Chapter 4 IntegralsExercises Pages 115-116

1. (a) Use the corresponding rules in calculus to establish the following

rules when w (t) = u (t) + iv (t) is a complex-valued function of a real variablet and w�(t) exists:

ddtw (�t) = �w0 (�t) ; where w0 (�t) denotes the derivative of w(t) with

respect to t, evaluated at -t.

(b) Use the corresponding rules in calculus to establish the following ruleswhen w (t) = u (t) + iv (t) is a complex-valued function of a real variable t andw�(t) exists:

ddt [w (t)]

2= 2w (t)w0 (t) :

Solution a:A function ! (t) is said to be di¤erentiable at t whenddt! (t) = lim

�t!0

!(t+�t)�!(t)(t+�t)�t : Replacing t by -t we get

ddt! (�t) = lim

�t!0

![�(t+�t)]�![�t][�(t+�t)�(�t)]

= lim�t!0

![�(t+�t)]�

= lim�t!0

![�(t+�t)]�![�t]��t

= � lim�t!0

![�(t+�t)]�![�t]�t

= �!0 (t) :Solution b:! (t) = u (t) + iv (t) = u+ iv

[! (t)]2= (u+ iv)

2= u2 + 2iv + (iv)

2

ddt [! (t)]

2= d

dt

�u2 + 2iuv + i2v2

�2u0 + 2i (uv0 + u0v) + i2 (2vv0)= 2

�u0+iuv0 + iu0v + i2vv0

�= 2 (u0 + iv) (u+ iv) = 2! (t) � !0 (t) :

Solution:(a) We have w�(t)=u�(t)+iv�(t) and so w�(-t)=u�(-t)+iv�(-t). Now we haveddtw (�t) =

ddt [u (�t) + iv (�t)]

= ddr [u (�t)] + i

ddt [v (�t)]

= u0 (�t) ddt (�t) + iv0 (�t) d

dr (�t)= �u0 (�t)� iv0 (�t) = � [u0 (�t) + iv0 (�t)] = �w0 (�t) :(b) We have [w (t)]2 = [u (t) + iv (t)]2

= [u (t)]2 � [v (t)]2

= [u (t) + iv (t)]2

= [u (t)]2 � [v (t)]2 + i2u(t)v(t): Therefore

ddt [w (t)]

2= d

dt

n[u (t)]

2 � [v (t)]2 + 2iu(t)v(t)o

1

= ddt [u (t)]

2 � ddt [v (t)]

2+ 2i � [u(t)v(t)]

= 2u(t)u0(t)� 2v(t)v0(t) + 2i [u0(t)v(t) + u(t)v0(t)] :Again we have w�(t) = u�(t) + iv�(t).Therefore 2w(t)w�(t) = 2[u(t)+iv(t)][u�(t)+iv�(t)]= 2u(t)u�(t)-2v(t)v�(t)+2i[u�(t)v(t)+u(t)v�(t)].Thus we have seen that d

dr [w (t)]2= 2w(t)w0(t):

2. Evaluate the following integrals:

(a)2R1

�1t � i

�2dt;

(b)�=6R0

ei2tdt;

(c)1R0

e�ztdt (Re z > 0) :

Solution (2b): The derivative of e2it=2i = e2it. Therefore�=6R0

e2itdt =

12i

�1=2 + i

p3=2� 1

�= i=4 +

p3=4:

Solution (2a):2R1

�1t � i

�2dt =

2R1

�1t2 � 1�

2it

�dt =

2R1

�1t2 � 1

�dt+ i

2R1

2t dt =�

� 1t � t

�j21 = i2 ln (t) j21 =

�� 12 � 2

�� (�1� 1) + i2 ln (2) = � 1

2 + i2 ln (2) :

Solution (2b):�=6R0

ei2tdt = 12ie

i2tj�=60 = 12i

�ei�=3 � 1

�= 1

2i

�� 12 + i

p32

�=

p34 + i

4 :

Solution (2c):1R0

e�ztdt = limN!1

NR0

e�ztdt = limN!+1

1�z e

�ztjN0 =

limN!+1

1�z�e�zN � 1

�: As N ! +1, we have Re (�zN) = �N Re z ! �1;

so��e�zN �� = eRe(�zN) ! 0: Thus e�zN ! 0 as N ! +1. So we have

1R0

eztdt = limN!+1

1�z�ezN � 1

�= �1

�z =1z :

3. Show that if m and n are integers,2�R0

eim�e�in�d� =

�0 when m 6= n;2� when m = n.

