Introduction to Complex Analysis

A. Bathi KasturiarachiKent State University, Stark Campus

December 5, 2007

Abstract

Complex Analysis is a rich area of mathematics. Its applications arenumerous and can be found in many other branches of mathematics, rang-ing from �uid dynamics, number theory, electrodynamics, and engineer-ing, to computer science. The purpose of this course is to introduce themain ideas of complex analysis to an undergraduate audience. It is as-sumed that the reader has familiarity with standard mathematical proofsand possesses a forti�ed knowledge of advanced calculus. Some back-ground in matrix theory may be useful but is not essential. The coursenotes are designed as a reader with many of the examples worked out indetail.

Contents

1 Complex Numbers 21.1 The Universe of Complex Numbers . . . . . . . . . . . . . . . . . 21.2 Geometrical Representation of Complex Numbers . . . . . . . . . 31.3 Complex Numbers as Vectors . . . . . . . . . . . . . . . . . . . . 5

1.3.1 Polar representation of a complex number . . . . . . . . . 61.4 Euler�s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.5 Roots of Unity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.6 Some Topological considerations . . . . . . . . . . . . . . . . . . 16

1.6.1 The Stereographic Projection . . . . . . . . . . . . . . . . 21

2 Analytic Functions 252.1 Complex Valued Functions . . . . . . . . . . . . . . . . . . . . . 252.2 Limits and Continuity . . . . . . . . . . . . . . . . . . . . . . . . 272.3 What is analyticity? . . . . . . . . . . . . . . . . . . . . . . . . . 29

3 Elementary Functions 323.1 Polynomials and Rational Functions . . . . . . . . . . . . . . . . 32

1

4 Complex Integration 414.1 Contours . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

4.1.1 Some Standard Parametrizations . . . . . . . . . . . . . . 424.2 Contour Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 434.3 Independence of Path . . . . . . . . . . . . . . . . . . . . . . . . 484.4 Cauchy�s Integral Theorem . . . . . . . . . . . . . . . . . . . . . 514.5 Cauchy�s Integral Formula . . . . . . . . . . . . . . . . . . . . . . 52

A Afterword 56

1 Complex Numbers

1.1 The Universe of Complex Numbers

Anyone who is comfortable with the real number system and the laws it�s ele-ments satisfy such as, commutative and associative laws for addition and multi-plication, distributive law, and inverse and identity properties, quickly realizesthe necessity to expand the universe when confronted with the simple equation

x2 = �1. (1)

Since the solution to equation (1) cannot be formulated using real numbers,we de�ne the symbol i =

p�1 and extend the universe of real numbers to

complex numbers in a natural way.

De�nition 1 A complex number has the form a + ib, where a and b are realnumbers, called the real and imaginary parts respectively. If b = 0; we say thenumber is real and if a = 0, we say the number is purely imaginary. Karl Friedrich

Gauss (1777-1855) was the�rst to intro-duce complexnumbers. RenéDescartes usedthe terms�real� and�imaginary� in1637.

Example 1.1 Write the following complex numbers in a+ ib form.

(a). (3 + i)� (2� 2i)

(b). (2 + i)(3� 5i)

(c). 1+i3�2i

Solution

(a). (3 + i)� (2� 2i) = 3 + i� 2 + 2i = 1 + 3i

(b). (2 + i)(3� 5i) = 6� 10i+ 3i� 5i2 = 11� 7i

(c). 1+i3�2i =

1+i3�2i

3+2i3+2i =

1+5i13 = 1

13 +513 i

Example 1.2 Let n be an integer. Show that

2

i4n = 1; i4n+1 = i; i4n+2 = �1; i4n+3 = �i.SolutionIt is easy to see that i2 = �1; i3 = �i; i4 = 1. Therefore, for any integer n;

i4n = 1

i4n+1 = i4ni = i

i4n+2 = i4ni2 = �1i4n+3 = i4ni3 = �i.

Example 1.3 Find all the solutions of z4 � 81 = 0.Solution

z4 � 81 = 0

(z � 3)(z + 3)(z + 3i)(z � 3i) = 0

z = �3;�3i

1.2 Geometrical Representation of Complex Numbers

Following the work of René Descarte on the Cartesian coordinate system, CasperWessel (1797) and Jean Pierre Argand (1806), independently introduced a methodby which we could represent complex numbers as a point on a plane.In what follows we will refer to a complex number as the point z = a+ ib or

simply the ordered pair (a; b). The following two de�nitions immediately arisefrom the geometrical representation.

x

y

z = a+ib

a­ib

a

b

|z|

3

Figure 1: Geometric representation of complex numbers

De�nition 2 The modulus of a complex number z = a + ib is denoted by jzjand is given by

jzj =pa2 + b2.

De�nition 3 The complex conjugate of a complex number z = a + ib isdenoted by z and is given by

z = a� ib.

Example 1.4 Prove the following very useful identities regarding any complexnumber z and its conjugate z.

(a). z1 + z2 = z1 + z2, z1 � z2 = z1 � z2

(b). z1z2 = z1 z2

(c). z1z2= z1

z2(here z2 6= 0)

(d). Re z = z+z2 and Im z = z�z

2i

(e). z = z

(f). zz = jzj2

(g). 1z =

zjzj2

SolutionIn what follows let z1 = x1 + iy1; z2 = x2 + iy2; z = x+ iy. First note that

(d), (e) are trivial.

(a). z1 + z2 = (x1 + x2) + i(y1 + y2) = (x1+x2)� i(y1+ y2) = x1� iy1+x2�iy2 = z1 + z2.

z1 � z2 = (x1 � x2) + i(y1 � y2) = (x1�x2)� i(y1�y2) = x1� iy1�(x2�iy2) = z1 � z2.

(b). z1z2 = (x1 + iy1)(x2 + iy2) = (x1x2 � y1y2) + i(x1y2 + y1x2) = (x1x2 �y1y2)� i(x1y2 + y1x2) = (x1 � iy1)(x2 � iy2) = z1 z2.

(c). Observe �rst that�z1z2

�= z1z2

z2z2= z1z2

jz2j2 =z1z2jz2j2 .

Next consider z1z2= z1z2

z2z2= z1z2

jz2j2 . Therefore,�z1z2

�= z1

z2.

(f). zz = (x+ iy)(x� iy) = x2 + y2 = jzj2. Part (g) follows from (f) easily.

4

1.3 Complex Numbers as Vectors

We can think of each complex number z = a+ ib as a vector in the plane: it isthe vector obtained by choosing the directed line segment that joins the originto the point (a; b). Sometimes we write v = ha; bi. The modulus jzj representsthe length of the vector v.

x

y

u

v

w = u + v

Figure 2: Illustration of the Triangular Inequality

Let u and v denote the vectors whose end points are the complex numbersz1 = x1 + iy1, z2 = x2 + iy2. We can �nd the vector sum w = u+ v using theparallelogram law as illustrated in Figure 2. Note that the vector w correspondsto the complex number z1 + z2. The sum of the lengths of any two sides of atriangle is always greater or equal to the third side. This fact leads to thefollowing very important law governing the magnitudes of complex numbers.

Proposition 4 Triangular Inequality Let z1 and z2 be any two complexnumbers. Then

jz1 + z2j � jz1j + jz2j. (2)

From equation (2) we can derive another useful inequality.Since z2 = z1 + (z2 � z1), by triangular inequality,

jz2j � jz1j + jz2 � z1j =) jz2j � jz1j � jz2 � z1j.Similarly, since z1 = z2 + (z1 � z2),

jz1j � jz2j + jz1 � z2j =) jz1j � jz2j � jz2 � z1j.

5

Combining the last two results we have,

j jz1j � jz2j j � jz2 � z1j. (3)

1.3.1 Polar representation of a complex number

If we use polar coordinates (r; �) to represent the complex number z = x + iythen we have the following identities.

x = jzj cos � (4)

y = jzj cos � (5)

r = jzj =px2 + y2 (6)

� = tan�1�yx

�(whenever de�ned): (7)

Notice that the angle � is multi valued. It is de�ned up to a multiple of 2�.We call � the argument or phase of z, and denote it by arg z. However, westill need to give a clear de�nition of the principal value we must select for it.

z = x+iy

x

yr

q

Figure 3: Polar Representation and branch cut

Start with z = x + iy = r(cos � + i sin �). Since one and only one value ofarg z, say ', satis�es the inequality

�� < ' � �,we shall call it the principal value of the argument of z, and write it as Arg

z. Speci�cally,

6

arg z = Arg z + 2n�,

where n is any integer. So �nally we have,

jzj =px2 + y2

tan(Arg z) =y

xz = x+ iy = r(cos � + sin �) = r cis(�),

where the operator cis refers to �cosine plus i sine.�The principle value has discontinuities, in that it jumps by 2� each time it

crosses the negative real axis. This line of discontinuities is known as the branchcut (see Figure 3).

Example 1.5 Let z1 = r1(cos �1+ i sin �1) and z2 = r2(cos �2+ i sin �2). Com-pute the following in polar form and comment on the modulus and arg of each.

(a). z1z2

(b). z1z2

(c). z1

(d). z�11

Solution

(a).

z1z2 = r1r2(cos �1 + i sin �1)(cos �2 + i sin �2)

= r1r2[(cos �1 cos �2 � sin �1 sin �2) + i(cos �1 sin �2 + sin �1 cos �2)]= r1r2[cos(�1 + �2) + i sin(�1 + �2)].

Therefore, jz1z2j = jz1j � jz2j and arg(z1z2) = arg(z1) + arg(z2).

(b).

Assume (z2 6= 0):z1z2

=r1(cos �1 + i sin �1)

r2(cos �2 + i sin �2)=r1r2

(cos �1 + i sin �1)

(cos �2 + i sin �2)

(cos �2 � i sin �2)(cos �2 � i sin �2)

=r1r2

[(cos �1 cos �2 + sin �1 sin �2) + i(cos �1 sin �2 � sin �1 cos �2)](cos2 �2 + sin

2 �2)

=r1r2[cos(�1 � �2) + i sin(�1 � �2)].

