper integral , wri t ten "

/P f(x) dx = +b f(x)dx =a 'a

.'s I I^"-" r(x)dx + flEt*la*J

- 56-

ro evaluate f #:*_oO

t . i 'd , "R

- iz ' -f f f i =f^ f f i+ j-R ' fz_al=e

f . i 'd ,

{" y=* f r = o ' The last

integral vanishee as

oo iz-Then f ? oi =- ie

J 7o -zc-6

d*, consider the contour

iz.eoza2 -zz r

of f igurs

iz.e d,z;Gr +

00.

ion

eiEe-- e igdg

. io[ z.-e. igJ

ry=T

FTGURE. 16. i0 . -

- i . i " fo eiEe--eiodero

. i0[ -2"_e" i0]

as E*0 Ingene

{,*"1=,

f t+

iafIro

. - ia-TE T

-?a

if t(zl has

then lim8*0

in f igure 1?.

55. The

tegrand has a

simple pole at

the contour C

f f.@)dz = l imc 6*0

where 0 is

where C is

C. If . z Ls

Hence i f zr

)ryrr1"1 =

The rule is, I # =arc of angle

Cauchy Principal Value of

simple pole can be defined

z-zrand

a given point,

is the arc shown

ie@)S.

0EIGURE 17 €

a contour intsgral whose

similarly. I f f(zl has a

a simple pole and residue of "- l

at

J t(z) dz = ia_1 0, where the contour

passes through zr,

, rzt-S. ioL J r J '* '" '+

r ' (z)dzl '

slope of C at

a c losed contou

inside C, I =

is on C,

f t@,) i f z+(-Lo i f z+

zr. Consider I (z) =

r and t(z) is analytic

t(z). I f z is outside

from inside

from outside.

6#)t " 9-z

on and

c, I

dz'

inside

=Q.

zt

z l

From contour in f igures,

- r I -

L f ( ( ) Ar -rf fr, oL = i t(zt) , the arithmetic

rnean of the two limiting values.

56. Integrals with branch points: The funct ion f(z la-1zz+L

a-1,-

has a

r ' r

a- llz l

func-

contour

branch point at 2= 0,

" i (a- l )arg

z , where

t ion with a branch cut

of f igure 18, I f .@)dzc

6 a- lf

_ + " i (a-r)Zn d* =

'R r+x

a-1R, ia, i l# l =o, ; i . " ,

= O(64). These integrals

lirnits R * oo,

i f a is not an integer. Def ining

0 1 atg z < Zt , g ives a s ingle valued

along the posit ive real axis. Using the

=f r , (z)d,z- f t@)dz+;n a- l- rl"l:" -- J1, i=i '6 ft; dx *

Zzr i le izr(a- l ) r . on the c i rc le of radius

On the circ le of radius 6, la, | | # l

Ltr,

a

contribution when 0 < a < I in the

figure 18

g + 0 respect ively. Then 1t- . i (a- t ; " , I i # dx = ?ni " iz(a- l ) ,

6 a-Iand f

-dx=-J3-

=4, for O<a<1.,,0 r+x sln 77 (a-I) sin za'

58. Principal value inteeral with branch points: consideroo a- l

t 'xI l - -r- dx. The contour is as shown in f igure 19. once

circ les of radi i R and o, centered at the or ig in, g ive no

gLve zeto

again, the

contribution.

- 58_

^I- 6 __a_l RThen 0=(f +dx* f-6 r-x 1+o

dz. For the smal. l arcs about z

# dx)(r-eizz(a-r),

f-, t=u - 1, put z = l+

arc above

i(a-1)arg z

From the

f6 xa- ld-Edx

figure 19

Then ,^ ' l = l t + o" ie1a- l " i (a- l )arg

z = L + o(6) on rhe

axis. On the arc below the axis , u" 'L = l l+6eigl"- t "

= "hr i (a ' t ) + 0(6). From the top, / r

tu. : l d0 = iz.un

-6"1u

bottom, f -t tu"t,t

-hr](a-L)' Jo -F

e-"r* ^ ' dO = l reZr i (a-L) . Thus,

. t+.?d(a-L)= -L1l

, - "Z2i(a- i f = a cotz (a-1) = - r cotv4, for 0<a<1.

However, this assumes that a is a real number. S,rp-pose that a is comprex. The function -n cot n a is anarytic for0(Rea( 1. Further, theintegralconvergesi f 0(Rea( IIt can be proved that this integrar defineg an analytic function of ain this str ip. Then by analyt ic continuation, ,^o *a-l

Jo fr- ctx = -z cot

for al l a with 0 ( Re a ( l , when *" = "aln

x

59. Summation of inf inite series

Series, I ike I * , can sometimes be surnrned by con-1"

tour integrat ion. I f a funct ion, f (z l , could be constructed that is

analyt ic except for poles at z = nt each pole having residue one,-eqlthe ser ies, I # , could be represented as f_

f (z_)az . uponT '- Jc, zz

_59_

f igure

I'zz

On the. contour of f igure

the contour of f igure 21.