�Solution: When n 6= m then an antiderivative is e(m�n)i�= (m� n) i and

when one plugs in 2�i and 0 we get the same thing namely, 1(m�n)i and so when

we subtract we get 0. When n = m then the integrand is 1 and so integral is2�:

Solution (3):2�R0

eim�e�in�d� =2�R0

ei(m�n)�d�: If m = n, then the right

hand side is2�R0

1� = 2�: If m 6= n, then the right hand side is equal to

1m�ne

i(m�n)�j2�0 = 1m�n

�ei(m�n)2� � e0

�= 1

m�n (1� 1) = 0:Solution

2

This is a very important integral that occurs frequently in analysis. It gives,e.g., a way to de�ne the Dirac delta function if you are familiar with that fromprevious studies. The computation is not particularly di¢ cult. We simply usethat d

d� eim� = imeim�;m 2 Z: Thus, if m 6= n;

2�R0

eim�e�in�d� =2�R0

ei(m�n)�d� = ei(m�n)�

i(m�n) j2�0 = 1

i(m�n)�ei(m�n)2� � 1

�: How-

ever, since m�n 2 Z; we have ei(m�n)2� = 1 and thus this integral vanishes. If,however, m = n, then the integral we did above is incorrect as we are dividing by

zero. Indeed, in this case, things are even simpler2�R0

eim�e�in�d� =2�R0

1d� = 2�:

4. According to de�nition (2), Section 37, of integrals of complex-valued

functions of a real variable,�R0

e(1+i)xdx =�R0

ex cosxdx+ i�R0

ex sinxdx: Evaluate

the two integrals of the right here by evaluating the single integral on the leftand then using the real and imaginary parts of the value found.

Solution: This is a nice trick as you may remember that the integrals onthe right are rather involved (you have to integrate by parts twice and recognizethat you return to where you began but with the opposite side ... thus you cansolve the desired equation).The formula follows easily from the de�nition of complex exponents and

Euler�s formula. Evaluating the left side, we have�R0

e(1+i)xdx = 1(1+i)e

(1+i)xj�0 =1

(1+i)

�e(1+i)� � 1

�= 1

(1+i) (e� cos� + ie� sin� � 1) :

Since 1(1+i) =

(1�i)2 ; we have that this is = 1

2 (1� i) (�e� � 1) :

Thus,�R0

ex cosxdx = � (1+e�)2 and

�R0

ex sinxdx = (1+e�)2 :

7. Apply the inequality���R ba w (t) dt��� � R ba jw (t)j dt (a � b) to show that

for all values of x in the interval �1 � x � 1; the functionsPn (x) =

1�

R �0

�x+ i

p1� x2 cos �

�nd� (n = 0; 1; 2; : : :) satisfy the in-

equality jPn (x)j � 1:

Solution:

Pn (x) =1�

Z �

0

�x+ i

p1� x2 cos �

�nd�

jPn (x)j =���� 1�Z �

0

�x+ i

p1� x2 cos �

�nd�

����jPn (x)j � 1

�

Z �

0

����x+ ip1� x2 cos ��n��� d�Consider x+ i

p1� x2 cos � = ei�

=

qx2 +

�p1� x2 cos �

�2=px2 + (1� x2) cos2 �

3

� = tan�1hp

1�x2 cos �2

i= tan�1 �h

Let �0 =p1�x2 cos �

x

i:��x+ ip1� x2 cos ���n = �� ei���n = j jn ��ein��� = j jn�

)��ein��� = 1�

) The equation jPn (x)j � 1�

Z �

0

���px2 + (1� x2) cos2 ����n d�) jPn (x)j �

���px2 + (1� x2) cos2 ����n d�= 1

�

Z �

0

���px2 sin2 � + cos2 ����n d�= 1

�

Z �

0

d�

= 1:jPn (x)j � 1:

Exercises Pages 120-1221. Show that if w(t) = iv(t) is continuous on an interval a � t � b, then

(a)�aR�bw (�t) dt =

bRa

w (�) d� ;

(b)bRa

w (t) dt =�R�

w [� (�)]�0 (�) d� , where � (�) is the function in equation

(9), Section 38.Suggestion: These identities can be obtained by noting that they are valid

for real-valued functions of t.