Therefore,

����z1z2���� =

jz1jjz2j

and arg�z1z2

�= arg(z1)� arg(z2).

7

(c).

z1 = r1(cos �1 + i sin �1) = r1(cos �1 � i sin �1) = r1(cos(��1) + i sin(��1)).Therefore, jz1j = jz1j and arg(z1) = � arg(z1).

(d).

Assume (z1 6= 0):

z�11 =1

r1(cos �1 + i sin �1)=1

r1

1

(cos �1 + i sin �1)

(cos �1 � i sin �1)(cos �1 � i sin �1)

=1

r1

(cos(��1) + i sin(��1))(cos2 �1 + sin

2 �1)=1

r1(cos(��1) + i sin(��1)).

Therefore, jz�11 j =1

jz1jand arg(z�11 ) = � arg(z1).

We summarize the results of the previous example in the Proposition thatfollows.

Proposition 5 Let z1; z2, and z be complex numbers. Then the following holdtrue.

jz1z2j = jz1j � jz2j (8)

arg(z1z2) = arg(z1) + arg(z2) (9)����z1z2���� =

jz1jjz2j

(z2 6= 0) (10)

arg

�z1z2

�= arg(z1)� arg(z2) (11)

jzj = jzj (12)

arg(z) = � arg(z) (13)

jz�1j =1

jzj (z 6= 0) (14)

arg(z�1) = � arg(z). (15)

Example 1.6 Compute the following.

(a).��� 2i(1+i)5(2�3i)3

���(b). arg[(2 + i)(1� 3i)]

(c). arg�2�p3i

1+i

�Solution

8

(a). ����2i(1 + i)5(2� 3i)3

���� = j2i(1 + i)5jj(2� 3i)3j =

j2ij � j(1 + i)j5j2� 3ij3 =

2(p2)5

(p13)3

(b).

arg[(2 + i)(1� 3i)] = arg(2 + i) + arg(1� 3i)

= tan�1�1

2

�+ tan�1(�3) + 2k�, where k = 0;�1;�2; � � �

(c).

arg

2�

p3i

1 + i

!= arg(2�

p3i)� arg(1 + i)

= tan�1

�p3

2

!� tan�1(1) + 2k�

= tan�1

�p3

2

!� �4+ 2k�, where k = 0;�1;�2; � � � .

1.4 Euler�s Equation

We now de�ne the most important function in mathematics; the exponentialfunction, for complex numbers.For every complex number z we de�ne the exponential function by the for-

mula,

exp(z) =1Xn=0

zn

n!(16)

The series (16) converges absolutely. Indeed,

� = limn�!1

����an+1an

���� = limn�!1

���� zn+1(n+ 1)!� n!zn

���� = limn�!1

jzj(n+ 1)

= 0 < 1.

Therefore by ratio test, series converges absolutely. The absolute conver-gence of (16) shows that

exp(u) exp(v) =1Xk=0

uk

k!

1Xm=0

vm

m!=

1Xn=0

(u+ v)n

n!= exp(u+ v). (17)

In order to see (17), we note the following rearrangement and add termsdiagonally as indicated.

9

� 1 v1!

v2

2!v3

3! � � � vn

n! � � �

1 1%yv1!

�!v2

2!

yv3

3! � � � vn

n! � � �u1!

u1!#

. u1!v1!% u

1!v2

2!

.u1!v3

3! � � � u1!vn

n! � � �

u2

2!u2

2!

%u2

2!v1!. u2

2!v2

2!u2

2!v3

3! � � � u2

2!vn

n! � � �u3

3!u3

3!#

.u3

3!v1!

u3

3!v2

2!u3

3!v3

3! � � � u3

3!vn

n! � � �

......

......

......

... � � �un

n!un

n!un

n!v1!

un

n!v2

2!un

n!v3

3! � � � un

n!vn

n! � � �...

......

......

......

. . .

Diagonal summation gives:

1Xk=0

uk

k!

1Xm=0

vm

m!

= 1 +� u1!+v

1!

�+

�v2

2!+u

1!

v

1!+u2

2!

�+

�u3

3!+u2

2!

v

1!+u

1!

v2

2!+v3

3!

�+ � � �

= 1 +(u+ v)

1!+(u+ v)2

2!+(u+ v)3

3!+ � � �

=1Xn=0

(u+ v)n

n!.

Armed with the important multiplicative property

exp(u) exp(v) = exp(u+ v),

we proceed with our de�nition.Start with a complex number z = x+ iy. Then

ez = ex+iy = ex � eiy. (18)

In order to de�ne ez, according to equation (18), we need to only describehow eiy should be de�ned. In order to accomplish this task, we consider the fol-lowing second order di¤erential equation with initial conditions on the functionand its derivative.

d2f(y)

dy2= �f(y), df

dy(0) = i, f(0) = 1. (19)

It can be easily checked that the function f(y) = cos y + i sin y satis�es theinitial value problem given in (19). Next consider the function eiy.Since

10

d2(eiy)

dy2= �eiy,

d(eiy)

dyjy=0 = iei(0) = i, and

ei(0) = 1,

we see that the function eiy also satis�es (19). By the uniqueness of solutionto (19), we conclude that

eiy = cos y + i sin y. (20)

Equation (20) is the celebrated Euler�s equation. From equation (18) wehave the following de�nition.

De�nition 6 If z = x+ iy, then ez is de�ned to be the complex number

ez = ex(cos y + i sin y). (21)

Note that, the polar representation of a complex number can be rewrittenas,

z = x+ iy = r(cos � + i sin �) = rei�:

Example 1.7 Use Euler�s formula to derive Taylor series expansions for sinxand cosx, where x is a real number.

Solution We begin by substituting z = iy in the exponential function (16).

eiy = 1 + iy +(iy)2

2!+(iy)3

3!+ � � �

=

�1� y

2

2!+y4

4!� � � �

�+ i

�y � y

3

3!+y5

5!� � � �

�= cos y + i sin y,

thus,

cos y = 1� y2

2!+y4

4!� � � �

sin y = y � y3

3!+y5

5!� � � � .

Example 1.8 Show the following identities.

11

(a). e2�ki = 1 for all k = 0;�1;�2; � � � Part (c) is afascinating for-mula: e comesfrom calculus, �from geometry,and i from alge-bra. They areall connectedby one magicalequation.

(b). e(�=2)i = i; e�(�=2)i = �i

(c). e�i = �1

Solution

(a). For any k = 0;�1;�2; � � � ; e2�ki = cos 2�k + i sin 2�k = 1.

(b). e(�=2)i = cos(�=2) + i sin(�=2) = i; e�(�=2)i = cos(��=2) + i sin(��=2) =�i.

(c). e�i = cos(�) + i sin(�) = �1.

Example 1.9 Show De Moivre�s formula given below which was �rst shown byAbraham De Moivre in 1707.

(cos � + i sin �)n = cosn� + i sinn�, for n = 1; 2; 3; � � � (22)

Solution

(cos � + i sin �)n =�ei��n= ein� = cos(n�) + i sin(n�).

Example 1.10 Write each of the following complex numbers in the form a+ib.

(a). eei

(b). ei+e�i

2 .

Solution

(a). eei

= e(cos 1+i sin 1) = ecos 1(cos(sin 1) + i sin(sin 1)) = ecos 1 cos(sin 1) +iecos 1 sin(sin 1)

(b). ei+e�i

2 = cos 1+i sin 1+cos 1�i sin 12 = cos 1

Example 1.11 Write each of the following complex numbers in the form rei�.

(a). (1 + 2i)5

(b). 23e1+i

Solution

(a). (1 + i)5 = (p2)5h1p2+ ip

2

i5= 4

p5�ei�=4

�5= 4

p5ei5�=4

(b). 23e1+i =

23e�(1+i) = 2

p2

3

h�1p2� ip

2

i= 2

p2

3 e�3�=4i.

Example 1.12 Derive the following trigonometric formula.

12

sin 4� = 4 sin � cos � � 8 cos � sin3 �.Solution

sin 4� = Im[(cos � + i sin �)4]

= Im�cos4 � + 4i cos3 � sin � + 6i2 cos2 � sin2 � + 4i3 cos � sin3 � + i4 sin4 �

�= Im

�cos4 � � 6 cos2 � sin2 � + sin4 � + i(4 cos3 � sin � � 4 cos � sin3 �)

�= 4 cos3 � sin � � 4 cos � sin3 �= 4 sin � cos �[1� sin2 �]� 4 cos � sin3 �= 4 sin � cos � � 8 cos � sin3 �

1.5 Roots of Unity

In this section we will discuss the roots of unity. That is, we want to �nd then-th roots of 1, namely, 1

1n .

We start with a complex number z = rei�. Note that,

zk = rkeik� = rk(cos k� + i sin k�) for any positive integer k.

If k is a negative integer, then k = �m, where m 2 Z+.

zk = z�m =1

zm=

1

rmeim�= r�me�im�

= r�m(cos(�m�) + i sin(�m�)) = rk(cos k� + i sin k�).

The next step of the extension raising z to any number of the form k = 1n is

somewhat tricky. If we want to de�ne z1=n then it must be the complex number& which satis�es the equation

&n = z. (23)

Let & = �(cos�+ i sin�) and z = r(cos � + i sin �). Then upon substitutionin equation (23) we obtain

�n(cosn�+ i sinn�) = r(cos � + i sin �).

From the above equation it is clear that

�n = r =) � = r1=n and

n� = � + 2m� for m = 0;�1;�2; � � �� =

arg z

n.

Combining these results we have the following Proposition.

13

Proposition 7 If n 2 Z+ then the nth root of a complex number z is given by

z1=n = r1=n�cos

arg z

n+ i sin

arg z

n

�, (24)

where the n di¤erent solutions are obtained when arg z runs through thevalues Arg z; Arg z + 2�; Arg z + 4�; � � � ; Arg z + 2�(n� 1).From the Proposition we can derive the following de�nition for the roots of

unity.