+

cot rz is l"rounded on the con-

inspect ion, thefunct ionf f i isseentosat isfya11

quirements. Using the contour of f igure 20, I 9 d,z =c

+ ?f i ( residue at z = O)" The Laurent expansion of the

Cr Ls stuo.re

these re-

ns-l Izn L6.

K= -n

integrand is

r-trll zz + * tra zt +

"'e-L6@)

The residue at .z = 0 is

z0

trzlt3/-7

-z( - ?) .tour Cn.

and

60.

this way.

stead, use

$l+;r

Then, as n+ L r f ( : \ $ L -r !€' u iJc^Tdz* o=zl F-7 '

- tZ 6

The serie" p # cannot be done

20, i t is found that 0 = 0. In-

+' s r,co!{rz d, =Y 4 - +Zni t

"

(z+L)l u' - U, 13- - 7r

z= 0. As

for N* o

o

- 60-

where the terrn *

contribution frorn

left

f igure 21

ie the contribution frorn

AB, BC, and CD vanish

befors, the

. We have

1 rA t cot tz - ^€ / -11hlorffitsa"= [-;1ffi*4'u'

= * I,* ffiuv - * 4'"'dl,#l o" =

Hence i *' = z - 1- tH coth xv

1-#+ coth zrv dv.

dy.

6t-

5r. pr inciple of the Argument. consider a funct ion, f (z) , anaryt icwithin a contour c except for poles. I f f has p pores within

c - double poles being counted twice and sirnirarry for pores of

higher order - and N zeros within c - again, counting the mult i-pl icity of each zero - then

*t f f idz=p-P

I f f has a zero of order rn at z = e, f . (z) = 1z-u1m h(zy, where

h(c) + O. Thpn f ' . (4r = g- , h ' (z)' f(4

= A *

hpt , and the residue is rn. For

a pole of order p, the same argument

-p. The residue theorem then gives the

holds wi th m replaced by

resul t W

=*Logf.(z) .

I f f(z) * 0, Log f '(z) has a continuous branch. In part icular, i t iscont inuous on the contour C. Then

I" + dz = | log f (z) ] "

=

l lo l t tz l l l + i [argf . (z) lc. Then N-p= 1.r- \ - , r rc ' 41416 LtztJc. r [en N _ p = f iLatC f .@)Jc

Rouchet s Theorern: Let f , (z) and g(z) be analyt ic funct ions insideand on a closed contour C, with t(z) non vanishing on C. I f

lg(r) l < l r t r t l on C, then f(z) and [ t@) + g(z)J have the samenumber of zeros inside C. Suppose that f (z) has N zeros insidec and that [ tp) + g(z)] has Nl zeros inside c ( i t is crearthat nei ther funct ion can be zero on c. Then 2zrN = ["rg f (z) ]cand 2zr Nr = [ . rg ( f+g)]" = [arg {r t r+f} ]" = ["rg r ]" *

l r l , the points of 1r * f l must

and unit radius. Using

where - t < + <t .

[ " 's 1t+f l l " . Becaus. lg l <

al l l ie inter ior to the circ le with

a polar form of representat ion,

center at one

(1 +f)=p' i+

q cannot increase by Zt no matter what contour C is t raversed.

figure. 2g

Hence + rnust return

Zeros of polvnomialq;

-62-

to i ts or ig inal value, and ["rg ( l+F)]" = 0It is clear that the polynomial f .(z) =

z4 * z3 + 4zz + Zz + 3 cannot have posi t ive real roots. In orderto consider possible negat ive real roots, set z = _u. f (u) =u4 - u3 + 4u2 - 2u + 3. However (ua - u3) = u3(u - l) is posit ifor u ) 1, and (4uz - Zu + 3) is posi t ive for u real . Hencef(u) > 0 for u ) 1, . Further, [on - u3 * 4uz] = uzluz _ u + 4Jposi t ive for a l l real u, and (-zu + 3) is sursly posi t ive for a l lu ( l . In al l cases, then, f (u) is posi t ive for u real , and thercan be no real roots. Consider pure i rnaginary roots. z _ iy iplies that f(y) = ya - iy3 - 4y, + ziy + 3. f(y) = 0 implies thaty+ -4yz +3=0 andthat -y3 +Zy= O. Thef i rstrelat iongivesthat y=*I , - r , ^ [g, '^ [2. Thesecondgives y= 0or y=r, [2.Both relat ions cannot be sat isf ied at once" Then there are no pureimaginary roots.

Consider the contour shown in f igure 29.

[ ""s r( , ) ] t = [arg ,n]E r [arg ( t + O l l t l = ztr .

[ ""e f (z) ] : = o.

[ " 'e f(")] 3= tao-l t|ffft3 r. The denominator , y4 - 4yz + 3, vanishes at

Y=1 andat

There can be

mials having

y = ' [3. Then [""g f (z) ]* = - Z*,

no roots in the f irst quadrant. Since

rational coeff icients occut in complex

and [ . rg rkf lc

roots of polyno-

conjugate pairs

there are two roots in

quadrant.

