Arnold D. N. - Complex Analysis.pdf - Complex analysis - Kuya

COMPLEX ANALYSIS 1

Douglas N. Arnold 2

References:

John B. Conway, Functions of One Complex Variable, Springer-Verlag, 1978.

Lars V. Ahlfors, Complex Analysis, McGraw-Hill, 1966.

Raghavan Narasimhan, Complex Analysis in One Variable, Birkhauser, 1985.

CONTENTS

I. The Complex Number System ::::::::::::::::::::::::::::::::::::::::::::: 2

II. Elementary Properties and Examples of Analytic Fns. ::::::::::::::: 3

Dierentiability and analyticity ::::::::::::::::::::::::::::::::::::::::::: 4

The Logarithm::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::6

Conformality::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::6

Cauchy{Riemann Equations:::::::::::::::::::::::::::::::::::::::::::::::7

Mobius transformations ::::::::::::::::::::::::::::::::::::::::::::::::::: 9

III. Complex Integration and Applications to Analytic Fns. :::::::::::: 11

Local results and consequences ::::::::::::::::::::::::::::::::::::::::::: 12

Homotopy of paths and Cauchy's Theorem ::::::::::::::::::::::::::::::: 14

Winding numbers and Cauchy's Integral Formula:::::::::::::::::::::::::15

Zero counting; Open Mapping Theorem :::::::::::::::::::::::::::::::::: 17

Morera's Theorem and Goursat's Theorem ::::::::::::::::::::::::::::::: 18

IV. Singularities of Analytic Functions::::::::::::::::::::::::::::::::::::19

Laurent series::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::20

Residue integrals:::::::::::::::::::::::::::::::::::::::::::::::::::::::::23

V. Further results on analytic functions:::::::::::::::::::::::::::::::::26

The theorems of Weierstrass, Hurwitz, and Montel ::::::::::::::::::::::: 26

Schwarz's Lemma :::::::::::::::::::::::::::::::::::::::::::::::::::::::: 28

The Riemann Mapping Theorem ::::::::::::::::::::::::::::::::::::::::: 29

Complements on Conformal Mapping :::::::::::::::::::::::::::::::::::: 31

VI.

Harmonic Functions :::::::::::::::::::::::::::::::::::::::::::::::::::::: 32

The Poisson kernel:::::::::::::::::::::::::::::::::::::::::::::::::::::::33

Subharmonic functions and the solution of the Dirichlet Problem ::::::::: 36

The Schwarz Reection Principle:::::::::::::::::::::::::::::::::::::::::39

1 These lecture notes were prepared for the instructor's personal use in teaching a half-semester course

on complex analysis at the beginning graduate level at Penn State, in Spring 1997. They are certainly not

meant to replace a good text on the subject, such as those listed on this page.

2 Department of Mathematics, Penn State University, University Park, PA 16802.

Email: dna@math.psu.edu. Web: http://www.math.psu.edu/dna/.

I. The Complex Number System

R is a eld. For n > 1, R n is a vectorspace over R, so is an additive group, but doesn't

have a multiplication on it. We can endow R 2

with a multiplication by

(a;b)(c;d) = (acbd;bc + ad):

Under this denition R 2 becomes a eld, denoted C. Note that (a=(a 2 + b 2 );b=(a 2 + b 2 ))

is the multiplicative inverse of (a;b). (Remark: it is not possible to endow R n with a eld

structure for n > 2.) We denote (0; 1) by i and identify x 2R with (x; 0), so RC. Thus

(a;b) = a + bi, a;b 2R. Note that i 2 = 1. C is generated by adjoining i to R and closing

under addition and multiplication. It is remarkable that the addition of i lets us not only

solve the equation x 2 + 1 = 0, but every polynomial equation.

For a and b real and z = a + bi we dene Re z = a, Im z = b, z = a bi, and

jzj = (a 2 + b 2 ) 1=2 . Then

Re z = (z + z)=2; Im z = (z z)=(2i);

jzj 2 ;

zw = z w; zw = z w;

z = z

z=w = z= w; jz + wjjzj + jwj:

The map 7! (cos ; sin ) denes a 2-periodic map of the real line onto the unit

circle in R 2 . In complex notation this map is 7! cis := cos + i sin . Every nonzero

complex number can be written as r cis where r > 0 is uniquely determined and 2 R

is uniquely determined modulo 2. The number 0 is equal to r cis where r = 0 and

is arbitrary. The relation z = r cis determines the relations z 7! r which is simply the

function r = jzj and z 7! . The latter is denoted = arg . Note that for z 6= 0, arg is

determined modulo 2 (while arg 0 is arbitrary). We can normalize arg by insisting that

arg z 2 (;]. Note that if z 1 = r cis 1 and z 2 = r cis 2 then z 1 z 2 = r 1 r 2 cis( 1 + 2 ).

