February 26, 2013 Lecturer: Shih-Yuan Chen

1

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Complex Variables Chapter 17 Functions of a Complex Variable

Contents Complex numbers Powers and roots Sets in complex plane Functions of a complex variable Cauchy-Riemann equations Exponential and logarithmic functions Trigonometric and hyperbolic functions Inverse trigonometric and hyperbolic

functions 2

Complex Numbers Solve the quadratic equation

Define imaginary unit i by i2 = −1

Def. A complex number is any number of the form z = a + ib where a & b are real numbers and i is the imaginary unit. Real part Re(z) = a Imaginary part Im(z) = b

2131

231 012 −±−=

−±−=⇒=++ xxx

3

Complex Numbers Def. Complex numbers z1 & z2 are equal if

Re(z1) = Re(z2) & Im(z1) = Im(z2). A complex number z = 0 if Re(z) = 0 & Im(z) = 0.

If z1 = x1 + iy1 & z2 = x2 + iy2 Addition: z1 + z2 = (x1 + x2) + i(y1 + y2) Subtraction: z1 − z2 = (x1 − x2) + i(y1 − y2) Multiplication: z1z2 = (x1 + iy1)(x2 + iy2) = (x1x2− y1y2) + i(y1x2 + x1y2) Division:

22

22

212122

22

2121

22

11

2

1

yxyxxyi

yxyyxx

iyxiyx

zz

+−

+++

=++

=

Complex Numbers Familiar laws hold for complex numbers Commutative laws:

Associative laws:

Distributive law:

=+=+

1221

1221

zzzzzzzz

( ) 3121321 zzzzzzz +=+

( ) ( )( ) ( )

=++=++

321321

321321

zzzzzzzzzzzz

5

Complex Numbers Complex conjugate If z = x + iy, then the conjugate of z is It is very easy to show that

The sum & product of a conjugate pair are real

iyxz −=

2

1

2

12121

21212121

zz

zzzzzz

zzzzzzzz

=

=

−=−+=+

( ) ( ) ( )( )( ) ( )

+=−=−+==−++=+

212

22222

yxyixiyxiyxzzxiyxiyxzz

( ) ( )izzzzzz

2Im and

2Re −

=+

=

Complex Numbers Difference between a conjugate pair is imaginary

(1) & (3) yield two useful formulas

Division of complex numbers using (2)

Ex. If z1 = 2 − 3i & z2 = 4 + 6i , find &

( ) ( ) ( )32 iyiyxiyxzz =−−+=−

( ) ( )22

22

21212121

22

21

2

1

yxyxxyiyyxx

zzzz

zz

+−++

==

2

1

zz

1

1z

7

Complex Numbers Geometric interpretation A complex number z = x + iy is uniquely determined

by an ordered pair of real numbers (x, y). Ex. The ordered pair (2, −3) corresponds to the

complex number z = 2 − 3i. One can associate a complex number z = x + iy with a point (x, y) in a coordinate plane. The complex number can also be viewed as a vector from the origin to the terminal point (x, y). 8

Complex Numbers Def. Modulus or absolute value of z = x + iy,

denoted by |z| , is the real number

Sum of the vectors z1 & z2 is the vector z1 + z2. And we have

( )422zzyxz =+=

( )

−≥++++≤+++

+≤+

2121

2121

2121 5

zzzzzzzzzz

zzzz

nn

9

Polar form Rectangular coordinate (x, y) and polar coordinate

(r, θ) are related by x = r cosθ & y = r sinθ. A nonzero complex number z = x + iy can be

written as z = (r cosθ) + i(r sinθ) or

Polar form of complex number z

Powers and Roots

( ) ( )6sincos θθ irz +=

( )

=⇒=

=

xyz

zr

θθ tan arg10

Powers and Roots Argument of a complex number in the interval −π < θ ≤ π is called the principal argument of z and is denoted by Arg(z).

Ex. Arg(i) = π/2. Ex. Express in polar form.

iz 31−=

( ) ( )( )

+=∴

==⇒−=−=

=−+==⇒

−==

35sin

35cos2

35arg 313tan

231 3

1 22

πππθθ

iz

z

zryx

11

Powers and Roots The principal argument of z is

Thus, an alternative polar form of the complex number is

−+

−=

3sin

3cos2 ππ iz

( ) 3Arg πθ −== z

12

Powers and Roots Multiplication & division It is very convenient to use the polar form. Suppose

( )( )

+=+=

2222

1111

sincossincos

θθθθ

irzirz

( )[( )]( ) ( )[ ]

( ) ( ) ( )

+==

+++=∴++

−=⇒

2121

2121

21212121

2121

21212121

argargargor

sincos sincoscossin

sinsincoscos

zzzzzzzz

irrzzi

rrzz

θθθθθθθθ

θθθθ

13

Powers and Roots

It is NOT true that ( ) ( ) ( )( ) ( ) ( )