Example: Consider " . ,

On I" l = t , pf(z) l = Z,

-63-

the second quadrant and two roots in the third

ze

r, qr

-1"= 2z + L. Take f@) = ?z andandl. . " - r l . / I . " l . laul

= e - I < l . g. Since f(z) has exacf ly one zero (at zthe contour,

"" = ?z + I has exact ly ons root for l r l

g(z) =

,J

-o

<1.I

' rC SIMFLE functions

f(z) is sirnple in a domain D if f ,(21) =simple function is also called univalent ( i t takes'schl icAf. '

If f(z) is simple in D, it maps D conformally and r _ i into adomain in the w_plane. The mapping preserves sense. Either or bothdomains may be r:nbor:nded. The w-domain is unbo'nded when f(z) hasone simple pole in D' (By Picards theorem, i t cannot have an essentialsingularity. )

Theee are two useful cri teria for simple f irnctions. First, i f t(zlis s imple, then f ' (zr * o in D. por suppose fr(zo) = g. Then f(z) - t (zo)has a zeto of order n >- Z at zo . The zeros are isolated. Take 6 sothat f (z) - f (zo) and { t" l have no other zeros in l r - "of

_. 6. Takem = minl t ,z l ' f (zo) | on Iu - ,or = o. By Rouche,s theorem,f(z)- f@o)-o has n zerosinside

lu_"ol =O when lo l .* . Butf '(z) * O; hence the zeros are not mult iple. Thus f(z) is rot simple.The converse does not hold, i .e. f , , (z) * 0 dor(z) is sirnple' rust consider zz in the il; ;J;r."':: ;tttsketch.

within

f(zz) impl ies z1 = zz. A

each value only once) or

However, i f f , (zo) * O,

-63a-

f(z) is sirnple in sorne neighborhood of zo. For t(z) - f(zol has one

zero inside sorne region l, - uo | < o, and, i f rn = rninl! l . l - t(ro)l

lo - uo | = o, then by Rouche's theorem t(z) - f . (zs) - a has one and

only one zero inside this region when l" l . *. Thus f(z) is simple

the domain which rnaps into l* - *o | .r:rr, where wo = f@ol. This

proves the inverse f irnction theorern. rf w = f(z) and f i (zo) * o"

z = F(w) is analytic in neighborhood of wo.

For the second criterion, suppose that f(zl is analytic insid,e

and on c, and r:nivalent on c. Then c mapg into a sirnple closed

contour c' in the w-prane. suppose w6 is inside cr. Then

where + I i f c ' is described counterclocl,nvise and - I i f clockwise.

But f(zl and hence f,(zl - ws has no poles. Hence value must be + 1

and f(z) = q7o has one and only one zero. rf f(z) = qr, for z in c,

and w1 is outside c ' , the r .h.s. is zeto whi le the 1.h.s. is the o

of the zeto, which is a contrad,iction. Thus the inside of c is

I - I into the inside of c'. rf t(zl has a pole, apply the theorem to

F(z) = l f ( " | + K]t where K is chosen so that f (z) + K has no zeros

in C.

consideration of L/t shows that a sirnple firnction cannot have

a pole of order >- Z.

Extension to case where f is only continuous onto c or has a

simple pole on c is given in Titchmarsh p. z0l. The criterion may

be false i f f(z) has a higher order pole on C " . '

R iernann' s rnapping theorern.

c is any sirnpre closed contour, possibly'nbor:nded.

exists f(zl, aaalytic and simple inside C, which maps the

{c

Then

.t

ar= 6 a=,u*=== =rI ,y W-Woc'

inside of

-63b-

into the r:nit circle l* l .1. f need not be analyt ic on C. Indeed, i f

c has sharp corners, f rnust have singularit ies on c. For proof, see

Titchrnarsh p. 201. (Note: C cannot be circle at inf inity by Liouvil le's

theorem. ) The rnapping is conforrnal because dw = fi (zldz. Thus,

f awl = Jf ' f la" l and arg dw = arg dz i arg f ' . The mapping is r :n ique

i f f , (zs) = 0, f ' (zs ) > 0 for arbi t rary zo inside c.

Examples ( i )

into l* l . t .

( i i)

lower half plane.

w=i@-?)z'zo

kn zo ) Q tnaps upper half plane

L=z+t rnapssernic i rc le lu l . t , y>0 into

rnaps lower half plane into l*l . t.

Thus

z2 +iz+lw =z-7;T

maps semicircle into circle n s7 = sz would not work as mapping not

1- 1 (z = x and z = - *. rnap into satne point).

E-i

'64-

VllI. Analvtic Fr:nctions Defined bv Sequences and Integrals

52 " Uniforrn Convergence.

Let {s-(z)} be a sequence of functions analyt ic in a region R (e"g.-n 'I

{srr(zD = {# ; {") ; { f f i ) ) , The sequence is said to converge i f there is

a function s(z) such tihat srr(z) - s(zl as n + €. If, in addition, for each

e)0, thereisan N suchthat lsrr(z)-s(z l l . . foral l n)N anaforaU

z in R , the sequence is said to converge uniforrnly to s(zl.=-==:

one sequence of functions which does not converge r:niformly is

{ s,r(z)} = {#"} in the domain lo t t Dr order that lSl . . , i t is neces0.