De�nition 8 The nth roots of unity, denoted by 11=n is de�ned by

11=n = ei2k�=n = cos2k�

n+ i sin

2k�

nwhere k = 0; 1; 2; � � � ; (n� 1). (25)

Example 1.13 Find the square, cube, fourth, and �fth roots of unity and rep-resent them geometrically.

Solutionn = 2: 11=2 = cos k� + i sin k� where k = 0; 1:When k = 0 =) z1 = 1:When k = 1 =) z2 = �1:n = 3: 11=3 = cos 2k�3 + i sin 2k�3 where k = 0; 1; 2:When k = 0 =) z1 = 1:When k = 1 =) z2 = e

2�i=3 = � 12 +

p32 i:

When k = 2 =) z3 = e4�i=3 = � 1

2 �p32 i:

n = 4: 11=4 = cos k�2 + i sink�2 where k = 0; 1; 2; 3:

When k = 0 =) z1 = 1:When k = 1 =) z2 = e

�i=2 = i:When k = 2 =) z3 = e

�i = �1:When k = 3 =) z4 = e

3�i=2 = �1:n = 5: 11=5 = cos 2k�5 + i sin 2k�5 where k = 0; 1; 2; 3; 4:When k = 0 =) z1 = 1:When k = 1 =) z2 = e

2�i=5:When k = 2 =) z3 = e

4�i=5:When k = 3 =) z4 = e

6�i=5:When k = 4 =) z4 = e

8�i=5:De�ne wn = e2�i=n. Then clearly, wnn = 1. We note that, with the exception

of the number 1, all the other n-th roots of unity can be found by using wn(known as a primitive root) and its powers. That is, the n-th roots of unity aregiven by,

1; wn; w2n; w

3n; � � � ; wn�1n : (26)

The following Proposition holds true for wn = e2�i=n.

14

Proposition 9 1 + wn + w2n + w3n + � � �+ wn�1n = 0:

Proof. Observe that (wn � 1)(1 +wn +w2n + � � �+wn�1n ) = wnn � 1 = 0. Sincewn 6= 1,1 + wn + w

2n + w

3n + � � �+ wn�1n = 0:

Fifth roots of unity : 11/5e8πi/5

e6πi/5

e4πi/5

e2πi/5

1

Fourth roots of unity : 11/4­i

­1

i

1

Cube roots of unity: 11/3

e4πi/3

e2πi/3

1

Square roots of unity : 11/2

­1 1

Figure 4: Roots of unity

We end this section with the de�nition of the n-th root of any complexnumber.

De�nition 10 Let z = rei�. Then z1=n = r1=nei(�+2k�)=n, where k is an integersuch that k = 0; 1; 2; � � � ; n� 2; n� 1.

Example 1.14 Show that (1 + i)65 = 232(1 + i).

15

Solution

(1 + i)65 = (p2e�i=4)65 = 265=2e65�i=4

= 232p2ei�=4 (since

65

4= 2� � 8 + �

4)

= 232p2

�1p2+ i

1p2

�= 232(1 + i):

Example 1.15 Construct the 5-th roots of z = i.

SolutionFirst note that i = ei�=2.

z1=5 = i1=5 =he�=2+2k�)i

i1=5= e

15 (�=2+2k�)i, k = 0; 1; 2; 3; 4.

z1 = e�i=10; (quadrant I)

z2 = e�i=2 = i;

z3 = e9�i=10; (quadrant II)

z4 = e13�i=10; (quadrant III)

z5 = e17�i=10; (quadrant IV):

1.6 Some Topological considerations

In this section we lay the ground work for future study by introducing termi-nology (mainly topological) via several de�nitions. From here on we will denotethe complex plane by C.

De�nition 11 The set of points z 2 C that satis�es jz � z0j < r is called anopen disk or neighborhood of z0 in the complex plane. Here r > 0 and z0 2 Cis �xed.

De�nition 12 A point z0 is said to lie in the interior of a set S, if we can �ndsome open disk about z0 that is completely contained inside of S. The interiorof the set S is denoted by Int(S).

De�nition 13 A point z0 is said to lie in the boundary of a set S, if everyopen disk about z0 contains points both in S and points outside of S. Theboundary of the set S is denoted by @S.

Example 1.16 Consider the open disk jz�3j < 2. Note that the point z1 = 4+iis the interior of S, while the point z2 = 3 + 2i is on @S. (Refer to Figure 5.)

16

1 2 3 4 5

­2

­1

0

1

2

3

x

y

4+i

3+2i

Figure 5: The open disk jz � 3j < 2

De�nition 14 A set S is closed if it contains all of its boundary points.

For example, the set of points z described by jz � (2 + i)j < 1 is open whilejz � (2 + i)j � 1 is closed.

De�nition 15 A set S is bounded if there exists a positive real number � suchthat jzj < � for all z 2 S.

De�nition 16 A set S that is both closed and bounded is called a compact set.

Next we de�ne the important topological notion of connectedness, which willlead us to the de�nition of a domain in C.

De�nition 17 An open set S is called connected if every pair of points z1; z2 2S, we can �nd a polygonal path that connects z1 to z2 and lying entirely insideS.

17

Note that not all open sets are connected. For example, if S and T are opendisjoint disks, then S [ T is not connected. Another example of an open setthat is not connected is S = fz j Im(z) 6= 0g.

De�nition 18 An open connected set in C is called a domain.

x

y

Figure 6: A domain in C:

Example 1.17 Show geometrically that the following open sets are connected,hence are domains in C.

(a). jz � ij > 1

(b). 1 < jz � 1j < 2

(c). �5 < Im(z) < 3

Solution(a). The domain jz � ij > 1.

18

­2 ­1 1 2

­3

­2

­1

1

2

3

x

y

(b). The domain 1 < jz � 1j < 2.

­2 ­1 1 2 3

­3

­2

­1

1

2

3

x

y

(c). The domain �5 < Im(z) < 3.

19

­5 ­4 ­3 ­2 ­1 1 2 3 4 5

­6

­4

­2

2

4

6

x

y

Our next result is a generalization of a familiar result from calculus of asingle variable.

Theorem 19 Let u(x; y) be a real valued function de�ned on a domain D � C.If

@u

@x=@u

@y= 0,

for all points (x; y) 2 D, then u(x; y) is constant on D.Proof. Join any two points z1; z2 2 D by any polygonal path and then replaceit with one consisting only of horizontal and vertical line segments. Since @u@x = 0for all (x; y) 2 D, u remains constant on all horizontal line segments. On theother hand, since @u

@y = 0 for all (x; y) 2 D, u remains constant on all verticalline segments as well. By the continuity of the polygonal path, it follows that uis constant on this path. Because our choice of z1; z2 was arbitrary, u is constanton all of D.

Example 1.18 Let u(x; y) be a real valued function that satis�es

@u

@x= 2xy and

@u

@y= x2

at all points of a domain D. Determine u(x; y).SolutionIntegrate @u

@x = 2xy with respect to x to obtain,

u(x; y) = x2y + f(y);

20

for some function f(y). Then

@u

@y= x2 + f 0(y) = x2 =) f 0(y) = 0 =) f(y) = constant = c:

Therefore, u(x; y) = x2y + c.

1.6.1 The Stereographic Projection

Can an interval on the real number line be mapped to the unit circle S1 : jzj = 1?In order to to answer this question we need to understand the wrapping function,which wraps the interval [0; 2�) of real number line on to the unit circle (Figure7).The mapping is given by,W : [0; 2�) �! S1 de�ned by W (t) = eit = cos t+ i sin t for all t 2 [0; 2�).

| z| =1

W(t) = eit ⇒

[0,2π)(0,0)

(0,1)

Figure 7: The wrapping map

If we carry the same line of inquiry to the next dimension, we can ask, is therea way to identify the points of a 3-D sphere with points on a plane (namely thecomplex plane C)? The stereographic projection provides an a¢ rmative answerto this question. Let us denote the unit sphere in 3-D by S2 : x21 + x

22 + x

23 = 1.

Then the familiar xy-plane (or the x1x2 plane in our case, or C) is the equatorialplane. We can de�ne a map by the following procedure:Let z = x+ iy 2 C be a point on the equatorial plane. The line connecting

z to the north pole N = (0; 0; 1) meets the unit sphere S2 a pointbz. Conversely, any point bz on S2 can be mapped to a point z by using thesame line that connects the north pole to bz.It should be clear that all the points on the plane outside the unit circle

S1 get mapped to points on the upper hemisphere, all the points on the plane

21

inside the unit circle S1 get mapped to points on the lower hemisphere, whilethe points on the unit circle (the equator) remain invariant.What happens to the north pole itself? It gets mapped to in�nity! However,

this 1 refers to all points on C such that jzj �! 1.

x3

x2x1

zy

Figure 8: The Riemann sphere

Criterion 20 Let z = x+ iy 2 C. Then the stereographic projection of z on tothe unit sphere S2 is a point bz = (x1; x2; x3) given by

x1 =2Re z

jzj2 + 1 ; x2 =2 Im z

jzj2 + 1 ; x3 =jzj2 � 1jzj2 + 1 . (27)

Criterion 21 Let bz = (x1; x2; x3) 2 S2. Then the projection of bz on to thecomplex plane C is a point z = x+ iy 2 C given by

x =x1

1� x3; y =

x21� x3

. (28)

To see Criterion 20, we note that the line passing through the north poleN = (0; 0; 1) and the point (x; y; 0) in parametric form is given by,

x1 = tx; x2 = ty; x3 = 1� t; where �1 < t <1. (29)

The point at which this line intersects the unit sphere can be found bysubstituting (29) into x21 + x

22 + x

23 = 1. Indeed

22

t2x2 + t2y2 + (1� t)2 = 1

(x2 + y2 + 1)t2 � 2t = 0

t[((x2 + y2 + 1)t� 1] = 0

t = 0 (corresponds to N), t =2

x2 + y2 + 1=

2

jzj2 + 1 :

Substitute t = 2jzj2+1 in (29) we obtain (27).