−=+=

2121

2121

ArgArgArgArgArgArg

zzzzzzzz

( )[

( )]

( ) ( )[ ]

( ) ( )212

1

2

1

2

1

21212

1

2

1

2121

21212

1

2

1

argargarg and or

sincos

sincoscossin

sinsincoscos

zzzz

zz

zz

irr

zz

irr

zz

−=

=

−+−=∴

−+

+=⇒

θθθθ

θθθθ

θθθθ

Powers and Roots Powers of z Integer powers of the complex number z

Moreover,

For any integer n:

( ) ( )[ ]θθ 2sin2cos1 22

2 −+−== −− irz

z

( ) ( )[ ] ( )( )

θθ

θθθθθθ

3sin3cos2sin2cossincos

323

222

irzzzirirz

+==

+=+++=

[ ]θθ ninrz nn sincos +=15

Powers and Roots Roots w is said to be an n-th root of z if Let

As k = 0, 1, 2, …, n − 1, we obtain n distinct roots with the same modulus but different arguments.

zwn =( ) ( )θθφφρ sincos and sincos irziw +=+=

( ) ( )

==

=⇒=⇒

+=+=⇒

θφθφ

ρρ

θθφφρ

sinsincoscos

sincossincos1

nn

rr

irninwnn

nn

nkkn πθφπθφ 2 2 +

=∴+=

16

Powers and Roots For k = n + m, where m = 0, 1, 2, …. Then

To summarize, the n-th root of a nonzero complex number z = r(cosθ + i sinθ) are given by

( )

+

=

+

=⇒

++

=++

=

nm

nm

nm

nmn

πθφπθφ

ππθπθφ

2coscos ,2sinsin

222

1 , ,2 ,1 ,0 where

2sin2cos1

−=

+

+

+

=

nkn

kin

krw nk

πθπθ

17

Powers and Roots Ex. Find the three cube roots of z = i.

( ) ( ) ( )

( ) .2 ,1 ,0 ,3

22sin3

22cos1

2sin2cos 2arg

1

31 =

+

+

+

=⇒

+=∴

===

⇒

kkikw

izi

r

kππππ

πππθ

iiwk

iiwk

iiwk

−=+=⇒=

+−=+=⇒=

+=+=⇒=

23sin

23cos2

21

23

65sin

65cos1

21

23

6sin

6cos0

2

1

0

ππ

ππ

ππ

18

Powers and Roots Principal n-th root of z: the root w of z obtained

by using the principal argument of z with k = 0. The n roots of the nonzero z lie on a circle of

radius centered at the origin in the complex plane and are equally spaced on this circle.

Ex. Find the four fourth roots of z = 1 + i.

nr1

( )( ) ( )[ ]

( ) .3 ,2 ,1 ,0 ,4

24sin4

24cos2

4sin4cos2 4arg

2

41=

+

+

+

=⇒

+=∴

===

⇒

kkikw

izz

r

kππππ

πππθ

19

Sets in Complex Plane Since is the

distance between the points z = x + iy and z0 = x0 + iy0, the points z satisfying the equation

lie on a circle of radius ρ centered at z0. Ex. |z| = 1 Ex. |z − 1 − 2i| = 5

The points z satisfying |z − z0| < ρ lie within, but not on, a circle of radius ρ centered at z0.

( )0 ,0 >=− ρρzz

( ) ( )202

00 yyxxzz −+−=−

20

Sets in Complex Plane z0 is said to be an interior point of a set S of

the complex plane if there exists some neighborhood of z0 that lies entirely within S.

If every point z of a set S is an interior point, then S is said to be an open set. Ex. Re(z) > 1 defines a right half-plane.

21

Sets in Complex Plane

Open annulus 22

Sets in Complex Plane If every neighborhood of z0 contains at least

one point that is in S & at least one point that is not in S z0 is a boundary point of S.

Boundary of a set S is the set of all boundary points of S.

Ex. For Re(z) ≥ 1, the points on Re(z) = 1 are boundary points.

23

Sets in Complex Plane If any two points z1 & z2 in a set S can be

connected by a polygonal line that lies entirely in S S is a connected set.

An open connected set is called a domain. Ex. The set Re(z) ≠ 4 is open but not

connected. Region is a domain in the complex

plane with all, some, or none of its boundary points.

24

Functions of a Complex Variable Function f from a set A to a set B is a rule of

correspondence that assigns to each element in A one and only one element in B. b = f (a) ⇔ b is the image of a Domain & range of the function f

Ex. A set of real numbers A: 3 ≤ x < ∞ & the function given by

the range of f : 0 ≤ y < ∞ f is a function of a real variable x.