,1that n > +r . There can be noe lz l

N independent of z such that I srr(z) I < e

for al l n ) N and al l z in lu l t O.

63.. Cauchv criterion. The sequence {srr(zD converges uniforrnly if, for each

e ) 0, lsrr(z) - s*(z) l . . for a l l n,m > N(e) and for al l z. I f {srr(z)} is

a uniformly convergent sequence of continr:ous fr.rnctions in a region R , thc

the lirniting function, s(zl, is also conf,inuous, and for any contour, c, of

f inite length in R, l im f s_@ldz = f s@)d,2.n+6/cn'

uc

s(zr) - s(zo) ={s(zr l - srr(zr)} - ts(oo1 - srr(zoD +{srr(zr)- sn(zo)} . Since

the sequence { srr(z)} converges uniformly, there is an N such that

l " ( to1-s*(zo)1. . and l r (u, 1-s*(zr)1. . . Since srr(z) iscont inuous,

thereisa 6>0 suchthat ls*(zr1-s*(zo) l . . whenever lo, _"0 1.6.Then l"(r t ) - s(zo)J

-< 16 whenever lu, - ro I < 6, and s(z) is cont in

l l "s@)a" - I . "n{" ldzl = l l . { " t "1- s,r(zDdzt -< [" l " t " l - so(z)f dz -< eL

L is the length of c) for n ) N. Then ,tA

i s,.,-)dz = -f " @\d.2. supposc

that { srr(z)} is a sequence of fr:nctions analytic in a d.ornain D. If srr(z) *

s(z) r:niforrnly in any closed region R contained in D, then s(zl is

g.

it

b

Llr

-65-analytic in D and

"lr*"'uni-formly in any closed region R contained in D. Let cany contour in D. I s@fi.2 = rirn I s,.,)d,2. Because snkr is anaryticinside ;, cauchyf

""rn"otur' lr i"* " ' .

I sn@rdz = 0t and hencec

I s@p,z = 0" By Morera's theorem, s(z) is anaryt ic. By cauchyrs

t l reorern,s,(z)-s, , (z)=hd#., .Let6=min| tu-el |over

c. Then choose N so that ls(6t _ srr(6) l .+: i f n > N. Then

l" ' ( " ) -s i (z) f ce for n)N.

Example: zn*0 uni formlyin luf - . l -6, for l r f r_.( I -6)n<e when_ei'er

" tffifu1 . I,r a similar w.r, }'- o rrniformry in lul r-e > 0.

€. z s@)=furr(z)t

The sequence of part ial sums of a series is defined as 3sn(z) = f \ r '1.

u the sequence of part ial surns converges r:niformty to s(z), the series isuniforrnly convergent, sr(z) =

xurr(z), and I s@p,z = X/orr( zld,z. power

series are uniformly convergent within their circres of convergence.Laurent series are uniformly convergent within the annulus of convergence.weierstrass M-test: The series f urr(z) is uniformly convergent i f thereis a convergent series, )Mr, independent of z, such that lurr(z)f . Mrr.

nf srr(z) - s

sequence.

Example:-

orb) l 'x t rnn' . whenever n, , ' ) N. Thus {srr(z)} is a Gauchy

- r I r Ie f Fl

= F."

, the series converges in Re z ) l,

I

* is analytic for Re z ) 1. This is the Riernann

(, fi:nction.

00

Example: f (z)= r 1' - !4. By the Weierstrass M-test , the convergence

on=*'n

and the firnction Lkl

Sinc

@

-s- lJI

-66-

is uniform in lr l < a. I zz - n2 | > o. Then f(z) is analytic in

except at z = tn. Since the fi:nction lt(rl - ;r-+] i" analytic

thesingu1ar i t iesaresimplepo1es.Infact , f ' (z)=W.

Power series: The series f,(zl = X..rrt is urtiformly convergent for

lul -' R - 6. Sirnilarly, a Laurent series is r:niforrnly convergent for

Rs * 6 -< lu - o l -<

R. - 6.

66. Abel 's theorem. (Behavior on the circle of convergence). Suppose thatco

f(zl = I"rrut has a radius of coavergence 1. Gonvergence is gr:aranteed for0

6

lu l . t , but what happens for l" l = tZ Does f6) = Ia f I f f (zl =#,co

r-, n

f(z) = f {-t)tot, but the serieE is nottonvergent for z = L even though t(zl0

is well defined 1f1f1 =|)r and in fact analyt ic. Abel 's theorem states that60salrr .5if tr.r, is convergent, then lim X"-"t = f(I)" Define R S -ov"vv'Ee$u'Lgsr 'z*L?nrtr ' 'L 'er lneorr=#r"n ' iJ lnce

srL^x converg.es, an N can be found for any € such tfiat lRol . e when-

NN

Define s*(x) I X"o*t. Then s*(x) - S*(x) = X (R* ., - R_

o .r r \ r l l f f i t 'n-r n

,*"+*****t-n;*M+l. f s*tx) - s*(x) l . lar . r ; + lnyl

+ ,',ax Jn-f tG - x) -t*

. ,, x ( r, rhen X*" .*, andN+1-< rr-( rrr' 11 ' - '

r.fir r

l ' f ' n,at z=*N

ever n ) N.