As for Criterion 21, begin with a point (x1; x2; x3) 2 S2. Then the point(x; y; 0) 2 C can be found by solving (29):

t = 1� x3 =) x =x1

1� x3; y =

x21� x3

, which yields (28).

Example 1.19 Show that the stereographic projections of the points z and � 1z

are antipodal points on the Riemann sphere (i.e. diametrically opposite points).

SolutionLet z = x+ iy. Then � 1

z = �zjzj2 = �

xjzj2 � i

yjzj2 and

��� 1z

�� = 1jzj

The point (x1; x2; x3) 2 S2 that results from the stereographic projection ofz is given by equation (27):

x1 =2x

jzj2 + 1 ; x2 =2y

jzj2 + 1 ; x3 =jzj2 � 1jzj2 + 1 .

Let (X1; X2; X3) 2 S2 be the stereographic projection of the point � 1z on

the Riemann sphere. Then from equation (27) we obtain,

X1 =�2x=jzj21=jzj2 + 1 =

�2xjzj2 + 1 = �x1

X2 =�2y=jzj21=jzj2 + 1 =

�2yjzj2 + 1 = �x2

X3 =1=jzj2 � 11=jzj2 + 1 =

1� jzj21 + jzj2 =

�(jzj2 � 1)(jzj2 + 1) = �x3:

Thus, the stereographic projections of the points z and � 1z are antipodal

points on the Riemann sphere.

Example 1.20 What are the projections on the Riemann sphere of the follow-ing sets in C. Describe them geometrically.

(a). The disk fz j jzj < 1=2g.

(b). The line y = x (extending to 1)

Solution

23

(a). Pick the point x = 12 ; y = 0; with jzj = 1

2 on the boundary of the diskfz j jzj < 1=2g. Equation (27) gives,

x1 =2(1=2)14 + 1

=4

5; x2 = 0; x3 =

14 � 114 + 1

= �35:

Therefore, all points in the interior of the disk get mapped to southern polarcap with x3 < � 3

5 .

(b). Pick any point x = x; y = x;with jzj =px2 + x2 =

p2jxj on the line

y = x. Equation (27) gives,

x1 =2x

2x2 + 1; x2 =

2x

2x2 + 1; x3 =

2x2 � 12x2 + 1

:

The points of y = x get mapped to points on the sphere with coordinatesx1 = x2; and �1 � x3 � 1. The last inequality can be obtained by simplylooking at the graph of the function f(x) = 2x2�1

2x2+1 . Therefore, on the Riemannsphere we obtain a great circle that intersects C at argument �4 and �1 � x3 � 1.

24

2 Analytic Functions

2.1 Complex Valued Functions

The calculus of complex valued functions di¤er from real valued functions in onemajor way. A complex valued function has two components - a real part andan imaginary part, each a two variable function. For instance, in de�ning thederivative, we have to bring to bear the two dimensional nature of the situation.In this Chapter we will formulate the derivative of a complex function.

De�nition 22 A complex valued function f(z) de�ned from a domain inthe z-plane in to a range in the w-plane is complex valued in the sense that it hasan real part u(x; y) and an imaginary part v(x; y). We write such a function as

f(z) = w = u(x; y) + iv(x; y).

w

f

z

w ­p lanez ­p lane

Figure 9: f(z) = w = u(x; y) + iv(x; y)

We note that one must take caution when handling complex valued functionsdue to the fact that we now have two real valued functions u; v; where each is afunction of two variables. The input z; however, is to be considered as a singleentity. We will deal with this matter shortly.

Example 2.1 For each of the following complex valued functions �nd the realand imaginary parts u; v.

(a). f(z) = e2z:

(b). f(z) = 2z2�3jz+1j :

25

(c). f(z) = 1z :

Solution

(a). f(z) = e2z = e2x+i2y = e2x(cos 2y + i sin 2y) = e2x cos 2y + ie2x sin 2y:

Therefore, u(x; y) = e2x cos 2y and v(x; y) = e2x sin 2y:

(b). f(z) = 2z2�3jz+1j =

2(x+iy)2�3jx+iy+1j = 2(x2�y2)�3+i4xyp

(x+1)2+y2= 2(x2�y2)�3p

(x+1)2+y2+i 4xyp

(x+1)2+y2:

Therefore, u(x; y) = 2(x2�y2)�3p(x+1)2+y2

and v(x; y) = 4xyp(x+1)2+y2

:

(c). This is an important function, known as the inversion mapping.

f(z) = 1z =

zjzj2 =

x�iyx2+y2 =

xx2+y2 � i

yx2+y2 :

Therefore, u(x; y) = xx2+y2 and v(x; y) = �

yx2+y2 :

Observe that,Arg(f(z)) = tan�1

�vu

�= tan�1

��yx

�= � tan�1

�yx

�= �Arg(z) and

jf(z)j =pu2 + v2 =

qx2

(x2+y2)2+ y2

(x2+y2)2= 1p

x2+y2= 1

jzj :

A few interesting consequences follow from these results.Suppose z = ei� is any point on the unit circle. Then f(z) = e�i� is also on

the unit circle with argument ��:If z = rei� is any point outside the unit circle. Then f(z) = 1

r e�i� is inside

the unit circle with argument ��:Similarly, if z = rei� is any point inside the unit circle. Then f(z) = 1

r e�i�

is outside the unit circle with argument ��:

Example 2.2 Consider the inversion mapping. Show the following.

(a). Any point in the z-plane satisfying jz � 1j = 1 is mapped to the verticalline x = 1

2 in the w-plane.

(b). Any point in the z-plane satisfying��z � (p2 + ip2)�� = 1 is mapped to the

circle���z � � 1

2p2� i 1

2p2

���� = 16 in the w-plane.

Solution

(a). First observe that jz � 1j = 1 can be written as,

j(x� 1) + iyj = 1

(x� 1)2 + y2 = 1

x2 + y2 = 2x

r2 = 2r cos �

r = 2 cos � (since r 6= 0).

26

f(z) =1

rei�=

1

2 cos �(cos � + i sin �)

=(cos � � i sin �)

2 cos �(cos � + i sin �)(cos � � i sin �)

=(cos � � i sin �)

2 cos �=1

2� i

2tan �; for � �=2 � � � �=2:

As � ranges from ��=2 to �=2; f(z) = 12 �

i2 tan � represents the vertical

line x = 12 :

(b). First observe that��z � (p2 + ip2)�� = 1 can be written as,

���(x�p2) + i(y �p2)��� = 1

(x�p2)2 + (y �

p2)2 = 1

x2 + y2 � 2p2(x+ y) + 3 = 0

r2 � 2p2r(cos � + sin �) + 3 = 0

r =2p2(cos � + sin �)�

p8(cos � + sin �)2 � 4(3)2

r =p2(cos � + sin �)�

p2(cos � + sin �)2 � 3

r =p2(cos � + sin �)�

p4 sin � cos � � 1

r =p2(cos � + sin �)�

p2 sin(2�)� 1:

By setting 2 sin(2�)� 1 = 0 and solving for � we can obtain the two valuesof the Arg that give the two tangent lines to the circle

��z � (p2 + ip2)�� = 1from the origin. Indeed,2 sin(2�)� 1 = 0 =) � = �

12 ;5�12 :

Let z1; z2 be the two points of tangency. Then

z1 =p3ei�=12; z2 =

p3ei5�=12 and (30)

f(z1); f(z2) will be the points of tangency of the image:

f(z1) =1p3e�i�=12; f(z2) =

1p3e�i5�=12 (31)

INSERT PICTURE HERE.

2.2 Limits and Continuity

In this section we provide the de�nitions of limits and continuity for complexvalued functions.Let fzng1n=1 be a sequence. Intuitively, z0 be a limit point if the disk

jz � z0j < � contain and in�nite number of points of the sequence fzng1n=1.The familiar formal de�nition follows.

27

De�nition 23 A sequence fzng1n=1 converges to a limit z0 2 C, written limn�!1 zn =z0 if for any � > 0 there exists an integer N such that jzn � z0j < � for all n > N:

De�nition 24 Let f be a function de�ned in some neighborhood of z0: We saythat the limit of f(z) as z �! z0 (in ALL directions) is a complex number w0such that

limz�!z0

f(z) = w0;

if for any � > 0 there exists a positive number � > 0 such that

jf(z)� f(w0)j < � whenever 0 < jz � z0j < �:INSERT PICTURE HERE.We now explore the de�nition of continuity of a complex variable function.

De�nition 25 Let f(z) be a function de�ned in a neighborhood of z0. Then fis said to be continuous at z0 if

limz�!z0

f(z) = f(z0):

We can extend the idea to continuity on a set S, by requiring that f becontinuous at every point of S.

Theorem 26 Suppose limz�!z0 f(z) = w1 and limz�!z0 g(z) = w2: Then

limz�!z0

[f(z)� g(z)] = w1 + w2

limz�!z0

f(z) � g(z) = w1w2

limz�!z0

f(z)

g(z)=

w1w2

(w2 6= 0):

From the above theorem it follows that if f; g are continuous functions thenso are f � g; f � g; fg (g 6= 0). In particular, polynomial functions and rationalfunctions (where de�ned) are continuous.

Example 2.3 Prove that zn �! z0 if and only if zn �! z0 as n �!1:

SolutionWe observe that jzn � z0j = jzn � z0j = jzn � z0j :Let � > 0.Suppose zn �! z0. Then there exists a positive integer N such that

jzn � z0j = jzn � z0j < � for all n > N: Thus, zn �! z0.Conversely, suppose zn �! z0. Then there exists a positive integer M such

that jzn � z0j = jzn � z0j < � for all n > M: Thus, zn �! z0.

28

Example 2.4 Determine if each of the following sequences converge. If so, �ndthe limit.