( ) 3−= xxf

25

Functions of a Complex Variable When the domain A is a set of complex

numbers z → f is said to be a function of a complex variable z.

The image w of z will be complex, too.

where u = Re(w) & v = Im(w) are real-valued functions.

Ex.

( ) ( ) ( ) ( )7,, yxivyxuzfw +==

( ) ( ) ( ) ( ) ( )yxyixyxiyxiyxzf 4244 222 −+−−=+−+=⇒

( ) zzzzf ∀−= 42

Functions of a Complex Variable A complex function w = f (z) can be interpreted

as a mapping or transformation from the z-plane to the w-plane.

27

Ex. Find the image of the line Re(z) = 1 under the mapping f (z) = z2.

Functions of a Complex Variable

( )( )( )

41

21

1Re2,

,

2

2

22

vu

yvyu

xzxyyxv

yxyxu

−=∴

=−=

⇒

==

=−=

⇒

28

Functions of a Complex Variable Complex function w = f (z) is completely

determined by real-valued functions u(x, y) & v(x, y), even though u + iv may not be obtainable via familiar operations on z alone.

Ex. f (z) = xy2 + i(x2 − 4y3)

29

30

We also may interpret w = f (z) as a 2-D fluid flow by considering f (z) as a vector based at the point z.

Functions of a Complex Variable

Functions of a Complex Variable If x(t) + iy(t) is a parametric representation for

the path of the flow,

must coincide with

When f (z) = u(x, y) + iv(x, y), the path of the flow satisfies

( ) ( )dt

tdyidt

tdxT += ( ) ( )( )tiytxf +

( ) ( )( ) ( )

=

=

yxvdt

tdy

yxudt

tdx

,

,

31

Functions of a Complex Variable Find the streamlines of the flows.

( ) ( )( ) 21

2

11 ccxy

ectyectx

ydtdy

xdtdx

iyxzzf t

t

=⇒

==

⇒

−=

=⇒−==

−

( ) ( )

ycyxyx

xydxdy

xydtdy

yxdtdx

xyiyxzzf

222

22

22

2222

2 2

2

=+⇒−

=⇒

=

−=⇒

+−==

32

Functions of a Complex Variable Limit of a function Suppose f is defined in some neighborhood of z0,

except possibly at z0 itself. f is said to possess a limit at z0,

For each ε > 0, there exists a δ > 0 such that |f (z) − L| < ε whenever 0 < |z − z0| < δ.

( ) ( )8lim0

Lzfzz

=→

33

Functions of a Complex Variable Limit of sum, product, quotient Suppose Then ( ) ( ) ( )[ ]( ) ( ) ( )

( ) ( )( ) 0 ,lim

lim

lim

22

1

21

21

0

0

0

≠=

=

+=+

→

→

→

LLL

zgzfiii

LLzgzfii

LLzgzfi

zz

zz

zz

( ) ( ) 2100

lim and lim LzgLzfzzzz

==→→

34

( ) 0 ,012

21

1 ≠+++++= −− n

nn

nn aazazazazazf

Functions of a Complex Variable Continuity at a point A function f is continuous at z0 if

A function f defined by

where n is a nonnegative integer & ai (i = 0, 1, …, n) are complex constants, is called a polynomial of degree n. A polynomial is continuous everywhere.

( ) ( )00

lim zfzfzz

=→

35

Functions of a Complex Variable A rational function

where g & h are polynomial functions, is continuous except at points at which h(z) = 0.

Derivative Suppose f is defined in a neighborhood of z0.

( ) ( ) ( ) ( )

yixzz

zfzzfzfz

∆+∆=∆∆

−∆+=′⇒

→∆

where

9lim 00

00

( ) ( )( )zhzgzf =

36

Functions of a Complex Variable If f is differentiable at z0, then f is continuous at z0. If f & g are differentiable at a point z, and c is a

complex constant, then

( ) ( )

( ) ( )[ ] ( ) ( )

( ) ( )[ ] ( ) ( ) ( ) ( )zgzfzgzfzgzfdzd

zgzfzgzfdzd

zfczcfdzdc

dzd

′+′=

′+′=+

′== ,0

( )( )

( ) ( ) ( ) ( )( )[ ]

( )( ) ( )( ) ( )zgzgfzgfdzd

zgzgzfzfzg

zgzf

dzd

′′=

′−′=

2

1−= nn nzzdzd

37

Functions of a Complex Variable

Ex. Differentiate f (z) = 3z4 − 5z3 + 2z.

Ex. Differentiate

In order for f to be differentiable at z0, (9) must approach the same complex number from any direction.