Solution: Start by writing I =�aR�bw (�t) dt =

�aR�bu (�t) dt + i

�aR�bv (�t) dt:

The substitution s = -t in each of these two integrals on the right then yields

I = �aRb

u (s) ds� iaRb

v (s) ds =bRa

w (s) ds: That is, I =�aR�bw (�t) dt =

bRa

w (s) ds:

b) Start with I =bRa

w (�t) dt =bRa

u (t) dt + ibRa

v (t) dt; and this time the

substitution t = � (s) in each of the integrals on the right gives the result.

2. Let C denote the right-hand half of the circle jzj = 2, in the counter-clockwise direction, and note that two parametric representations for C are z =z (�) = 2ei�

���2 � � �

�2

�and z = Z (y) =

p4� y2+iy (�2 � y � 2) :

Verify that Z (y) = z [� (y)], where � (y) = arctan yp4�y2�

��2 < arctan t <

�2

�. Also show that this function � has a positive deriva-

tive, as required in the condition following equation (9), Section 38.

Solution:We check

4

z [� (y)] = 2ei arctan yp

4�y2 = 2

�cos arctan yp

4�y2+ i sin arctan yp

4�y2

�:

Notice that arctan y4�y2 is the angle made from the right triangle with basep

4� y2 and height y. Thus, the hypotenuse is 2. Using this triangle, wemay evaluate the cos and sin to see that z [� (y)] = 2

p4�y22 + i2y2 = Z (y) as

desired. Noting that �0 (y) = 1�1+

�yp4�y2

�2�p4�y2+ y2p

4�y2

4�y2 = 1p4�y2

we see

that the derivative is indeed positive for y 2 (�2; 2) :

5. Suppose that a function f(z) is analytic at a point z0 = z (t0) lying ona smooth arc z = z (t) (a � t � b). Show that if w (t) = f [z (t)], thenw0 (t) = f 00 [z (t)] z0 (t) when t = t0. Suggestion: Write f(z) = w(x,y)+iv(x,y)and z(t) = x(t)+iy(t), so that w (t) = u [x (t) ; y (t)] + iv [x (t) ; y (t)] : Thenapply the chain rule in calculus for functions of two real variables to writew0 = (uxx

0 + uyy0) + i (vxx

0 + vyy0), and use the Cauchy-Riemann equations.

Solution: f (z) = u (x; y) + iv (x; y) and z (t) = x (t) + iy (t) so that w (t) =u [x (t) ; y (t)] + iv [x (t) + iy (t)]. The chain rule says that w0 (t) = uxx

0 (t) +uyy

0 (t) + i [vxx0 (t) + vyy

0 (t)]. Since ux = vy and uy = �vx we get w0 (t) =uxx

0 (t)� vxy0 (t) + i [vxx0 (t) + uxy0 (t)] = uxz0 (t) + ivxz0 (t) = f 0 [z (t)] z0 (t) :

6. (a) Let y(x) be a real-valued function de�ned on the interval 0 � x � 1

by means of the equations y (x) =�x3 sin

��x

�when 0 < x � 1:

0 when x = 0.

�:

Show that the equation z = x + iy (x) (0 � x � 1) represents an arc Cthat intersects the real axis at the points z = 1=n (n = 1; 2; : : :) andz = 0.

(b) Let y(x) be a real-valued function de�ned on the interval 0 � x � 1 by

means of the equations y (x) =�x3 sin

��x

�when 0 < x � 1:

0 when x = 0.

�:

Verify that the arc C in part (a) is, in fact, a smooth arc. Suggestion: Toestablish the continuity of y(x) at x = 0, observe that0 �

��x3 sin ��x ��� � x3 when x > 0. A similar remark applies in �nding y�(0)and showing that y�(x) is continuous at x = 0.