The latter formula just encapsulates the formula for the sine and cosine of a sum, and gives

arg z 1 z 2 = arg z 1 + arg z 2 . In particular, ir cis = r cis( + =2), so multiplication by i is

just the operation of rotation by =2 in the complex plane. Multiplication by an arbitrary

complex number r cis is just rotation by arg followed by (or preceded by) dilation by a

factor r. Further, z n = r n cis(n). Every nonzero z 2 C admits n distinct nth roots: the

jzj 2 = zz;

nth roots of r cis are p r cis[( + 2k)=n], k = 0; 1;::: ;n.

Lines and circles in the plane. Circles given by jz aj = r where a 2 C is the center

and r > 0 is the radius. If 0 6= b 2 C then the line through the origin in the direction

b is the set of all points of the form tb, t 2 R, or all z with Im(z=b) = 0. If t 2 R

and c > 0 then (t + ci)b = tb + cib represents a point in the half plane to the left of

b determined by the line tb, i.e., fz : Im(z=b) > 0g is the equation of that half-plane.

Similarly, fz : Im[(z a)=b] > 0g is the translation of that half-plane by a, i.e., the half-

plane determined by the line through a parallel to b and in the direction to the left of

Stereographic projection determines a one-to-one correspondence between the unit

sphere in R 3

minus the north-pole, S, and the complex plane via the correspondence

z $ x 1 + ix 2

1 x 3 ;

x 1 =

2 Re z

1 + jzj 2 ; x 2 =

1 + jzj 2 ; x 3 = jzj 2 1

jzj + 1 :

If we dene C 1 = C[f1g, then we have a one-to-one correspondence between S and

C 1 . This allows us to dene a metric on C 1 , which is given by

d(z 1 ;z 2 ) =

(1 + jz 1 j 2 )(1 + jz 2 j 2 ) ; d(z;1) =

1 + jzj 2 :

II. Elementary Properties and Examples of Analytic Functions

n=0 z n

converges (to 1=(1z)) if jzj < 1. It clearly diverges, in fact its terms become unbounded,

if jzj > 1.

For z 6= 1,

n=0 z n = (1 z N+1 )=(1 z). Therefore the geometric series

M n < 1 and suppose

that f n : X ! C are functions on some set X satisfying sup x2X jf n (x)j M n . Then

n=0 f n (x) is absolutely and uniformly convergent.

R = lim supja n j 1=n :

Then (1) for any a 2 C the power series

n=0 a n (z a) n converges absolutely for all

jzaj < R and it converges absolutely and uniformly on the disk jzaj r for all r < R.

(2) The sequence a n (z a) n is unbounded for all jz aj > R (and hence the series is

certainly divergent).

Thus we see that the set of points where a power series converges consists of a disk

jzaj < R and possibly a subset of its boundary. R is called the radius of convergence of

its series. The case R = 1 is allowed.

Proof of theorem. For any r < R we show absolute uniform convergence on D r = fjzaj

rg. Choose r 2 (r;R). Then, 1=r > lim supja n j 1=n , so ja n j 1=n < 1=r for all n suciently

large. For such n, ja n j < 1=r n and so

sup

z2D r

ja n (za) n j < (r=r) n :

(r=r) n < 1 we get the absolute uniform convergence on D r .

If jz aj = r > R, take r 2 (R;r). Then there exist n arbitrarily large such that

ja n j 1=n 1=r. Then, ja n (za) n j (r=r) n , which can be arbitrarily large.

Since

2 Im z

2jz 1 z 2 j

Weierstrass M-Test. Let M 0 ;M 1 ;::: be positive numbers with

Theorem. Let a 0 ;a 1 ;2C be given and dene the number R by

Theorem. If a 0 ;a 1 ;::: 2 C and limja n =a n+1 j exists as a nite number or innity, then

this limit is the radius of convergence R of

a n (za) n .

a n z n has terms of increasing magnitude, and so cannot be convergent.

Thus jzj R. This shows that limja n =a n+1 j R.

Similarly, suppose that z < limja n =a n+1 j. Then for all n suciently large ja n j > ja n+1 zj

and ja n z n j > ja n+1 z n+1 j. Thus the series has terms of decreasing magnitude, and so, by

the previous theorem, jzj R. This shows that limja n =a n+1 j R.

Remark. On the circle of convergence, many dierent behaviors are possible.

z n diverges

z n =n diverges for z = 1, else converges, but not absolutely (this follows

from the fact that the partial sums of

z n =n 2

converges absolutely on jzj 1. Sierpinski gave a (complicated) example of a function

which diverges at every point of the unit circle except z = 1.