( ) 2151223543 2323 +−=+⋅−⋅=′⇒ zzzzzf

( )14

2

+=

zzzf

( ) ( )( ) ( )2

2

2

2

1424

144214

++

=+

⋅−⋅+=′⇒

zzz

zzzzzf

38

Functions of a Complex Variable

Ex. Show that f (z) = x + i4y is nowhere differentiable.

⇒ With ∆z = ∆x + i∆y, we have

Let ∆z → 0 along a line parallel to x-axis Let ∆z → 0 along a line parallel to y-axis

( ) ( )( )

( ) ( )yixyix

zzfzzf

yixiyxyyixxzfzzf

zz ∆+∆∆+∆

=∆

−∆+∴

∆+∆=−−∆++∆+=−∆+

→∆→∆

4limlim

44)(4

00

39

Functions of a Complex Variable Analytic functions A function f (z) is said to be analytic at a point z0

if f is differentiable at z0 & at every point in some neighborhood of z0.

f is analytic in a domain D if it is analytic at every point in D.

Ex. f (z) = |z|2 is differentiable only at z = 0. Ex. g(z) = z2 is differentiable at every point z in the

complex plane. A function that is analytic everywhere is said to be

an entire function. 40

Functions of a Complex Variable A number c is a zero of a polynomial function

f if and only if z − c is a factor of f (z). Ex. f (z) = z2 − 2z + 2 = (z − 1 − i)(z − 1 + i). Ex. Find the value of the limit.

( )( )( )[ ] ( )[ ]

( ) 21

122

11lim

1111lim

222lim

1

12

2

1

ii

iiziz

iziziziz

izzz

iz

iziz

+=

+=

+++−

=

−−−+−+−−−

=−+−

+→

+→+→

41

( )10 and xv

yu

yv

xu

∂∂

−=∂∂

∂∂

=∂∂

Cauchy-Riemann Equations A necessary condition for analyticity Suppose f (z) = u(x, y) + iv(x, y) is differentiable at a

point z = x + iy. Then at z the 1st-order partial derivatives of u & v exist and satisfy the Cauchy-Riemann equations

(Proof)

( ) ( ) ( )

( ) ( ) ( ) ( )yix

yxivyxuyyxxivyyxxuz

zfzzfzf

z

z

∆+∆−−∆+∆++∆+∆+

=

∆−∆+

=′

→∆

→∆

,,,,lim

lim

0

0

Cauchy-Riemann Equations Since the limit exists. ∆z can approach zero from

any direction. In particular, if ∆z → 0 horizontally,

If we let ∆z → 0 vertically,

( ) ( ) ( ) ( ) ( )

( )11

,,lim,,lim00

xvi

xu

xyxvyxxvi

xyxuyxxuzf

xx

∂∂

+∂∂

=

∆−∆+

+∆

−∆+=′

→∆→∆

( ) ( ) ( ) ( ) ( )

( )12

,,lim,,lim00

yv

yui

yiyxvyyxvi

yiyxuyyxuzf

yy

∂∂

+∂∂

−=

∆−∆+

+∆

−∆+=′

→∆→∆

43

Cauchy-Riemann Equations If f is analytic throughout a domain D, then the

real functions u & v must satisfy (10) at every point in D.

Ex. Polynomial f (z) = z2 + z is analytic for all z.

Ex. Show that f (z) = (2x2 + y) + i(y2 − x) is not analytic at any point.

( ) ( ) ( ) ( ) ( )

xvy

yu

yvx

xu

yxivyxuyxyixyxzf

∂∂

−=−=∂∂

∂∂

=+=∂∂

⇒

+=+++−=

2 and 12

,,222

44

Cauchy-Riemann Equations

We see that but that is

satisfied only on the line y = 2x. f is nowhere analytic.

−=∂∂

=∂∂

=∂∂

=∂∂

⇒1 and 1

2 and 4

xv

yu

yyvx

xu

xv

yu

∂∂

−=∂∂

yv

xu

∂∂

=∂∂

45

Cauchy-Riemann Equations Criterion for analyticity Suppose the real-valued functions u & v are

continuous and have continuous 1st-order partial derivatives in a domain D. If u & v satisfy the Cauchy-Riemann equations at all points of D, then f (z) = u(x, y) + iv(x, y) is analytic in D.

Ex. For we have ( ) 2222 yxyi

yxxzf

+−

+=

( ) ( ) xv

yxxy

yu

yv

yxxy

xu

∂∂

−=+

−=

∂∂

∂∂

=+

−=

∂∂

222222

22 2 and 46

( ) ( )13

yui

yv

xvi

xuzf

∂∂

−∂∂

=∂∂

+∂∂

=′

Cauchy-Riemann Equations (11) & (12) were obtained under the basic

assumption that f was differentiable at z.