Solution:(a) For a complex variable we take the x-axis as the real axis and the y-axis

as the imaginary axis. If an arc intersects the real axis. If an arc intersectsthe real axis, then y = 0. Given z = x+ iy (x) and y (x) = x3 sin

��x

�: When

y = 0, we have x3 sin��x

�:

) x3 = 0 or sin��x

�= 0�

) x = 0 or �x = n�; for n = 1; 2; � � �

5

) x = 1n

When x = 0, the equation z = x+ iy (x) becomes z = 0 + iy (0)) z = 0:When x = 1

n ; the equation z = x+ iy (x) becomes z =1n + i (0)) z = 1

n :

(b) y (x) =�x3 sin(�x );0<x�1

0 when x = 0.

y0 (x) = 3x2sin

��x

�� x cos

��x

�= x

�3x sin

��x

�� cos

��x

��jy0 (x)j � 0��x �3x sin ��x �� cos ��x ���� � 0jxj��3x sin ��x �� cos ��x ��� � 0

jxj � 0 or��3x sin ��x �� cos ��x ��� � 0)

0 ���3x sin ��x �� cos ��x ��� � 3x� 1 when x > 0:�

)��sin ��x ��� � 1 and ��cos ��x ��� � 1� :

Exercises Pages 128-130For the function f and contours C in Exercises 1 through 6, use parametric

representations for C, or legs of C, to evaluateRC

f (z) dz:

1. f (z) = (z+2)z and C is

(a) the semicircle z = 2ei� (0 � � � �) ;(b) the semicircle z = 2ei� (� � � � 2�) ;(c) the circle z = 2ei� (0 � � � 2�) :Solution (1a): z = 2ei� so z0 (�) = 2iei�d�: Thus the integral is�R0

2ei�+22ei�

2iei�d� =�R0

i�2ei� + 2

�d� = 2ei(��0) + 2i� = �4 + 2�i:

Solution (1a):RC

f (z) dz =�R0

2ei�+22ei�

2iei�d� =�R0

�2iei� + 2i

�d� =�

2ei� + 2i��j�0 = 2ei� � 2e0 + 2�i� 0 = �4 + 2�i:

Solution (1b):RC

f (z) dz =2�R0

2ei�+22ei�

2iei�d� =2�R0

�2iei� + 2i

�d� =�

2ei� + 2i��j2�� = 2ei2� � 2ei� + 4�i� 2�i = 4 + 2�i:

Solution (1c): This integration is the sum of the integrations in (a) and (b).So it is equal to (�4 + 2�i) + (4 + 2�i) = 4�i:2. f (z) = z � 1 and C is the arc form z = 0 to z = 2 consisting of(a) the semicircle z = 1 + ei� (� � � � 2�) ;(b) the segment 0 � x � 2 of the real axis.

Solution (2a): z0 = iei� so the integral isRf (z) dz =

2�R�

ei�iei�: The

antiderivative is e2i�=2 and when we plug in the terms cancel so we get 0.

4. f(z) is de�ned by the equations f (z) =�

1 when y < 0,4y when y > 0,

�and C is

the arc from z = -1-i to z = 1+i along the curve y = x3:Solution: Here z (t) = t + it3 so z0 (t) = 1 + i3t2. Here �1 � t � 1: Our

integral is0R�1

�1 + i3t2

�dt+

1R0

4t3�1 + i3t2

�dt = (1 + i) + (1 + 2i) = 2 + 3i:

6

Solution (4): C is parameterized by z (t) = t + it3;�1 � t � 1: If t< 0, Im z (t) = t3 < 0; so f (z (t)) = 1: If t > 0, Im z (t) = t3 > 0, sof (z (t)) = 4 Im z (t) = 4t3: We compute z0 (t) = 1 + i3t2: Thus

RC

f (z) dz =

1R�1f (z (t)) z0 (t) dt =

0R�1f (z (t)) z0 (t) dt+

1R0

f (z (t)) z0 (t) dt =0R�11�1 + i3t2

�dt+

2R1

4t3�1 + i3t2

�dt =

�t+ it3

�j0�1 +

�t4 + i2t6

�j10

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

Problem statement:

EvaluateRC

f (z) dz for f (z) =�

1 when y < 0,4y when y > 0,

�and C is the arc from z

= -1-i to z = 1+i along the curve y = x3:Solution: The curve can be represented parametrically by (x (t) ; y (t)) =�

t; t3�;�1 � t � 1: We call C1 the portion of C where �1 � t � 0 and

C2 the portion for 0 � t � 1: ThenRC

f (z) dz =RC1

f (z) dz +RC2

f (z) dz =

0R�11 ��1 = i3t2

�dt+

1R0

4t3�1 + i3t2

�dt =

�t+ it3

�j0�1+

�t4 + i2t6

�j10 = (1 + i) +

(1 + 2i) = 2 + 3i:6. f(z) is the branchz�1+i = exp [(�1 + i) log z] (jzj > 0; 0 < arg z < 2�) of the indicated power

function, and C is the positively oriented unit circle jzj = 1:

