Basic complex analysis 1. Complex differentiation

(September 5, 2013)

Basic complex analysis

Paul Garrett garrett@math.umn.edu http://www.math.umn.edu/ garrett/

[This document ishttp://www.math.umn.edu/garrett/m/mfms/notes 2013-14/00 basic complex analysis.pdf]

1. Complex differentiation2. Exponentials, trigonometric functions3. Differentiating power series: Abels theorem4. Path integrals5. Cauchys theorem6. Cauchys formula7. Power series expansions, Moreras theorem8. Identity principle9. Liouvilles theorem: bounded entire functions are constant10. Laurent expansions around singularities11. Residues and evaluation of integrals12. Logarithms and complex powers13. Argument principle

Complex analysis is one of the most natural and productive continuations of basic calculus, not addressingpathologies and pitfalls, but, instead, showing that natural, well-behaved functions are even better thanimagined.

Mercifully, very many functions arising in practice are indeed natural and well-behaved in the relevant sense,so avoid pathologies, and behave even better than we had hoped. Arguably, the first 150 years of calculus infact addressed such functions, thereby discovering the further remarkable usefulness of calculus-as-complex-analysis long before anyone thought to worry about the subtler distinctions and troubles highlighted in the19th century. Thus, arguably, Euler, Lagrange, and the most effective of their contemporaries, to somedegree inadvertently thought in terms we can now reinterpret as justifiable as complex analysis.

1. Complex differentiation

[1.1] Complex differentiation For complex-valued f on an open set C, the complex derivative f (z),if it exists, is

f (z) = limh0

f(z + h) f(z)h

(for complex h 0)

It is critical that the limit exist for complex h approaching 0. If the limit exists for all z , say f is complexdifferentiable on .

Given a function f on a region , when there is a complex-differentiable F with F = f , say that F is aprimitive of f on . This is obviously a kind of anti-derivative.

[1.2] Examples Polynomial functions in z are complex-differentiable, with the same differentiationformulas as in single-real-variable calculus, because those results are essentially algebraic:

(zn) = limh0

(z + h)n zn

h= lim

h0

zn + nhzn1 + h2(. . .) zn

h

= limh0

nhzn1 + h2(. . .)

h= lim

h0

(nzn1 + h(. . .)

)= nzn1

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Paul Garrett: Basic complex analysis (September 5, 2013)

[1.3] z

and z

and Cauchy-Riemann equation From the notation, and as applied to polynomials in z,it seems that complex differentiation is application of z . What about z? For the moment, with z = x+ iy,we simply declare

z= 12

( x i

y

)and

z= 12

( x

+ i

y

)The signs are explained/remembered by checking that

zzn = 12

( x i

y

)(x+ iy)n = 12

(n(x+ iy)n1 i i n(x+ iy)n1

)= 12

(n(x+ iy)n1 + n(x+ iy)n1

)= nzn1

Thus, yes, there is the perhaps-surprising outcome

zzn = 12

( x

+ i

y

)(x+ iy)n = 12

(n(x+ iy)n1 + i i n(x+ iy)n1

)= 12

(n(x+ iy)n1 n(x+ iy)n1

)= 0

The latter is correct, despite appearing to be in conflict with the idea that knowing z or z determines theother, but a partial derivative of one thing with respect to another being 0 means theyre independent.

Thus, complex-differentiable f satisfies

zf(z) = 12

(lim

real 0

f(x+ + iy) f(x+ iy)

+ if(x+ i(y + )) f(x+ iy)

)= lim

real 0

f(x+ + iy) f(x+ iy)

f(x+ iy + i) f(x+ iy)i

)= 12

(f (z) f (z)

)= 0

That is, complex-differentiable f satisfies

zf(z) = 0 (Cauchy-Riemann equation)

This can be written in various equivalent forms, refering to real and imaginary parts separately, and/orwriting out the apparent definition of z .

[1.3.1] Remark: The converse is also true: a nice-enough function satisfying the Cauchy-Riemann equationis complex-differentiable.

2. Exponentials, trigonometric functions

[2.1] The exponential function The exponential functions power series expansion

ex = 1 +x

1!+x2

2!+x3

3!+ . . .

arises from the idea that bx+y = bx by for any b > 0 and x, y R. [1] Thus,

d

dxbx = lim

h0

bx+h bx

h= lim

h0

bx bh bx

h= bx lim

h0

bh 1h

[1] As usual, bn is first defined for positive integers n as repeated multiplication, then for negative integer n bybn = 1/bn, then for rationals via b1/n = n

b, and then for real exponents by taking limits of rational exponents.