(a). zn = Arg��2 + i

n

�(b). zn = 2n

n! + in2n

(c). zn =npb+ i sin 1

n (b > 0)

(d). zn =�1�i4

�nSolution

(a). The sequence zn = Arg��2 + i

n

�converges to �: Indeed, limn�!1Arg

��2 + i

n

�=

Arg(�2) = �:

(b). The sequence zn = 2n

n! + in2n converges to 0: Indeed, using single variable

calculus limits we can show that

limn�!1�2n

n! + in2n

�= limn�!1

�2n

n!

�+ i limn�!1

�n2n

�= 0 + i0 = 0:

(c). The sequence zn =npb+ i sin 1

n converges to 1: Indeed,

limn�!1

�npb+ i sin 1

n

�= limn�!1

�npb�+i limn�!1

�sin 1

n

�= b0+i sin 0 =

1:

(d). We will show that the sequence zn =�1�i4

�nconverges to limit 0:

Let � > 0. Then�����1� i4�n���� = ���� j1� ijn4n

���� = p2n

4n=

1

23n=2:

Choose N such that 123N=2

< � =) 23N=2 > 1� =)

3N2 > ln(1=�)

ln(2) =) N >23ln(1=�)ln(2) .

Now for all n > N;��� 1�i

4

�n � 0�� = 123n=2

< �. By de�nition, limn�!1�1�i4

�n=

0.

2.3 What is analyticity?

In this section we explore which functions would be considered admissible ascomplex functions. One must understand why we even consider this as aninteresting question. The reason lies in the fact that the input z, while being asingle complex number, has two real components x and y. So when we applyf to the input z, are we allowed to treat x and y as two separate entities? Weallow as admissible, any function that treats z as a whole. For instance,

29

f(z) = z2 + 2z � 3;

f(z) =z2 + i

z � 3 ;

f(z) = x2 � y2 + i2xy (which = z2)

are all admissible functions. However,

f(z) = Re z;

f(z) = z = x� iy;f(z) = jzj =

px2 + y2;

are not admissible because of the presence of Re; conjugate, and modulusrespectively. These are all operations that apply only to parts of the input z.In order to formalize the allowable complex functions we use di¤erentiability

as the deciding criterion.

De�nition 27 Let f be a complex valued function de�ned in a neighborhood ofz0: Then the derivative of f(z) at z0 is de�ned by

df

dz(z0) = f

0(z0) = lim�z�!0

f(z0 +�z)� f(z0)�z

; (32)

provided the limit exists. Here �z �! 0 in all possible directions.INSERT ALL PROOFS IN THESE PAGES (AND IN THE EXAMPLES

BELOW)

Example 2.5 Show that f(z) = z is not di¤erentiable at every point of C.

Example 2.6 Show that f(z) = jzj2 is not di¤erentiable at every point of Crf0g.

Example 2.7 Show that f(z) = jzj is not di¤erentiable at every point of C.

Example 2.8 Show that for any positive integer n;

d

dz(zn) = nzn�1:

The familiar rules of derivatives hold true for complex valued functions.

Theorem 28 Let f; g be di¤erentiable at a point z. Then the following aretrue.

1. (f � g)0(z) = f 0(z)� g0(z)

30

2. (cf)0(z) = cf 0(z) for c 2 C

3. (f � g)0(z) = f(z) � g0(z) + g(z) � f 0(z)

4.�fg

�0(z) = g(z)�f 0(z)�f(z)�g0(z)

g2(z) (whenever g(z) 6= 0)

5. d(f(g(z))dz = f 0(g(z)) � g0(z)

De�nition 29 A complex valued function f(z) is said to be analytic on anopen set G if it had a derivative at every point of G.

De�nition 30 If f(z) is analytic on all of C we say it is entire.

Every polynomial function is entire. The exponential function de�ned insection (1.4) is entire.

31

3 Elementary Functions

In this section we will introduce the elementary complex functions. Our discus-sion of polynomials will make a pathway for us to tackle complex integrationin the next section. The exponential function and the logarithmic functions areintroduced in detail.

3.1 Polynomials and Rational Functions

We begin with the de�nitions of polynomial and rational functions.

De�nition 31 A polynomial function in z has the form

pn(z) = a0 + a1z + a2z2 + � � �+ anzn:

Its degree is de�ned to be n when an 6= 0: Solutions to pn(z) = 0 are calledzeros.

De�nition 32 A rational function in z has the form

Rm;n(z) =pm(z)

qn(z)=a0 + a1z + a2z

2 + � � �+ amzmb0 + b1z + b2z2 + � � �+ bnzn

; am 6= 0; bn 6= 0:

We will assume that m < n; otherwise we perform long division to get there.Rational functions have zeros (these are solutions of pm(z) = 0) and poles (theseare solutions of qn(z) = 0).

Example 3.1 Factor the following polynomials and �nd their zeros.

(a). p3(z) = z3 � 3z2 + 6z � 4

(b). p2(z) = z2 � (3� 2i)z + 1� 3i

(c). p(z) = 1 + z + z2 + z3 + z4

Solutions

(a). p3(z) = z3 � 3z2 + 6z � 4:

It is easy to see that z = 1 is a solution.p3(z) = z

3� 3z2+6z� 4 = (z� 1)(z2� 2z+4) = (z� 1)(z� (1+ ip3))(z�

(1� ip3)):

Zeros are: z = 1; z = 1� ip3:

(b). Use the quadratic formula to �nd the zeros.

z =(3�2i)�

p(3�2i)2�4(1�3i)2 = (3�2i)�

p9�12i�4�4+12i2 = 3�2i�1

2 = 2� i; 1� i:p2(z) = z

2 � (3� 2i)z + 1� 3i = (z � (2� i))(z � (1� i)):

32

(c). p(z) = 1 + z + z2 + z3 + z4 = z5�1z�1 :

In order to factor p(z);�nd all the �fth roots of unity (these are the zeros ofthe numerator z5 � 1) and avoid the trivial solution z = 1:The zeros of z5�1 = 0 are: z0 = 1; ! = ei2�=5; !2 = ei4�=5; !3 = ei6�=5; !4 =

ei8�=5:) p(z) = 1 + z + z2 + z3 + z4 = (z � !)(z � !2)(z � !3)(z � !4):How do we know that a given polynomial in z; with complex coe¢ cients

actually has zeros? This question was answered by Karl Gauss in 1799, in aresult that has now come to be known as the Fundamental Theorem of Algebra.We state below the theorem without proof.

Theorem 33 Every non-constant polynomial with complex coe¢ cients has atleast one zero in C:

Corollary 34 Every polynomial p(z) with real coe¢ cients can be expressed asthe product of linear and quadratic factors, each having real coe¢ cients.

Proof. We start by observing that the complex zeros occur in conjugate pairs.Suppose z0; z0; are a conjugate pair of zeros of p(z) and that the other zerosz1; � � � ; zn�2 are all real. Then

(z � z0)(z � z0)= z2 � (z0 + z0) + z0z0= z2 � 2Re(z0) + jz0j2

p(z) = (z2 � 2Re(z0) + jz0j2)(z � z1) � � � (z � zn�2): (a)

= (z � z0)(z � z0)(z � z1) � � � (z � zn�2): (b)

Expression (a) above has only real coe¢ cients and has both linear factorsand irreducible quadratic factors. On the other hand, expression (b) has alllinear factors with real and complex coe¢ cients.

Example 3.2 Find the zeros and poles of

R(z) =(3z + 3i)(z2 � 4)(z � 2)(z2 + 1)2 :

Solution

R(z) =(3z + 3i)(z2 � 4)(z � 2)(z2 + 1)2

=3(z + 2)

(z � i)2(z + i) :

Zero is: z = �2:

33

Poles are: z = i (mul = 2); z = �i:Recall, from Calculus, the de�nition of the Taylor series of a real valued func-

tion. This idea extends to complex variables as seen in the following de�nitionfor complex polynomials.

De�nition 35 The complex valued function pn(z) has a Taylor series expan-sion centered at z0 given by,

pn(z) =nXk=0

p(k)n (z0)

k!(z � z0)k

=pn(z0)

0!+p(1)n (z0)

1!(z � z0)1 +

p(2)n (z0)

2!(z � z0)2 + � � �+

p(n)n (z0)

n!(z � z0)n:

If z0 = 0 we call it a Maclaurin series.

Example 3.3 Find the Taylor series expansion of p5(z) = z5+3z+4 centeredat z0 = 2:

Solution

p5(z) = z5 + 3z + 4 =) p5(2)

0!=42

1= 42

p(1)5 (z) = 5z4 + 3 =) p

(1)5 (2)

1!=83

1= 83

p(2)5 (z) = 20z3 =) p

(2)5 (2)

2!=160

2= 80

p(3)5 (z) = 60z2 =) p

(3)5 (2)

3!=240

6= 40

p(4)5 (z) = 120z =) p

(4)5 (2)

4!=240

24= 10

p(5)5 (z) = 120 =) p

(1)5 (2)

5!=120

120= 1

) p5(z) = 42 + 83(z � 2) + 80(z � 2)2 + 40(z � 2)3 + 10(z � 2)4 + (z � 2)5:

Before we look at the next example, let us develop a notation to ease theunderstanding of how to identify constants in the partial fraction expansion ofrational functions.

Notation 36 A constants A(j)i will be used to refer to the numbers that appearin the numerators of partial fractions. The subscript i = 0; 1; 2; � � � ; d�1; whered is the multiplicity of the pole zj : The superscript (j) refers to the index of thepole under consideration.

34

Example 3.4 Decompose the following function into its partial fractions.