Solution: Let z = rei�: Thenz�1+i = e(�1+i)(log r+i�)

= e� log r��+i(log r��)

= e��

r [cos (log r � �) + i sin (log r � �)] :The correct branch is obtained by assuming 0 < � < 2�:The circle C can be parameterized as ei�; 0 � � < 2�: ThereforeRC

f (z) dz =2�R0

e�� (cos � � i sin �)�iei��d� (since r = 1 on contour)

= i2�R0

e��d� (by Euler�s formula)

= i�1� e�2�

�Problem statement: Evaluate

RC

f (z) dz for f(z) for the branch z�1+i =

exp [(�1 + i) log z] (jzj > 0; 0 < arg z < 2�)of the indicated power function, and C is the positively oriented unit circle

jzj = 1:Solution:First, z (t) = eit; 0 � t � 2�: ThenRC

f (z) dz =2�R0

exp�(�1 + i) log eit

�ieitdt

7

=2�R0

exp�(�1 + i) log eit

�ieitdt

= i2�R0

e�tdt = �ie�tj2�0 = i�1� e�2�

�:

10. Let C0 denote the circle jz � z0j = R, taken counterclockwise. Use theparametric representation z = z0 +Re

i� (�� � � � �) for C0 to derive thefollowing integration formulas:(a)

RC0

dzz�z0 dz = 2�i;

(b)RC0

(z � z0)n�1 dz = 0 (n = �1;�2; � � � ) :

Solution: Let z = z0 + Rei� so that dz = Riei� and z � z0 = Rei�. Then

the �rst integral is2�R0

id� = 2�i and the second integral is2�R0

iRneni� = 0:

Solution: (10a)RC0

dz(z�z0)dz =

�R��

1Rei�

Riei�d� = 2�i:

(10b) First possibility: use the antiderivative of (z � z0)n�1 : Second pos-sibility: parameterize as in a).

11. Use the parametric representation in Exercise 10 for the oriented circleC0 there to show that

RC0

(z � z0)n�1 dz = i 2Ra

a sin (a�), where a is any real

number other than zero and where the principal branch of the integrand andthe principal value of Ra are taken. [Note how this generalizes Exercise 10(b).]

Solution:RC0

(z � z0)a�1 dz

=RC0

exp [(1� 1)Log (z � z0)] dz�R��exp

h(a� 1)Log

�Rei�

�iiRei� d�

=�R��exp [(a� 1) (lnR+ i�)] iRei� d� = iR

�R��Ra�1e(a�1)i�ei�d�

= iRa�R��eia�d� = Ra

a eia�j��� = Ra

a

�eia� � eia(��)

�= Ra

a [2i sin (a�)] :

At the last step, we are using sin � = (ei��e�i�)2i :

Exercises Pages 133-134

1. Without evaluating the integral, show that����RC

dzz2�1

���� � �3 when C is the

same arc as the one in Exercise 1, Section 41.Solution: On the circle

��z2 � 1�� � 3 so the integral is bounded by 13

RC

jdzj :

The integral is arclength which is 4�=4 = �:

8

Solution (1): On the contour C, jzj = 2, so��z2 � 1�� � ��z2��� 1 = 22� 1 = 3,

which implies��� 1z2�1

��� � 1=3: We already know that L (C) = �, so

����RC

dzz2�1

���� �13L (C) =

�3 :

2. Let C denote the line segment from z = i to z = 1. By observing that,of all the points on that line segment, the midpoint is the closest to the origin,

show that

����RC

dzz4

���� � 4p2 without evaluating the integral.Solution: The closest point, the midpoint, is (1 + i) =2 at distance 1p

2away

from 0. Thus jzj4 � 1=4 and so the integral is bounded by 4RC

jdzj = 4p2:

3. Show that if C is the boundary of the triangle with vertices at the points0, 3i, and -4, oriented in the counterclockwise direction (see Figure 47), then����RC

(ez � z)���� � 60:

Solution:The triangle is a 3-4-5 right triangle, and thus, its length is 12. Note that

jez � zj � jezj+ jzj � jezj+4 on C since it is easy to see that -4 is the point onthe triangle farthest from the origin. Note also that Re z � 0 for all points onthe triangle. Thus, jezj = jexj

��eiy�� � 1 on the triangle. Thus, the integrand is� 5 on the triangle. Hence,

����RC

(ez � z) dz���� � 5 (12) = 60:

4. Let CR denote the upper half of the circle jzj = R (R > 2), taken

in the counterclockwise direction. Show that

����RC

2z2�1z4+5z2+4dz

���� � �R(2R2+1)(R2�1)(R2�4) .