As an application, we see that the series

z n are bounded for z 6= 1 and 1=n # 0).

z n

n=0

converges absolutely for all z 2C and that the convergence is uniform on all bounded sets.

The sum is, by denition, exp z.

Now suppose that

n=0 a n (za) n has radius of convergence R, and consider its formal

derivative

n=1 na n (za) n1 =

n=0 (n+1)a n+1 (za) n . Now clearly

n=0 a n+1 (za) n

n=0 a n (za) n since

has the same radius of convergence as

(za)

a n+1 (za) n =

N+ X

a n (za) a 0 ;

n=0

and so the partial sums on the left and right either both diverge for a given z or both

converge. This shows (in a roundabout way) that lim supja n+1 j 1=n = lim supja n j 1=n =

1=R. Now lim(n+1) 1=n = 1 as is easily seen by taking logs. Moreover, it is easy to see that if

lim sup b n = b and lim c n = c > 0, then lim sup b n c n = bc. Thus lim supj(n + 1)a n+1 j 1=n =

1=R. This shows that the formal derivative of a power series has the same radius of

convergence as the original power series.

Dierentiability and analyticity. Denition of dierentiability at a point (assumes

function is dened in a neighborhood of the point).

Most of the consequences of dierentiability are quite dierent in the real and complex

case, but the simplest algebraic rules are the same, with the same proofs. First of all,

dierentiability at a point implies continuity there. If f and g are both dierentiable at a

point a, then so are f g, f g, and, if g(a) 6= 0, f=g, and the usual sum, product, and

quotient rules hold. If f is dierentiable at a and g is dierentiable at f(a), then g f

is dierentiable at a and the chain rule holds. Suppose that f is continuous at a, g is

continous at f(a), and g(f(z)) = z for all z in a neighborhood of a. Then if g 0 (f(a)) exists

and is non-zero, then f 0 (a) exists and equals 1=g 0 (f(a)).

Proof. Without loss of generality we can suppose that a = 0. Suppose that jzj >

limja n =a n+1 j. Then for all n suciently large ja n j < ja n+1 zj and ja n z n j < ja n+1 z n+1 j.

Thus the series

for all jzj = 1.

Denition. Let f be a complex-valued function dened on an open set G in C. Then f

is said to be analytic on G if f 0 exists and is continuous at every point of G.

Remark. We shall prove later that if f is dierentiable at every point of an open set in C

it is automatically analytic; in fact, it is automatically innitely dierentiable. This is of

course vastly dierent from the real case.

If Q is an arbitrary non-empty subset of C we say f is analytic on Q if it is dened and

analytic on an open set containing Q.

We now show that a power series is dierentiable at every point in its disk of convergence

and that its derivative is given by the formal derivative obtained by dierentiating term-

by-term. Since we know that that power series has the same radius of convergence, it

follows that a power series is analytic and innitely dierentiable in its convergence disk.

For simplicity, and without loss of generality we consider a power series centered at zero:

f(z) =

zz 0

g(z 0 )

is satised for all z suciently close to z 0 , where g(z) =

n=1 na n z n1 . Let s N (z) =

n=0 a n z n , R N (z) =

n=N+1 a n z n . Then

f(z) f(z 0 )

s N (z) s N (z 0 )

zz 0

g(z 0 )

zz 0

s 0 N (z 0 )

R N (z) R N (z 0 )

=: T 1 + T 2 + T 3 :

+ js 0 N (z 0 ) g(z 0 )j +

zz 0

Now s 0 N (z 0 ) is just a partial sum for g(z 0 ), so for N suciently large (and all z), T 2 =3.

Also,

R N (z) R n (z 0 )

zz 0

a n z n z 0

zz 0 :

n=N+1

Now jz 0 j < r < R for some r, and if we restrict to jzj < r, we have

a n z n z 0

zz 0

= ja n jjz n1 + z n2 z 0 + + z n 0 j a n nr n1 :

a n nr n1 is convergent, we have for N suciently large and all jzj < r then

T 3 < =3. Now x a value of N which is suciently large by both criteria. Then the

dierentiability of the polynomial s N shows that T 1 =3 for all z suciently close to z 0 .

We thus know that if f(z) =

a n z n , then, within the disk of convergence, f 0 (z) =

n(n1)a n z n2 , etc. Thus a 0 = f(0), a 1 = f 0 (0),

a 2 = f 00 (0)=2, a 3 = f 000 (0)=3!, etc. This shows that any convergent power series is the sum

of its Taylor series in the disk of convergence:

na n z n1 , and by induction, f 00 (z) =

f(z) =

f n (a)

(za) n :

In particular, exp 0 = exp.

n a n z n . Suppose that the radius of convergence is R and that jz 0 j < R. We must

show that for any > 0, the inequality

f(z) f(z 0 )

Since

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