R(z) =4z + 4

z(z � 1)(z � 2)2 :

SolutionThe poles are: z1 = 0; z2 = 1; z3 = 2 (mul = 2):

R(z) =4z + 4

z(z � 1)(z � 2)2

=A(1)0

z+

A(2)0

(z � 1) +A(3)0

(z � 2)2 +A(3)1

(z � 2) :

When the degree of a linear factor in the denominator is equal to its multi-plicity (i.e. any constant that has 0 subscript), we use the cover-up rule to �ndthe constants. For instance, we can �nd A(1)0 ; A

(2)0 ; and A(3)0 using the cover up

rule. suppose the pole is at z = zj :The idea is to cover-up the factor in thedenominator of R(z) that gives rise to the pole zj . Then substitute z = zj intothe resulting expression. The value obtained is A(j)0 :More formally,

A(1)0 = lim

z�!0zR(z) = lim

z�!0

4z + 4

(z � 1)(z � 2)2 =4

�4 = �1:

A(2)0 = lim

z�!1(z � 1)R(z) = lim

z�!1

4z + 4

z(z � 2)2 =8

1= 8:

A(3)0 = lim

z�!2(z � 2)2R(z) = lim

z�!1

4z + 4

z(z � 1) =8 + 4

(2)(1)= 6:

Finding the constant A(3)1 cannot be done using the cover-up rule since anextra factor of (z � 2) will remain in the denominator. However, if we multiplyR(z) by (z�2)2 and then di¤erentiate with respect to z once and then substitutez = zj in the resulting equation, we can �nd A

(3)1 : Indeed,

35

(z � 2)2R(z) =4z + 4

z(z � 1)

=�1z(z � 2)2 + 8

(z � 1)(z � 2)2 + 6 + (z � 2)A(3)1

d

dz[(z � 2)2R(z)] =

d

dz

�4(z + 1)

z(z � 1)

�=

d

dz

��1z(z � 2)2 + 8

(z � 1)(z � 2)2 + 6 + (z � 2)A(3)1

�4[z(z � 1)� (z + 1)(2z � 1)]

z2(z � 1)2 = (�)(z � 2) +A(3)1 ; where (�) is an expression in z.

limz�!2

�4[z(z � 1)� (z + 1)(2z � 1)]

z2(z � 1)2

�= lim

z�!2

h(�)(z � 2) +A(3)1

i4[(2)(1)� (3)(3)]

(4)= A

(3)1

A(3)1 = �7:

Therefore, the partial fractions are:

R(z) =4z + 4

z(z � 1)(z � 2)2

=�1z+

8

(z � 1) +6

(z � 2)2 +�7

(z � 2) :

De�nition 37 Let R = PQ be a rational function with degP < degQ: If � is a

pole of R (i.e a zero of Q) then the coe¢ cient of 1(z��) in the partial fraction

expansion of R is called the residue of � and is denoted by Res(�).

In the above example

Res(0) = �1;Res(1) = 8; and Res(2) = 6:

The complex exponential function is the most important function in all ofmathematics. It can be used to de�ne complex trigonometric and hyperbolicfunctions. Recall that for a complex number z = x+iy; the complex exponentialfunction is de�ned by,

ez = ex(cos y + i sin y). (33)

The complex exponential function is entire with derivative equal to ez: Ob-serve that

jezj = ex; arg(ez) = y + 2k� (k = 0;�1;�2; � � � ):

36

Since arg(ez) is multiple valued (depends on the value of k), unlike the realexponential function, the complex exponential function is not one to one. Thefollowing theorem elaborates this fact.

(i) The equation ez = 1 is true if and only if z = 2k�i (k = 0;�1;�2; � � � ):

(ii) The equation ez1 = ez2 is true if and only if z1 = z2 + 2k�i (k =0;�1;�2; � � � ):

Proof. (i). Suppose ez = 1. Then jezj = ex = 1 =) x = 0:Therefore, ez = e0+iy = cos y + i sin y = 1 =) cos y = 1; sin y = 0 =) z =

iy = 2k�i (k = 0;�1;�2; � � � ):Conversely, suppose that z = 2k�i (k = 0;�1;�2; � � � ):Then ez = e2k�i = cos(2k�) + i sin(2k�) = 1:(ii). Proof of this part follows from the fact that ez1 = ez2 () ez1�z2 = 1.

Now use part (i).By part (ii) of the above Theorem, ez+2k�i = ez for k = 0;�1;�2; � � � . This

means that ez is periodic. This allows us to divide the z-plane into in�nitehorizontal strips Sn, on each of which ez behaves identically the same. Indeed,for n = 0;�1;�2; � � �

Sn = fx+ iy j �1 < x <1; (2n� 1)� < y � (2n+ 1)�g:

x

y

πι

−πι

3πι

−3πι

z=x+iy

It is interesting to note that ez can be equal to every complex number exceptzero, as the following Proposition indicates.

Proposition 38 If w is a non zeros complex number, then there exists a com-plex number z such that ez = w:

37

Proof. Let w = rei�: Then

eln r+i� = eln r � ei� = rei� = w:

Therefore, the choice for z is z = ln r + i�:Next we move to the de�nitions of the de�nitions of the complex trigono-

metric functions. The derivative rules are the same as in real variables.

De�nition 39 Let z 2 C: We de�ne

sin z =eiz � e�iz

2i(34)

cos z =eiz + e�iz

2(35)

tan z =sin z

cos z

sec z =1

cos z

csc z =1

sin z

cot z =cos z

sin z:

The corresponding derivatives are given by,

d

dzsin z = cos z

d

dzcos z = sin z

d

dztan z = sec2 z

d

dzsec z = sec z tan z

d

dzcsc z = � csc z cot z

d

dzcot z = � csc2 z

The complex hyperbolic functions are de�ned as follows.

De�nition 40 Let z 2 C: We de�ne

38

sinh z =ez � e�z

2

cosh z =ez + e�z

2

tanh z =sinh z

cosh z

sech z =1

cosh z

csch z =1

sinh z

coth z =cosh z

sinh z:

Example 3.5 Write the following complex numbers in the form a+ ib:

(a). exp(2 + i�=3)

(b). cosh(2 + �i)

Solution

(a). exp(2 + i�=3) = e2+i�=3 = e2ei�=3 = e2(cos�=3 + i sin�=3) = e2

2 + ip3e2

2

(b).

cosh(2 + �i) =e2+�i + e�(2+�i)

2

=�e2 � e�2

2= � cosh 2:

Example 3.6 Explain why the function f(z) = cos(z2) + ez + iz is entire.

SolutionEach of the functions cos(z2); ez; and z are entire. Hence a linear combina-

tion of these will also be entire.

Example 3.7 Is Re�sin zez

�harmonic?

SolutionConsider the function f(z) = sin z

ez . Since both sin z and e�z are entire, f(z)

is entire. Therefore, its real part Re�sin zez

�and imaginary part Im

�sin zez

�are

harmonic.

Example 3.8 Find all numbers z (if any) that satisfy

(a). e2z = 1

39

(b). cos z = i sin z:

Solution

(a). e2z = 1 = e2k�i =) z = k�i (k = 0;�1;�2; � � � )

(b).

cos z = i sin zeiz + e�iz

2= i

eiz � e�iz2i

eiz + e�iz = eiz � e�iz

e�iz = 0 =) no such numbers z

40

4 Complex Integration

Our main goal in this section is to prove Cauchy�s Integral Theorem. Roughlythe theorem states that:

If f is analytic on and inside a loop C thenIC

f(z)dz = 0.

4.1 Contours

What is a smooth curve? Intuitively, a smooth curve is a continuous curvethat can be drawn with uniform speed, without self intersections, and withoutmaking any abrupt interruptions in the sketch (i.e. speed 6= 0). Such curves arein�nitely di¤erentiable. Examples of such curves are given in Figure 11.

De�nition 41 A set of points 2 C is a smooth curve if it is the range of acontinuous function z(t); a � t � b satisfying

(i) z(t) has a continuous derivative on [a; b] (with one sided derivatives atthe end points)(ii) z0(t) 6= 0 for all t 2 [a; b](iii) z(t) is one to one on [a; b]A smooth curve is closed if we replace the interval [a; b] by [a; b) and require

z(b) = z(a) and z0(b) = z0(a):

41

De�nition 42 A set of points in C is a smooth curve if it is the image ofa continuous function z(t); a � t � b satisfying

(i) z(t) has a continuous derivative on [a; b] (one sided derivative at endpoints)(ii) z0(t) 6= 0 for all t 2 [a; b](iii) z(t) is one to one on [a; b]:A smooth curve is closed if we can replace the interval [a; b] by [a; b).

4.1.1 Some Standard Parametrizations

Following are some parametrizations that one often encounters. Knowing thesehelps facilitate the evaluation of contour integrals.

1. Horizontal line from z = a to z = b.

z(t) = t ; a � t � b:

2. Vertical line from z = a+ ib1 to z = a+ ib2.

z(t) = a+ it ; b1 � t � b2:

3. The line segment joining z1 and z2:

z(t) = z1 + t(z2 � z1) ; 0 � t � 1:

4. The circle jz � z0j = r:

z(t) = z0 + reit ; 0 � t � 2�:

5. The parabola y = x2 for a � x � b:

z(t) = t+ it2 ; a � t � b:

De�nition 43 A contour � is either a single point or a �nite sequence ofdirected smooth curves 1; 2; � � � ; n such that the terminal point of k equalsthe initial point of k+1 for k = 1; 2; � � � ; n� 1. We write

� = + 2 + � � �+ n:

We say that a closed curve is positively oriented, if when you traverse thecurve counterclockwise the interior is on the left hand side. Refer to Figure 12for examples of counters. Pay close attention to the directions indicated in each�gure.

42

De�nition 44 Let in C is a smooth curve with parametrization z(t) = x(t)+iy(t); a � t � b then the length of is

l( ) =

Z b

a

s�dx

dt

�2+

�dy

dt

�2dt =

Z b

a

����dzdt���� dt:

Example 4.1 The circle x2 + y2 = r2 is a smooth curve since z(t) = r cos t+ir sin t ; 0 � t � 2� is an admissible parametrization.

Example 4.2 (page 160 HW #8) Parametrize the contour � and �� and �ndits length.