N=(x-1) In

MTl

Mo if x = t. Then t(t - *) X

-"]N-+t

for o -(

x -<

l. Then X"*** is

lirn Ia *t = I*"1,lr-rr^

' /-tan'

67. Functions defined bv definite integrals.

Consider f (z) = I"F@,i lat , e.g. f , (z l = [^"ut"- tdt . I f F(z, t ) is

M

{(t - *) I *t} =I\ITl

3e i f R*<e,

0-<x-(1, and

4 1, and lS*tx) - S*(x) l

trnifo rrnly c onvergent for

-67-

continuous in t and analytic in z f.or aLl

of finite length, then t(z) is analytic for

z in some dornain D, and if C is

z inD, and f rAtr' (z) = l tat .

Let f(x, y) be continuoua in both x and y in some region bounded.

byasimplecurver i .e. f (x+hr)r*k)* f (xry) as h*0 and k*0 forx and y in the given region. Then

Utr",y)dxdy = ,[Ayt IrE,y)dxJ =

,f u*t I te,y)dyl .By cauchyrs theorern, F(z,tl =

{ff1l}qE when r is a ctosed curve

in D' rhen r(zl = I.*{r##dE = { #t ,[ "rg,

r)dt] = $rff{{. Hence,

r(zl is anarvt ic in D. Further, r ' | (z) = h $rf f i = #. L" t" \W= d#ur.consequently, f,(zl = 1a "'t"'tat

is an entire function. since the0

Tavlor ser ies .ot=L+ zt-W +.. . converges 'n i formly, f . (z)={t ,T#1e-tat

= I,{ *"-tt'dt) =

}# { Io*.-tt'dt1 =

f."# where

",, = .,[-"-tt'dt.

However, 8n a# , and, the series converges for all z. In this caee,

f(zl = *-tr "(z-l)a - rl . The singurarity at z = L is rernovable, and the

function is entire. Note that result fails if a = €.

- 68-

IX Erfinite &rtegrals: Improper Integrals

An integral is terrnedtrimproperil either if its range is not finite or if

the integrand is singular in the range of integration or both. Given a seq

of firrrctions { srr(zD converging to s(z), does lim .,i[€srr{=)d o = I s(z)dzn+€

apply? Can an infinite series be integrated terrn by terrn over an infinite.€S S,€ € 6

i . e. is I ()u,r(z))d = l.r l

u,r(z)dz)? Does { tf ,"r,tutld.z = f,({u,r(zFzl

urr(O) = o? First consider t("1 = IC F(z,t)dt, where C. is a contour of@

finite length. Define f*(z) = .I"*"tr,t)dt where cn is that part of the coa-

tour G. lyrng within lt l = n. suppose f*(z) * f(zl as R *o. Then the

finite integral is convergent. Furtherr+if there is an Ro, depending on €,

sucht ihat l t ( " | - f " (z) | . . whenever R)Ro andforal l z inthedornain

then the integral /c-r1zrt1at is unifor'mlv coavergent for all z in D.€

Construct a sequence of firnctions f.n@l by letting n = R. Since

f.n@l - f.(zl r,rniformly, t(zl is analytic in D and f,i@l * f'(zl. However,

fn@l = /"_#ur, and, _tim \(zl = f, (zl = Ic #u,

= * h Fdt. Thus, r:ai-

n n+€ _€ _6

{o.rm convetgence of the infinite integral is gufficient to allow differentiation

rrnder the integral sign.

suppose that the integrand becornes infinite at gome point, F(2,0) =

The integral is then interpreted as f,(zl = lim - 1 flr,t1at. If the integral is

6 - 0-6

uniformly convergent, t(zl is an analfiic firnction.

58. Exarnple (i) (Garrrrna Functionl tr(z) = 1t.-tt"-lat. Integration by parts

gives the recurrence relation f(z) = (z - f)tr@ - L) for z)I. The garnrna

function is a generalization of the factorial fi:nction, since by suecessive

uses of the recurrence relition, f (n + l) = n! . Substituting t= v2 givesl -

trQ) = Vzt. Uni-forrn convergence of the gamrna function follows frorn the

-69_

comparison test: l f there is a f irnction M(t) such that lr(", t) l < M(t), and

"€ ^@I M(t)dt converges, then I F(z,t)dt is r:niforrnly convergent, since

00

. ntz - ntzI J. F@,t)dtl < /-

-rrnltyAt. For the garnrna firnction, f tz-l.-t l . t*-l"-t,t1 tr

^coand I--r"'lu-tdt converges for x ) 0. Then the garnrna firnction integral

0

converges uniformly for Re z ) 6 ) 0, and I.(z) iq an analwic fi:nction for

z ) 0"