Thus,

Ex. f (z) = z2 is differentiable for all z. ( )

( )( ) zyixzf

yxvxyyxv

xxuyxyxu

222 22,

2, 22

=+=′∴

=∂∂

⇒=

=∂∂

⇒−=

47

Cauchy-Riemann Equations If the real-valued functions u & v are

continuous & have continuous 1st-order derivatives in a neighborhood of z and if u & v satisfy (10) at z, then f (z) = u(x, y) + iv(x, y) is differentiable at z & f′ (z) is given by (13).

Ex. f (z) = x2 − iy2 is nowhere analytic.

(10) are satisfied only when y = −x. On that line (13) gives f′ (z) = 2x = − 2y.

0 ,0 ,2 ,2 =∂∂

=∂∂

−=∂∂

=∂∂

⇒xv

yuy

yvx

xu

48

Cauchy-Riemann Equations Harmonic functions A real-valued fx φ(x, y) that has continuous 2nd-

order partial derivatives in a domain D & satisfies Laplace equation is said to be harmonic in D.

If f (z) = u(x, y) + iv(x, y) is analytic in a domain D, then u & v are harmonic functions.

(proof) Assume u & v have continuous 2nd-order partial

derivatives. Since f is analytic, (10) are satisfied.

xyv

yu

xv

yu

yxv

xu

yv

xu

∂∂∂

−=∂∂

⇒∂∂

−=∂∂

∂∂∂

=∂∂

⇒∂∂

=∂∂ 2

2

22

2

2

and

Cauchy-Riemann Equations Adding these two equations gives

This shows that u(x, y) is harmonic. Similarly, we

can obtain

If u(x, y) is a given fx harmonic in D, it is sometimes possible to find another fx v(x, y) that is harmonic in D so that u(x, y) + iv(x, y) is an analytic fx in D. v is called a conjugate harmonic function of u.

02

2

2

2

=∂∂

+∂∂

yu

xu

02

2

2

2

=∂∂

+∂∂

yv

xv

50

Cauchy-Riemann Equations Ex. (a) Verify that u(x, y) = x3 − 3xy2 − 5y is

harmonic in the entire complex plane. (b) Find the conjugate harmonic fx of u.

(a)

−=∂∂

−−=∂∂

=∂∂

−=∂∂

xyuxy

yu

xxuyx

xu

6 ,56

6 ,33

2

2

2

222

0662

2

2

2

=−=∂∂

+∂∂

⇒ xxyu

xu

51

Cauchy-Riemann Equations (b) Since v must satisfy (10), we must have

( ) ( )

( )

( )

( ) ( )( ) Cxyyxyxv

Cxxhxh

xyxhxyxv

xyyu

xv

xhyyxyxvyxxu

yv

++−=∴

+==′⇒

+=′+=∂∂

⇒

−−−=∂∂

−=∂∂

+−=⇒−=∂∂

=∂∂

53,5 ,5

566

56

3, 33

32

3222

52

Cauchy-Riemann Equations Suppose u & v are the harmonic fxs forming

the real & imaginary parts of an analytic fx f (z). The level curves u(x, y) = c1 & v(x, y) = c2 form two orthogonal families of curves.

Ex. f (z) = z = x + iy → x = c1 & y = c2.

53

Exponential & Logarithmic Functions Exponential function In real variables, f (x) = ex has the properties For Euler’s formula,

For z = x + iy, it is natural to expect that

The exponential fx of a complex variable z is

defined as

( ) ( ) ( ) ( ) ( )2121 and xfxfxxfxfxf =+=′

( ) ( )14sincos yiyeee xiyxz +== +

number real a ,sincos yyiyeiy +=

( )yiyeeee xiyxiyx sincos +==+

54

Exponential & Logarithmic Functions Ex. Evaluate e1.7+4.2i.

Re(ez) = u(x,y) = excos y & Im(ez) = v(x,y) = exsin y are continuous & have continuous 1st partial derivatives at every point z of the complex plane. Moreover, (10) are satisfied at all points.

xvye

yu

yvye

xu xx

∂∂

−=−=∂∂

∂∂

==∂∂

⇒ sin and cos

( ) iieeyx

i 7710.46837.22.4sin2.4cos2.4 and 7.1

7.12.47.1 −−=+=⇒

==⇒+

55

Exponential & Logarithmic Functions Thus, f (z) = ez is analytic for all z; in other

words, f is an entire function. The derivative of f can be obtained via (11).