SolutionThe contour � consists of of two smooth curves - the straight line piece 1

and the semi-circle 2:�

� 1 : z1(t) = (�2 + 2i) + t(�1 � (�2 + 2i)) = �2 + 2i + t(1 � 2i) ;0 � t � 1

� 2 : z2(t) = e�i�t ; 1 � t � 2

��

� � 2 : z2(t) = ei�t ; 0 � t � 1

� � 1 : z1(t) = �1 + (t� 1)(�1 + 2i) ; 1 � t � 2

To �nd the length of �, we add up the lengths of 1 and 2: Although it iseasy to compute the length of � using geometry, we will do it using the formula. 1 : z(t) = (t� 2) + i(2� 2t) ; 0 � t � 1 ; dzdt = 1� 2i =)

��dzdt

�� = p5:l( 1) =

R 10

p5dt =

p5:

2 : z(t) = e�i�t ; 1 � t � 2 ; dzdt = �i�e�i�t =)

��dzdt

�� = �:l( 2) =

R 21�dt = �:

) l(�) = l( 1) + l( 2) =p5 + �:

Example 4.3 Read example 2 on page 156.

4.2 Contour Integrals

The de�nition of a de�nite integral of a complex-valued function follows theparallel development of that of a real-valued function. Such integrals are knownas contour integrals, since the curve that we partition is a contour.We begin the discussion with an example that will serve as an important

reference in analyzing various contour integrals.

43

Example 4.4 Compute the integralICr

(z � z0)ndz;

where n is an integer and Cr : jz � z0j = r is the circle centered at z0 withradius r, traversed once clockwise.

SolutionWe parametrize Cr as z�z0 = reit. Then (z�z0)n = rneint and dz = rieitdt.I

Cr

(z � z0)ndz =

Z 2�

0

rneint � rieitdt

= irn+1Z 2�

0

ei(n+1)tdt

=

(i(2�) if n = �1

irn+1hei(n+1)t

i(n+1)

i2�0= irn+1

h1

i(n+1) �1

i(n+1)

iif n 6= �1

=

�2�i if n = �10 if n 6= �1

Let us now tackle the de�nition of contour integrals. Consider a directedsmooth curve with starting point � and ending point �.

Partition the curve into n parts, each of length �zk; k = 1; 2; � � � ; n. Letbzk be a point on lying between the points zk�1 and zk. Iflimn!1

nXk=1

f( bzk)�zkexists on every partition of , then we call it the integral of f along and

we write Z

f(z)dz = limn!1

nXk=1

f( bzk)�zk:Some properties are immediate:

44

�R [f(z)� g(z)] dz =

R f(z)dz �

R g(z)dz

�R [cf(z)] dz = c

R f(z)dz

�R� f(z) dz = �

R f(z)dz:

Suppose we can write a complex function f : R �! C as f(t) = u(t) + iv(t)where t 2 [a; b] then Z b

a

f(t)dt =

Z b

a

u(t)dt+ i

Z b

a

v(t)dt:

If f(t) possesses an antiderivative F (t) [ i.e. F 0(t) = f(t) ],Z b

a

f(t)dt = [F (t)]ba = F (b)� F (a):

This is the Fundamental Theorem of Calculus.The next theorem describes how to integrate a complex function on a smooth

curve when the function can be parametrized.

Theorem 45 Let f be a continuous function de�ned on a smooth curve . Ifz = z(t); t 2 [a; b] is a parametrization of thenZ

f(z)dz =

Z b

a

f(z(t)) � z0(t)dt:

We will omit the proof of the theorem but encourage the reader to observethat dz = z0(t)dt. If the parametrization is changed the value of

R f(z)dz

remains unchanged.The de�nition of a contour integral is given below.

De�nition 46 Suppose � = + 2+ � � �+ n, where i are smooth curves. Letf be continuous on �. Then the contour integral of f(z) along � is de�nedby, Z

�

f(z)dz =

Z 1

f(z)dz +

Z 2

f(z)dz + � � �+Z n

f(z)dz:

Example 4.5 Evaluate the following integrals.

(a). Z 0

�2(2 + i) cos(it)dt:

(b). Z 2

0

t

(t2 + i)2dt:

45

Solution

(a).

Z 0

�2(2 + i) cos(it)dt

= (1 + i)sin(it)

i

����0�2= � (1 + i)

isin(2i)

=(1 + i)

ii sinh(2) = (i+ i) sinh 2:

(b). Z 2

0

t

(t2 + i)2dt ; (u = t2 + i; du = 2tdt; limits: u = i to u = 4 + i)

=1

2

Z 4+i

u=i

u�2du

=

�� 1

2u

�4+ii

=�12

�1

4 + i� 1i

�=

i2

2� �4i(4 + i)

=�2i4 + i

=�217� 8i

17:

Example 4.6 Let C be the contour which is the perimeter of the square withvertices z = 0; z = 1; z = 1 + i; and z = i, traversed counterclockwise. EvaluateRCezdz.

SolutionWe will o¤er two methods of solution. The �rst method uses parametriza-

tion.The contour C consists of of four straight lines.C

� 1 : z(t) = t ; 0 � t � 1

� 2 : z(t) = 1 + i(t� 1) ; 1 � t � 2

� 3 : z(t) = (3� t) + i ; 2 � t � 3

� 4 : z(t) = i(4� t) ; 3 � t � 4

46

ZC

ezdz

=

Z 1

ezdz +

Z 2

ezdz +

Z 3

ezdz +

Z 4

ezdz

=

Z 1

0

etdt+

Z 2

1

e1+i(t�1)(i)dt+

Z 3

2

e(3�t)+i(�1)dt+Z 4

3

ei(4�t)(�i)dt

= (e� 1) + (e1+i � e) + (ei � e1+i) + (1� ei)= 0:

The second solution involves computing the integrals using antiderivativesand FTC. Z

C

ezdz

=

Z 1

ezdz +

Z 2

ezdz +

Z 3

ezdz +

Z 4

ezdz

=

Z 1

0

etdt+

Z 1+i

1

etdt+

Z i

1+i

etdt+

Z 0

i

etdt

= (e� 1) + (e1+i � e) + (ei � e1+i) + (1� ei)= 0:

Example 4.7 Evaluate the integralR�(3z�2)dz, where � is the circular contour

arc z = eit;��=2 � t � 0 joining z = �i to z = 0.

SolutionUsing the parametrization z = eit;��=2 � t � 0 together with z0(t) = ieitdt

we obtain,Z�

(3z � 2)dz =

Z 0

��=2(2eit + 1) � ieitdt

= i

�2eit

2i+eit

i

�0��=2

= i

�1

i+1

i���1i� 1��

= i

�3

i+ 1

�= 3 + i:

Example 4.8 Evaluate the following integral where C : jz � ij = 4 traversedcounterclockwise. I

C

�10

(z � 2i)5 +3

(z � 2i) + 7� (z � 2i)3

�dz:

47

SolutionUsing the results of the example at the beginning of this section,I

C

�10

(z � 2i)5 +3

(z � 2i) + 7� (z � 2i)3

�dz

=

IC

10

(z � 2i)5 dz +IC

3

(z � 2i)dz +IC

7dz �IC

(z � 2i)3dz

= 0 + 3(2�i) + 0 + 0 = 6�i:

4.3 Independence of Path

One of the most fundamental results in complex analysis states that the integralof a function is independent of the particular path joining the initial and terminalpoints. A discussion of this result is the goal of this section.

Theorem 47 Suppose that a function f(z) is continuous in a domain D andhas an antiderivative F (z) throughout D [ i.e F 0(z) = f(z) ]. The for anycontour � lying in D, with initial point z0 and terminal point z1 we have

Z�

f(z)dz = F (z1)� F (z0):

Corollary 48 Suppose that a function f(z) is continuous in a domain D andhas an antiderivative throughout D; then

R�f(z)dz = 0 for all closed loops �

lying in D.

Theorem 49 Let f be continuous in a domain D. Then the following areequivalent:

(i). f has an antiderivative in D:

(ii).R�f(z)dz = 0 for all closed loops � lying in D.

(iii). If �1 and �2 are two contours in D that have the same initial and terminalpoints, then

Z�1

f(z)dz =

Z�2

f(z)dz:

Proof. (i) =) (ii) :Suppose f has an antiderivative F in D. Let z0 = z1 be the initial and

terminal point on any closed loop � lying in D.Z�

f(z)dz = F (z1)� F (z0) = 0, which is (ii).

48

(ii) =) (iii) :Suppose (ii) is true. Let �1 and �2 are two contours in D that have the

same initial (z0) and terminal point (z1). Then �1 � �2 = � is a closed loop.We have,

0 =

Z�1��2

f(z)dz =

Z�1

f(z)dz �Z�2

f(z)dz

)Z�1

f(z)dz =

Z�2

f(z)dz.

(iii) =) (i) :Assume (iii) holds true. Pick a �xed point z0 2 D. Let � be any path that

connects z0 to z 2 D.Let F (z) =

R�f(w)dw. This integral remains the same for all paths � that

connects z0 to z by (iii). We need to show that F 0(z) exists and equals f(z).Consider F (z+�z)�F (z) =

R�1f(w)dw; where �1 is an in�nitesimal path

connecting z to z +�z. Parametrize �1 byw(t) = z + t�z ; 0 � t � 1 ; dw = �zdt. Then

F (z +�z)� F (z)�z

=

R�1f(w)dw

�z

=

R 10f(z + t�z)�zdt

�z=

Z 1

0

f(z + t�z)dt

F 0(z) = lim�z!0

F (z +�z)� F (z)�z

= lim�z!0

Z 1

0

f(z + t�z)dt

=

Z 1

0

f(z)dt = f(z):

Example 4.9 (page 178 #1 c, d, g)

(c).R�1zdz where � is any path that connects �3i to 3i in the right half

plane. The principal branch cut of log(z) avoids the right half plane.Z�

1

zdz = log zj3i�3i = log(3i)� log(�3i)

= log j3ij+ iArg(3i)� (log j�3ij+ iArg(�3i))= log(3) + i

�

2� (log(3)� i�

2) = i�.