Example (i i): f.(zl = I* "oj tur. Near t = 0, the integral behaveg like +,uo

tz -- 'vE-!- v ' l

{

whichconverges i f Re z ( l . Near t = o, lgo" t t - L .1

Tl 'F ' which converges i f

Re z ) l. It appears that this integral never defines an analytic function. How-

ever. f€cos t41 = 1l co=s t &+ I_6 "oj ,ur . As before, . [ t "or"

rua converges'"otztotzJt tzotu

for Re z <1. For Re z> 0, / "" j ,ur= .Tt l -* .uf t"r l , rut=

- s in l+' t tz tz ' l

-Jr ,z lL

ulr* ffur.

since {- ";?ltdt converges uniformly for Re z ) 0,

f.(zl = l- "oj ' u, is analytic for 0 < Re z lL.' '/r tz

This same idea of integrating by parts is helpfur for series.

Exarnple (i i i): X co?ne.

This series converges for Re s ) I by cornparison

with the serie" X +. In order to investigate the case 0 < Re s ( l, defineNll lNN

"r ,=Xcosn0, so=0. Thencosn0=B-.-s , , and y.cosn0 -y"t-sn-t-- -n -n-I ' lJ s_- - lJ- =l _

l n- 1 n

$", , $ " '=*- \ -=i"-+-4,* ; : : " :* : : ; . rhesul : r , en, isLr-Tno i1n+t1s (N+t)s u-t ' r rE

Ff"6 ' ' , rborrnded if e is not zelo or zkr. rn particurar, i "o"rre

= ["hfri)O 1.,r -@_a

(this is just the resurt of substituting "

= .i0 into the identity t - :Tt =

L*z+.. .+rN).

€-€.-?0-rhen i+=i . , , { - ;*r" 1{- ;* t = t* l . l r - F*=\ ' ;

I n r n (n*1) ' 'ne (n+1)" (n+1)-

-+ " This converges if Re s ) 0. Hence, this series is r:niforrnly con-

l t t"*t Ivergentasaf i rnct ionof e i f lg l"6>O and Res)0.

69" Beta function, B(p,e) = LtrP-l(t - t)9-1dt, orr lett ing g =

fh ,

F6 -q-1e(p,q) = J^

-gdr.

B(p,e) is an analyt ic f i .rnction of P and q for-0 (1 + 71r ' :

Rep>0andReq>0..WhenB(p,q) isana1yt ic,B(p,q)=f f i .

Initially, suppose that p > I and q > 1. r(p)rtet = {t'"-ttP-lat11/-"-rt9-1u

Let L = xt. r(p)r(e) = (,ft"-ttP-tuo (f "-xtr9*9-td*).

Since the integrands

are continuous and exponentially small at € I inversion of the order of

integration is perrnitted, and

r (p)r (e) = ,r-*o- tu*t'

"- (x+l)t n+q- lat

Take (x + l)t = [

= f#-,u*t'"-6'n+e-1ag= B(P,e)tr(P + q)

Theresul tsoestabl ishedfor p>I and q>I rnustholdfor p>0 and q>0

by analytic continuation (to be e:<plained later).

?0. It still rernains to consider under what ^€6conditions 1[ (f,urr(z))dz =0

oo oo

Xf /* u(zldzl. Suppose that s(z) = luo(z) converges uniformly in z on any00

finite part of the contour (if there is a singularity at z = 0, tmiforrn conver-

gence in lul .- 6 > o is enough). It is then sufficientthat A = ,f ti lortrll

ooo

or "

= Xt/ f u,"(z) l la" l ) should converge. The convergence of e i ther A or0

B irnplies the convergence of the other and also the eqr:a1ity of A and B.

t -1.'o

rds

-7L_

consider a = f f ,a rr a ich^i-ro:r =

nQg ftgnn' rf

"*rr is monotone increasing in both rn and

n' and if "-",

) 0, then the existence of lirn ( lim a__) implies the11a+€ n+6

l ru l '

existence of lirn ( lirn a r) and that

n+€ rn+€ ' 'u ***"r- I*-mlt ' r l+oo rn+€

lim a rn the case of A and B, the ind,ex n plays the role of oner:n + € rnn

n+6

€sumrnation index and dz prays the role of the other. Then lI /-"-dz | _<

:-€ oR;

L(J^ lorr l . f auf * g, since the surnmation converges. Thus rirn Xt f u_dz) =oR6

R *oo7 t* n

sln '6

L(J ur"dz). By r:niforrn convergence of the series s(z), L I^ undz = [^ (furr)dz,o i-

_rn vR

and the regult is established. Exarnple: The integrand of /t roglr{1a* is0

singular at x = 1. For x ( l , fog f -*= x * f t * i t + . . . , a ser ies that is

r:niforrnly convetgent for 0 ( x ( I - 6, but not convergent at x = I. f 'dd* =uo n

r3€n6#, and I #m converses absolutely. rhen f

tt i 4,u* = i f 'du* =

I - - \ - - - , to.sn,_-- ?to

n-€ootr I sr , l I r I Ia,fr l f f l i= iq - ;FT) = (1 - 2l n q- i l * . . . = r.l l