If z1 = x1 + iy1 & z2 = x2 + iy2, we can have

( ) ( ) ( )

zz

xx

eedzd

zfyeiyexvi

xuzf

=∴

=+=∂∂

+∂∂

=′ sincos

( ) ( ) ( ) ( )221121 sincossincos 21 yiyeyiyezfzf xx ++=

Exponential & Logarithmic Functions

Similarly, one can prove that 21

2

1zz

z

z

eee −=

( ) ( ) ( )[( )]( ) ( )[ ]

( )2121

21

21

21

2121

2121

212121

sincos

sincoscossin sinsincoscos

zzzz

xx

xx

eeezzf

yyiyyeyyyyi

yyyyezfzf

+

+

+

=∴

+=+++=

++−=

57

Exponential & Logarithmic Functions Periodicity f (z) = ez is periodic with the complex period 2πi.

Divide the complex plane into

( ) ( )zfizfzeeee

ieziziz

i

=+∴==⇒

=+=+

π

ππππ

π

2 allfor

12sin2cos22

2

( ) ( ),2 ,1 ,0 where1212

±±=+≤<−

nnyn ππ

( ) ( ) ( ) =±=±= izfizfzf ππ 42 58

Exponential & Logarithmic Functions The strip −π < y ≤ π is called the fundamental

region for f (z) = ez. The flow over the fundamental region.

59

Exponential & Logarithmic Functions Polar form of a complex number Using (6), z = r(cosθ + i sinθ).

Ex. Find the steady-state current I(t) in an RLC series circuit.

( )tjeEqC

RIdtdIL

dtdqItEq

CdtdqR

dtqdL

ω

ω

0

02

2

Im1

and sin1

=++⇒

==++

θ

θ θθi

i

rezie

=∴

+= sincos

60

Exponential & Logarithmic Functions

Assume I(t) = Im(I0 e jωt).

( )

=∴

−+==

=

−+

=++

=∴

=

++⇒

−

−−

tjj

RC

Ljj

eeZEtI

eC

LReZZ

ZE

CLjR

E

CjRLj

EI

EICj

RLj

ωθ

ωω

θ

ωω

ωω

ωω

ωω

0

1tan22

0000

00

Im

1 where

11

1

1

61

Exponential & Logarithmic Functions Logarithmic function Logarithm of a complex number z (z ≠ 0) is defined

as the inverse of the exponential function.

To find the real & imaginary parts of ln z

( )15 if ln

wezzw ==

( )

==⇔=

=∴=⇒=+⇒

==⇒

+===+= +

zvvxy

zuezeyx

veyvexviveeeiyxz

euu

uu

uivuw

arg tan

log

sin and cossincos

22222

θ 62

( ) ( )16 ,2 ,1 ,0for 2logln ±±=++= nnizz e πθ

Exponential & Logarithmic Functions For z ≠ 0 and θ = arg(z)

Note that there are infinitely many values of the

logarithm of a complex number z. Ex. Evaluate (a) ln(−2) and (b) ln(−1 − i).

( ) ( )( ) ( )

( ) ( )( ) ( )ππ

πθ

ππ

πθ

niiiib

nia

ee

e

2453466.01ln 3466.02log1log and 451arg

26932.02ln 6932.02log and 2arg

++=−−∴

==−−=−−=

++=−∴

=−=−=

63

( )17 Arg logLn zizz e +=

Exponential & Logarithmic Functions Principal value As a consequence of (16), the logarithm of a

positive real number has many values. With the principal argument of a complex number,

Arg(z), in the interval (−π, π], we can define the principal value of ln z as

Ex. Evaluate (a) Ln(−2) & (b) Ln(−1 − i). ( ) ( ) ( )( ) ( ) ( ) ( )433466.01Ln 431Arg

6932.02Ln 2Arg ππθ

ππθiiib

ia−=−−∴−=−−=

+=−∴=−=

Exponential & Logarithmic Functions (16) can be interpreted as an infinite

collection of logarithmic functions. Each fx in the collection is called a branch of ln z.

f (z) = Ln z is called the principal branch of ln z or the principal logarithmic function.

Some familiar properties hold in the complex case: ( )

−=

+=

212

1

2121

lnlnln

lnlnln

zzzz

zzzz

65

Exponential & Logarithmic Functions Ex. For z1 = i & z2 = −1 + i,

( ) ( )

( )21

21

21

453466.0

433466.0

20

433466.01

zzi

iizz

iizz

Ln

LnLn

LnLn

≠+=

++

+=+

−=−−=

π

ππ

π

66

Exponential & Logarithmic Functions Analyticity f (z) = Ln z is not continuous at z = 0 since f (0) is

not defined. f (z) = Ln z is discontinuous at all points of the

negative real axis because Im[f (z)] = v = Arg(z) is discontinuous at these points. For x0 on the negative real axis, as z → x0 from the

upper half-plane, Arg(z) → π, whereas as z → x0 from the lower half-plane, Arg(z) → −π.

Thus, f (z) = Ln z is NOT analytic on the nonpositive real axis.