(d).R�csc2 zdz where � is any closed path that avoids all multiples of �.

The poles of csc2 z are at all multiples of �, which our closed path � avoids.Z�

csc2 zdz = 0 since csc2 z has an antiderivative.

49

(g).R�z1=2dz where � is the spiral given in the �gure that starts at � and

ends at i.

Z�

z1=2dz

=

�2

3z3=2

�i�

=2

3

hi3=2 � �3=2

i=

2

3

h(ei�=2)3=2 � �3=2

i=2

3

hei3�=4 � �3=2

i=

2

3

"�p2

2+ i

p2

2� �3=2

#:

50

4.4 Cauchy�s Integral Theorem

In this section we will introduce the famous Cauchy�s Integral Theorem. Al-though Cauchy was the �rst to publish this result, it was already known toGauss in 1811. We will begin with a statement of the theorem and a relatedextension that is a consequence of the theorem.

Theorem 50 Cauchy�s Integral Theorem If f is analytic in a simplyconnected domain D and � is any loop (closed contour) in D, then

Z�

f(z)dz = 0: (1)

Proof. It can be shown that in a simply connected domain, a loop � can bedeformed into a single point. We then evaluate the integral at this single point,which of course, would yield zero.It can be shown via topological methods that if � is a simple closed contour

and f is analytic at each point inside and on �, then f must be analytic in somesimply connected domain containing �. Therefore, we can restate Cauchy�sTheorem as �If f is analytic inside and on �, then

R�f(z)dz = 0.�

We combine what we have gained from Cauchy�s Theorem with our resultsfrom the previous section and summarize as:

Theorem 51 In a simply connected domain, an analytic function has an anti-derivative. As a consequence, its integrals are independent of path, and its loopintegrals are zero.

Example 4.10 EvaluateR�1zdz where � is the ellipse 9x

2 + y2 = 1 traversedtwice in the counterclockwise direction.

The ellipse can be written as x2

(1=3)2 +y2

(1)2 = 1. It is easily deformed to theunit circle jzj = 1 traversed counterclockwise, without passing through z = 0.Therefore, f(z) = 1

z is analytic in Cr f0g. HenceZ�

1

zdz =

Zjzj=1

1

zdz = 2�i.

Example 4.11 Evaluate

Ijzj=1

z2 + ez

z2 � 9 dz:

Note that f(z) = z2+ez

z2�9 is analytic everywhere except at z = �3. Sincethese two poles are outside jzj = 1, we can deform � : jzj = 1 to a single point.Therefore, I

jzj=1

z2 + ez

z2 � 9 dz = 0.

51

Example 4.12 Compute the following integral where � is the simple closedcontour indicated in the �gure on page 189 of text.

Z�

3z � 2z2 � z dz

First observe that the partial fractions are:3z�2z2�z =

3z�2z(z�1) =

2z +

1z�1 .

The contour can be deformed to a barbell. Let C0 be the circle around z = 0traversed counterclockwise, C3 be the straight line path, and C1 be the circlearound z = 1. ThenZ

�

3z � 2z2 � z dz

=

ZC0

3z � 2z2 � z dz +

ZC3

3z � 2z2 � z dz +

ZC1

3z � 2z2 � z dz +

Z�C3

3z � 2z2 � z dz

=

ZC0

2

zdz +

ZC0

1

z � 1dz +ZC1

2

zdz +

ZC1

1

z � 1dz

= 2(2�i) + 0 + 0 + 2�i = 6�i.

4.5 Cauchy�s Integral Formula

We will begin this section with a very surprising result that is true for complex-valued functions but not necessarily for real-valued functions. Its proof will bedemonstrated at the end of the section.

Theorem 52 If f is analytic in a domain D, then all its derivatives f (1); f (2); f (3); � � �exists and are analytic in D.

This result is true only for complex-valued functions. For example, the real-valued function

f(x) =px

has a derivative at all values of x but f (1)(x) = 12pxdoes not have a derivative

at x = 0. Likewise, the real-valued function

g(x) = x4=3

has a �rst derivative g(1)(x) = 43x

1=3 which exists for all x, but g(2)(x) =4

9x2=3does not have a derivative at x = 0.

Next we focus on the primary result of this section, namely, Cauchy�s IntegralFormula.

Theorem 53 (Cauchy�s Integral Formula) Let � be a simple closed pos-itively oriented contour. If f is analytic in some simply connected domain D

52

and z0 is a point inside �, then

f(z0) =1

2�i

Z�

f(z)

(z � z0)dz or equivalentlyZ

�

f(z)

(z � z0)dz = 2�if(z0):

Proof. The function f(z)z�z0 is analytic everywhere in D except at z0. So we can

reduce � to a circle Cr : jz � z0j = r by a continuous deformation.Z�

f(z)

(z � z0)dz

=

ZCr

f(z)

(z � z0)dz

=

ZCr

f(z0)

(z � z0)dz +

ZCr

f(z)� f(z0)(z � z0)

dz

= f(z0)2�i+ limr!0+

ZCr

f(z)� f(z0)(z � z0)

dz (terms are independent of r)

Now let us estimate the last term. Let Mr = maxfjf(z)� f(z0)j : z on Crg.Then for z on Cr we have����f(z)� f(z0)(z � z0)

���� =jf(z)� f(z0)j

r� Mr

r;����Z

Cr

f(z)� f(z0)(z � z0)

dz

���� � Mr

rl(Cr) =

Mr

r2�r = 2�Mr:

limr!0+

����ZCr

f(z)� f(z0)(z � z0)

dz

���� � limr!0+

2�Mr = 0;

) limr!0+

ZCr

f(z)� f(z0)(z � z0)

dz = 0:

The next theorem is useful in understanding generalizations of Cauchy�sIntegral Formula.

Theorem 54 Let g be continuous on the contour �, and for each z =2 �, letG(z) =

R�g(z)��zd�. Then the function G is analytic at each point not on �, and

53

its derivatives are given by,

G(1)(z) =

Z�

g(z)

(� � z)2 d�

G(2)(z) = 2

Z�

g(z)

(� � z)3 d�

G(3)(z) = (2)(3)

Z�

g(z)

(� � z)4 d�

...

G(n)(z) = n!

Z�

g(z)

(� � z)n+1 d�:

The following results are consequences of Cauchy�s Integral Formula and theabove discussion.

Theorem 55 If f is analytic in a domain D, then all its derivatives f (1); f (2); f (3); � � �exists and are analytic in D.

Theorem 56 If f = u+iv is analytic in a domain D, then all partial derivativesof u and v exist and are continuous in D.

We already know that if f is analytic then its real and imaginary parts u; vare harmonic. Now we can extend this and say that the real and imaginaryparts of all the functions f (1); f (2); f (3); � � � are harmonic.

Theorem 57 (Morera�s Theorem) If f is continuous in a domain D and ifR�f(z)dz = 0 for every closed contour � in D, then f is analytic in D.

Proof. Morera�s theorem follows from the fact that the condition �R�f(z)dz = 0

for all closed loops � lying in D� is equivalent to f having an antiderivative inD, from which analyticity follows.

Theorem 58 Let � be a simple closed positively oriented contour. If f is ana-lytic inside and on �, and z0 is a point inside �, then

f (n)(z0) =n!

2�i

Z�

f(z)

(z � z0)n+1dz or equivalentlyZ

�

f(z)

(z � z0)n+1dz =

2�i

n!f (n)(z0):

Example 4.13 Compute the integralR�ez+cos z

z dz where � is the circle jz � 1j =2 traversed twice counterclockwise.

SolutionNote that f(z) = ez +cos z is analytic on and inside �. Hence we can apply

Cauchy�s Integral Formula twice to obtain,Z�

ez + cos z

zdz = 2 � 2�if(0) = 2 � 2�i � 2 = 8�i:

54

Example 4.14 Evaluate

Ij =

Z�j

sin z

z2 � iz + 2dz for j = 1; 2; 3;

where �1 is the circle jz � 2j = 1, �2 is square centered at 2i with a sidelength 4 units, and �3 is any rectangle that encloses both 2i and �i. All contoursare traversed once counterclockwise.SolutionThe singular points of sin z

z2�iz+2 =sin z

(z�2i)(z+i) are at z = 2i and z = �i.Since the singular points of sin z

z2�iz+2 lie outside of �1, by Cauchy�s TheoremI1 = 0.To compute I2 we set f(z) = sin z

z+i .

I2 =

Z�2

sin z

z2 � iz + 2dz

=

Z�2

�sin z

z + i

�� 1

(z � 2i)dz

= 2�i � f(2i) = 2�i � sin 2i3i

= i2�

3sinh 2.

To compute I3 we �rst deform the contour into a barbell. Let the two circularcurves be C13 : jz � 2ij = 1=2 and C23 : jz + ij = 1=2. Then

I3 =

Z�3

sin z

z2 � iz + 2dz

=

ZC13

�sin z

z + i

�� 1

(z � 2i)dz +ZC23

�sin z

z � 2i

�� 1

(z + i)dz

= 2�i

�sin z

z + i

�z=2i

+ 2�i

�sin z

z � 2i

�z=�i

= i2�

3sinh 2 + i

2�

3sinh 1:

Example 4.15 Evaluate I =IC

e2z

z4 dz where C : jzj = 1.

SolutionBy the general Cauchy�s Integral Theorem,

I =

IC

e2z

z4dz

=2�i

3!

�d3

dz3e2z�����

z=0

=�i

3

�23e2z

���z=0

=8�i

3:

55

References

[1] Sa¤, E. B. & Snider, A. D. Fundamentals of Complex Analysis with Appli-cations to Engineering and Science. Third Edition, Prentice Hall, 2003.

A Afterword

A more complete version of the notes will be placed on the web. A specialthanks to the �ve students Herb Dover, Carrie Foradis, Amber Kirkpatrick,and Wen Liu who braved a tough fall 2007 semester and successfully navigatedthe course.

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