For sequences, the existence of I s@)d,z is guaranteed if there isCoo

a f i :nct ion M(z) such thar for a l l n lso(z)f . Vt(u) and I M@) lazl < o.Co

(Remernber that, in this notation, s(z)= l im*srr(z). ) since every rnember of the

seguence {srr(z)} is bounded' by M(zl, so also is the lfunit firnction bounded byM(zl. Then the lirnit fi:nction, s(z), is integrabre over the contour.

rn order to avoid the misconception that the processes of sumrnation andintegration can always be interchanged, consid,er f(x) =

f,u.(x) =

f {""-r"*+1-- I

b.-tb*). The series converges uniformry for 0 < 6 -(

x-( R (or aod

f(x)

I*r:

.72-

- a b- cL-\ : - ^^-r : - . -^.-- : f r /n\ : ^ J^ c2-^) &^ L^ (b -

a)= -;|.a - ffi.

f(x) is continuous if f(0) is defined to be T

I . -nax -nbx-o 3. "-urr(x) = - l [ . - " -o - . -""o] l = 0, and L(J, uo(x)dx) = 0. Ffowever,

6o.srr6ab(fru,r)dx = J^ qi,- - tr)dx ) 0 whenever b ) a. Irr this example,'1 - e - I ._ - I -

Xt 1* loo ld*) = -

.f,. -o

l

Infinite Integrals

'We ask under what conditions

f6 F6 ,"6 ^€J ayU f(x, y)dx) = J d*U f(x, y)dy)

00;, !00

where the integrals are either over the infinite range, or irnproper at

x=y=0r orboth.

It is sufficient that

^€ rQ

J f(x, y)dx and J f(x, y)dy00

be uni forrnly convergent for 0 < 6 -< y< A (o and 0 < 6

-( x-( A (o,

respectively, and that either

I* urr l* Jr laxl or I* u*r l- l r layl exist.0000

rf Lebesgue integration is used instead of Riemann integration, the re-

quirement of r:niforrn convergence can be dropped. This is because

Lebesgue integration can be defined for non-continuous frrnctions, whereas

Riernann integration is e.ffectively restricted to continuous functions.

As a cor:nter example, consider

1t

/-$I,/-€+#ff,= *r -

7L.

-73'

X Analvtic Conti:ruation

Let \ (zl be an analytic firnction

in a dornain Dl . Let fz@l be analytic in

a domain D2 . If \ (z) = fz@l throughout

Drf) D2 (see f igure 22), then fz@l is cal led

tihe analytic continrration of \ (zl into Dz .

Notice that Drn Dz has no holes in it, it FIGURE 22

is sirnply connected.

Exarnple; \ (z l=L+ z+ z2 +. . . for l r l . f . fz@l =+ *f f i1,

\ (z l and fz@l

two different

which converges for lffi; . t. Suppose, for definiteness, that a =

fz@l converges within l " - i l . l i - l l = " fZ.

Observing that

are power series representations of the fr:nction # about

points, it ig clear that the two series must be equal in the region of overlap.

Thus fz@) is the analyt ic continrration of fr@l into the circle l , -\ . rfZ.

Uniqueness theoretn:

Let \ (z l , fzk), " t9

fs@l be analyt ic

in dornains Dr, D2 and D3 respdbtively.

Suppose that \ = fz in Dln Dz and that fr = fr

in DlnD3 . I f Drn Dz and Dln D3 have an

infinite number of points in cornrnon, including

a point of accumulation, then fz = fg in Dzn D3.

(A11 sets Ooo D: are assumed to be corurected). FIGURE 23

of Dzn D3, theySince f.z and f3 are analytic in Dzn D3 and equal in part

are equal ever)rwhere in Dzn D3.

If Dr , Dz, ?nd D3 do not overlap, f.z and f3 rnay not be equal in

Dzn D3. This would occur i f there were a

the paths of continuation (see figute 241.

singularity or branch point between

, ( r -a)2,' ' ( i ;y '

i . Then

FIC'TIREZI.

-74-

Examplez f .1@)=r*Z' nZt * . . "3-Log ( l -z l for [ " f . r

fz@)= - Log (- I - i ) * ( t , - .2 - , , t ) * . ( ' - 2 - 112, .TT:-i---ZT_fiIY r -o

rs@r = - LoB (- r+ i r* f f i r f f i# * . . . F,q zs:In DrnDz, \=fz =-Log(I-z l . ID DrnD3, f i=fr=-Log(I-o. f , - -

However, on l. ( x -<

3, fz - fz = Zri. Dln Dz and Dln D3 have no co

point where f is analytic.

It is usually difficult to construct an analytic continr:ation although i:r

principle it is easy to continue along a path. Consider a firnction analytic in

a domain D. Choose sotne point a in D and

expand f in a power series about a. The

result ing power series has a radius of conver-

gence Re. Choose a point a1 inside the

radius Rs about a and perforrn anotherFI€URE.26

power series expansion. Usrrally, each expansion will converge in a circle

that contains points not in D. In this w4y, continuation can, in principle, be

perforrned along any path not passing too close to a singularity(recaIl that

there is a singularity on the circle of convergence of every power series.).