67

Dzz

zdzd in allfor 1Ln =

Exponential & Logarithmic Functions However, f (z) = Ln z is analytic throughout the

domain D consisting of all the points in the complex plane except the nonpositive real axis.

Since f (z) = Ln z is the principal branch of ln z, the nonpositive real axis is referred to as a branch cut for the function.

(10) is satisfied throughout D. Also,

68

The figure shows w = Ln z as a flow.

Exponential & Logarithmic Functions

69

( )18 0for ln≠= zez zαα

Exponential & Logarithmic Functions Complex powers Define complex powers of a complex number. If α is a complex number & z = x + iy,

Since ln z is multiple-valued, zα is multiple-valued. However, when α = n (integer), (18) is single-

valued since there is only one value for z2, z3, z−1… Ex. Suppose α = 2 & z = reiθ

( )( )

2

242log22log2ln2

1

zrereeereeeee

ii

iikiirkirz ee

=⋅=

⋅=== ++

θθ

θθπθπθ

70

Exponential & Logarithmic Functions If we use Ln z in place of ln z, (18) gives the

principal values of zα. Ex. Evaluate i2i.

i2i is real for every value of n. Since Arg(z) = π/2, we obtain the principal value of

i2i for n = 0.

( )[ ] ( ) ,2 ,1 ,0 ,

2 ,2arg , 41221log22 ±±===⇒

===+−++ neei

iziznniii e πππ

απ

043.02 ≅=⇒ −πei i

71

2cos and

2sin

sincos and sincosixixixix

ixix

eexieex

xixexixe−−

−

+=

−=⇒

−=+=

Trigonometric & Hyperbolic Functions Trigonometric functions For a real variable x,

Similarly, for a complex number z = x + iy, ( )19

2cos and

2sin

iziziziz eezieez

−− +=

−=

zz

zz

zz

zzz

sin1csc ,

cos1sec ,

tan1cot ,

cossintan ====

Trigonometric & Hyperbolic Functions Analyticity Since eiz & e−iz are entire functions, it follows that

sin z & cos z are entire functions. Note that sin z = 0 only for z = nπ & cos z = 0 only

for z = (2n + 1)π/2. Thus, tan z & sec z are analytic except at z = (2n + 1)π/2, and cot z & csc z are analytic except at z = nπ.

73

Trigonometric & Hyperbolic Functions Derivatives Since (d/dz)ez = ez, we have (d/dz)eiz = ieiz and

(d/dz)e−iz = −ie−iz.

zeeiee

dzdz

dzd iziziziz

cos22

sin =+

=−

=⇒−−

( )20

cotcsccsctansecsec

csccotsectan

sincoscossin

22

zzzdzdzzz

dzd

zzdzdzz

dzd

zzdzdzz

dzd

−==

−==

−==

74

Trigonometric & Hyperbolic Functions Identities Same in the complex case.

( )( )

( )( )

zzzzzz

zzzzzzzzzzzz

zzzzzz

22

212121

212121

22

sincos2coscossin22sin

sinsincoscoscossincoscossinsin

1sincoscoscos

sinsin

−=

==±

±=±=+

=−−=−

75

( )21 sinhsincoshcoscossinhcoscoshsinsin

−=+=

∴yxiyxzyxiyxz

Trigonometric & Hyperbolic Functions If y is real, the hyperbolic sine & cosine are

From (19) & Euler’s formula 2

cosh and 2

sinhyyyy eeyeey

−− +=

−=

( ) ( )

−+

+=

−=⇒

−−

+−+

2cos

2sin

2sin

yyyy

iyxiiyxi

eexieex

ieez

76

Trigonometric & Hyperbolic Functions From (21),

( )

( )( )( )24sinhcoscos

23sinhsin sinhcossinh1sin

sinhcoscoshsinsin

22sinhcosh1

222

22

2222

22222

22

yxz

yxyxyx

yxyxz

yy

+=

+=

⋅++=

⋅+⋅=

−=

77

Trigonometric & Hyperbolic Functions Zeros A complex number z is zero iff |z|2 = 0. To have sin z = 0, we must have sin2x + sinh2y = 0.

from (23). This implies that sin x = 0 & sinh y = 0, and so x = nπ & y = 0.

Zeros of sin z are z = x + iy = nπ, where n = 0, ±1, ±2, …

Similarly, zeros of cos z are z = (2n + 1)π/2, where n = 0, ±1, ±2, …

78

Trigonometric & Hyperbolic Functions Ex. Evaluate sin(2 + i).

Ex. Solve the equation cos z = 10.