2

Paul Garrett: Basic complex analysis (September 5, 2013)

because multiplication is continuous. Presuming that the limit exists, a constant depending on b > 0, wed likethe simplest outcome, namely, that this constant limit is 1, and find out a little later that b = e = 2.71828 . . .in that case. Presume that the differential equation

d

dxf(x) = f(x)

with f(0) = 1 has a convergent power series expansion f(x) = 1 + c1x+ c2x2 + . . ., solve iteratively for cns

by differentiating term-wise [2] and equate coefficients

c1 + 2c2x+ 3c3x2 + 4c4x

3 + . . . = 1 + c1x+ c2x2 + c3x

3 + . . .

giving the pattern n cn = cn1, so cn = 1/n!, giving the power series for ex. Since the factorial n! growsfaster than any power xn, for all x, the power series for ex converges absolutely for all real x, and, similarly,for all complex x. [3]

To see what number the base e is, use e1 = e:

e = 1 +1

1!+

1

2!+

1

3!+ . . . = 2.71828 . . .

We can check that this has achieved the desired effect:

[2.1.1] Claim:ez+w = ez ew (for z, w C)

[2.1.2] Remark: For non-real complex z, there is no need to try to define ez as a limit of simpler things,apart from the value of the power series as a limit of its finite partial sums.

Proof: This will follow from the binomial theorem

(x+ y)n = xn +

(n

1

)xn1y +

(n

2

)xn2y2 + . . .+

(n

n 2

)x2yn2 +

(n

n 1

)xyn1 + yn

with the usual binomial coefficients(nk

)= n!k! (nk)! . Compute directly

ez+w =n0

(z + w)n

n!=n,k

(n

k

)znkwk

n!=n,i

n!

k! (n k)!znkwk

n!

The n!s cancel. Letting ` = n k, this gives

ez+w =`,k

1

k! `!z`wk =

(`

z`

`!

)(k

wk

k!

)= ez ew

as desired. ///

[2.1.3] Corollary: The complex conjugate ez of ez is ez = ez, and |eix| = 1 for real x.

[2] To know that a convergent power series really can be differentiated term-wise is believable, but not completely

trivial to prove: this is Abels theorem below.

[3] The convergence of the power series for ex is also uniform on compact (closed and bounded) subsets of R or of C.

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Paul Garrett: Basic complex analysis (September 5, 2013)

Proof: Since complex conjugation is a continuous map from C to itself, respecting addition andmultiplication,

ez = 1 +z

1!+z2

2!+ . . . = 1 +

z

1+z2

2!+ = ez

Then|eix|2 = eixeix = eixeix = e0 = 1

for real x. ///

[2.2] Trigonometric functions Similarly, sinx and cosx both satisfy f = f , in radian measure:making this differential equation hold determines what radian measure must be. The two trig functions aredistinguished from each other by the initial conditions

cos 0 = 1, cos 0 = 0 and sin 0 = 0, sin 0 = 1

Presuming existence of power series expansions f(x) = c0 + c1x+ c2x2 + . . . for solutions of f (x) = f(x)

and differentiating term-wise, the differential equation gives

2c2 + 6c3x+ 12c4x2 + 20c5x

3 . . . = (c0 + c1x+ c2x

2 + . . .)

son(n 1)cn = cn2

The first two coefficients c0, c1 determine all: c2n = (1)nc0/(2n)! and c2n+1 = (1)nc1/(2n+ 1)!. Thus,

cosx = 1 x2

2!+x4

4! x

6

6!+ . . . sinx = x x

3

3!+x5

5! x

7

7!+ . . .

These power series do converge for all x R and, in fact, for x C, so really do produce solutions to thedifferential equations.

[2.3] Eulers identity The power series for ex, cosx, and sinx suggest Eulers identity

eix = cos x+ i sin x

at least for real x, and then for complex x, extending the definition of cosine and sine to complex numbersby their power series expansions:

eix = 1 +ix

1!+

(ix)2

2!+

(ix)3

3!+

(ix)4

4!+ . . . = 1 + i

x

1! x

2

2! ix

3

3!+x4

4!+ i

x5

5!+ . . .

=(

1 x2

2!+x4

4!+ . . .

)+ i(x x

3

3!+x5

5!+ . . .

)= cosx+ i sinx

It is amusing to use Eulers relation, coming from power series, to prove identities seemingly related totriangles and circles:

[2.3.1] Corollary: For real or complex z

cos2 z + sin2 z = 1

Proof: For z real or complex,

1 = eizeiz = (cos z + i sin z)(cos z i sin z) = cos2 z + sin2 z

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Paul Garrett: Basic complex analysis (September 5, 2013)

as desired. ///

Information about , implicit in the differential equations satisfied by cosx and sinx in radian measure, canbe certified to give the same we thought it should:

[2.3.2] Lemma: The function cos x defined by the previous power series has least positive zero (namely,/2) between 32 and 2.

Proof: From elementary estimates: noting that everything is real,

cos(3

2

)= 1

(32

)22!

+

(32

)44!(32

)66!

+ . . . 1 12> 0

Indeed, further, for any