73. Monodromv theorem: The r:niqueness theorem irnplies the following. Let

f(z) be analytic in a dornai-r Do, which is contained in a sirnply connected

dornain D. If f(zl can be continued analytically along every path in D,1

then f(zl is a single valued analytic firnction in D. Exarnple z f.(z) = zE .

Expanding t(z) about z = L, i t is found that f . (z l =, n " ; ' - $e- t )z | " . . ,

a convergent series in l , - I l o f . I f z = reio, this expansioq c.qrresponds

to the branch f(z) = ";1.o" 9n t

" i r l t where

- lr . g .t . Analyt ic continuation is possible

along any path except those through z = O. In

particular, the contiauation rnay be"done along

f. e". tT

f ' e" 3

, te.

=FT= -t

FIGURE A7

-7 5_

the polygonal path shown in figure 2?. since continuation is possible alongall paths not passing through the origin, the result, on co tinuation onceabout the originr rnust atways give fr= , i" ig,

#., .y.-- ;r . single-valued fr:nction

I cannot be continued through z = 0t so the impossibilityof continuation need not imply multi-valuedness.

4' Natqral Bor:ndaries: A curve across which a firnction cannot be continued

' - ' -#.Forexample, f , (z|=L",ncanrrotbecont inuedacrogslu l = t .

Al1 values of a multivalued function can be obtained by analytic con-tinr:ation, just as they wete in the example of f,(z) = 2tr . In effect, if f(zlis given in a domain D, it is specified everywhere. All branches of multi-valued fi:nctions are obtained upon continrration along paths about the branchpoints.

. : Suppose that F(f1 (z), gr (z)) = 0 for allz in a domain D' Let tzkl and gz@l be anarytic contin.ations of fi andgr into D2 ' rf F is an anal'tic firnction of its argumeirts, the rrniqueness

theorem guarantees that F(fz rgz ) = 0.Example (Gontin'ation of r(z)). For n" f> o, r(z) = .f*"-r1u-lur.Consider F(z) =- t f r t , ,z- ! r - - , ta-

uo=A*?;,f

"oou-ldo (Hankel,s integral) where c is asshown in figure Zg. This is a meromorphicfirnction in the finite z plane, possibly havingsimple poles at 2 = *n. ot- l = . (z- l )Log u

. Take z=x)0 anddeformintegral into negative real axis. put u = _ t.Then o*-l = ,x-1"* zri(x-l)

and, "o

= .-t. ThenF(x) = zr#, 4'"-t."-11-.oi1*-l) *

"- ri(x-r\ u,

= tr(x). Then E(zl is aa analytic continr:ation of

. , ' -r_ rr-r"f i (r-r)

-Ul- l - t r - r -

FlcuRE a8 -

-7 6-

f (z) to Re z < 0" The property t r (z + Ll = zT(zl is preserved as can be

verif ied directly doing integration by parts. AIso notice that T(zl is a

meromorphic funct ion, having single poles at z = 0, -1, -2, . . . Further,

s ince f (z) t r ( I - z l =# for 0 < Re z <L, i t is t rue for al l z.

?6. Asvrnptotic Expansion of the F Function. (Stir1ing's !"orrnula). Two

expressions,f(z|andg(z| ,aresaidtobe@asZ*Zo

if lirn i{3{ = r. This relationship is usually written f,(zl - e@1., _ ,o Elr)

As an exarnple, consider the asyrnptotic expansion of l(x + 1) =

fQ - t x. . 'oo L

Jo e-utodt = Jo e-u+xlntur.

The exponent, f(t) = - t *xlnt, has a maximr:ro

at t = x. As the exponent changes, the exponential decreases so rapidly that

the behavior near t = x is the dorninant feature when x is large. Take

t = gx. Then 1. (x + I ) = xx+l 1€u+*(-g+lng)dE = , .x*1"- * l*"x(- f+lng+1)a6.Jo E -- - Jo

Since the region near 6 = I is irnportant, a series expansion neat I = 1 is

appropriate. 1- 6+ lnl i - |rc- U' , and the range of integration can be

A general result, Watsonrs lemma, gr:arantees that substitution of such a

power series expansion gives. a true asyrnptotic expansion. ) Then

f*"*(-g+rn6+1)d*- I ;8rc't)za6 = E. Thus 1" (x + t) - ^,ff i **"-*,Jo

_€

or n! - xlffi rrtu-t. Further developrnent of this series would give the

result that tr(x + 1) - rE?E "-=**

(t * # - tBb=, + . . .) . The series diver

Nevertheless, a finite number of terms of the asyrnptotic expansion (one te

yields a surprisingly good approxirnation for large n.

tic terrn

3,62g, go0 3,599,695"6210

z

1

z

I

1.919

o.9zz

1. 008

1.04

1.08