( ) iii 4891.04031.11sinh2cos1cosh2sin2sin −=+=+⇒

( )( )

,2 ,1 ,0for 11310log2

,2 ,1 ,0for 211310log

11310

0120

102

cos

2

±±=+=∴

±±=+±=⇒

±=⇒

=+−⇒

=+

=⇒−

ninz

niniz

e

ee

eez

e

e

iz

iziz

iziz

π

π79

( )25 2

cosh and 2

sinh

zzzz eezeez−− +

=−

=

Trigonometric & Hyperbolic Functions Hyperbolic sine & cosine For any complex number z = x + iy,

Also,

( )26

sinh1csch

cosh1sech

tanh1coth

coshsinhtanh

zz

zz

zz

zzz

==

==

80

( )27 sinhcosh and coshsinh zzdzdzz

dzd

==

Trigonometric & Hyperbolic Functions Hyperbolic sine & cosine are entire functions. Functions of (26) are analytic except at points

where the denominators are zero. From (25), it is easy to see that

Trigonometric & hyperbolic functions are related in complex calculus.

( ) ( ) ( )( ) ( ) ( )29 coscosh ,sinsinh

28 coshcos ,sinhsin

izzizizizziziz

=−==−=

81

Trigonometric & Hyperbolic Functions Zeros Zeros of sinh z & cosh z are pure imaginary and are

respectively,

Also, note that

( ) ( )( ) ( )[ ]

[ ]( )

( )31sinsinhcoscoshcosh Similarly,30sincoshcossinhsinh

sinhcoscoshsin sinhcoscoshsin

sinsinsinh

yxiyxzyxiyxz

xyixyixyixyi

ixyiiziz

+=+=∴+−−=

−+−−=+−−=−=

( ) ,2 ,1 ,0for 2

12 and ±±=+== ninzinz ππ

Trigonometric & Hyperbolic Functions Periodicity From (21),

From (30) & (31), ( ) ( )

( ) ( )( ) ziz

zyxiyxiiyxiz

cosh2coshsinh2sincosh2cossinh

2sinh2sinh

=+=+++=

++=+

πππ

ππ

( ) ( )( ) ( )

( ) zzzyxiyx

yxiyxiyxz

cos2cossinsinhcoscoshsin

sinh2coscosh2sin 2sin2sin

=+=+=+++=

++=+

π

ππππ

83

wzzw sin if sin 1 == −

Inverse Trigonometric & Hyperbolic Functions Since the inverse of these analytic functions

are multiple-valued functions, they do NOT possess inverse functions in its strictest interpretation.

Inverse sine Def.

( ) ( )[ ]2121212

2

1lnsin 1

012 2

zizizzize

izeeziee

iw

iwiwiwiw

−+−=∴−+=⇒

=−−⇒=−

⇒

−

−

Inverse Trigonometric & Hyperbolic Functions Inverse cosine

Inverse tangent

( ) ( )[ ]2121212

2

1lncos 1

012 2

zizizzze

zeezee

iw

iwiwiwiw

−+−=∴−+=⇒

=+−⇒=+

−

−

( ) ( )

zizii

ziziiz

izize

eizezeeiee

iw

iwiwiwiw

iwiw

−+

=+−−

=∴−+

=⇒

+=−⇒=+−

⇒

−

−

−

ln2

ln2

tan 11

11

12

22

85

Inverse Trigonometric & Hyperbolic Functions Ex. Find all values of

5sin 1−

( )[ ] ( )[ ]

( )

( )25log22

,2 ,1 ,0 ,22

25log

25ln25ln

515ln5sin2121

++=

±±=

++±−=

±−=±−=

−+−=−

e

e

in

nini

iiiii

ii

ππ

ππ

86

Inverse Trigonometric & Hyperbolic Functions Derivatives To find the derivative of w = sin−1z, we begin by

differentiating z = sin w: ( ) ( )

( ) 212

1

212212

11sin

11

sin11

cos1 sin

zz

dzd

zwwdzdww

dzdz

dzd

−=∴

−=

−==⇒=

−

( ) 21

212

1

11tan and

11cos

zz

dzd

zz

dzd

+=

−

−= −−

87

Inverse Trigonometric & Hyperbolic Functions Ex. Find the derivative of w = sin−1z at

( ) ( ) 22

141

51

121212

5

iidz

dw

z

=±

=−

=

−

=⇒=

5=z

88

( )[ ] ( )( )[ ] ( )

211

212

12121

212

12121

11tanh

11ln

21tanh

11cosh1lncosh

11sinh1lnsinh

zz

dzd

zzz

zz

dzdzzz

zz

dzdzzz

−=

−+

=

−=−+=

+=++=

−−

−−

−−

Inverse Trigonometric & Hyperbolic Functions Inverse hyperbolic functions & derivatives

Ex. Find all values of cosh−1(−1) ( ) ( ) ( )

( ) ,2 ,1 ,0 ,12 21log1ln1cosh 1

±±=+=++=−=−−

ninine

πππ

89

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