Then, by dividing the numerator and denominator on the right here by R4,show that the value of the integral tends to zero as R tends to in�nity.

Solution (4): If z 2 CR, then jzj = R; so��2z2 � 1�� � ��2z2��+ j�1j = 2 jzj2 +

1 = 2R2 + 1, and��z4 + 5z2 + 4�� = ��z2 + 1�� ��z2 + 4�� � �jzj2 � 1��jzj2 � 4� =�

R2 � 1� �R2 � 4

�> 0: Thus

��� 2z2�zz4+5z2+4

��� = j2z2�1jz4+5z2+4 �

2R2+1(R2�1)(R2�4) : We may

parametrize CR by z (�) = Rei�; 0 � � � �: Then L (CR) =�R0

jz0 (�)j d� =

�R0

��Riei��� d� = �R0

Rd� = R�: Thus

����� RCR 2z2�zz4+5z2+4dz

����� � 2R2+1(R2�1)(R2�4)L (CR) =

�R(2R2+1)(R2�1)(R2�4) :

Solution: The contour C in the upper half of the circle jzj = R:On the contour,

��2z2 � 1�� � ��2z2��+1 = 2R2+1: Similarly, ��z4 + 5z2 + 4�� ���z4�� = ��5z2��� 4 = R4 � 5R2 � 4: We have9

���� (2z2+1)(z4+5z2+4)

���� � (2R2+1)(R4�5R2�4)

for z on the contour C.

Therefore the contour integral is bounded by �R�(2R2+1)

(R4�5R2�4)

�:

The limit of this bound is 0 as R!1:

5. Let CR be the circle jzj = R (R > 1), described in the counterclockwise

direction. Show that

����RC

Logzz2 dz

���� < 2� ��+lnRR

�, and then use l�Hospital�s rule

to show that the value of this integral tends to zero as R tends to in�nity.

Solution:

Let us compute the integral. We use the expression z = Rei� for complex

numbers belonging to CR: We note thatZ 2�

0

ei�d� =

Z 2�

0

e�i�d�:Zjzj=R

[Log(z)]z2 dz = 2i

R

Z 2�

0

[lnR+ i (� + 2k�)] e�i�d�

= 2iR (lnR+ 2k�i)

Z 2�

0

e�i�d� � 2R

Z 2�

0

�e�i�d�

= � 2R

Z 2�

0

�e�i�d�

= � 4�iR :

The limR!1

�4�iR = 0:

6. Let C� denote the circle jzj = � (0 < � < 1), oriented in the counterclock-wise direction, and suppose that f(z) is analytic in the disk jzj � 1. Show thatif z�1=2 represents any particular branch of that power of z, then there is a non-

negative constant M, independent of �, such that

����RC

z�1=2f (z) dz

���� � 2�Mp�.Thus show that the value of the integral here approaches 0 as � tends to 0.Suggestion: Note that since f(z) is analytic, and therefore continuous,

throughout the disk jzj � 1, it is bounded there (Section 17).

Solution:

Since f(z) is analytic, and therefore continuous throughout the disk jzj � 1;it is bounded there. This implies the existence of a constant M such that

jf (z)j � �2M for all jzj � 1: We have

����Z RC

z�1=2f (z) dz

���� � �M2

����RC

z�1=2dz

���� :The latter integral is calculated as follows.RC

z�1=2dz =

Z 2�

0

��1=2e�[i(�+2k�)=2]�iei�d�

= i�1=2e�ik�Z 2�

0

ei�=2d�

10

= 2�1=2e�ik��ei� � 1

�= �4p�e�ik�:Here k = 0 or k = 1 depending on the branch of the function z�1=2: In both

cases we have����RC

z�1=2f (z) dz

���� � �M2

����RC

z�1=2dz

���� � 2�p�M:Hence, as �! 0 the value of the integral tends to zero as w

Exercises Pages 141-1422. By �nding an antiderivative, evaluate each of these integrals, where the

path is any contour between the indicated limits of integration:

(a)i=2Ri

e�zdz;

(b)�+2iR0

cos�z2

�dz;

(c)3R1

(z � 2)3 dz:

Solution (2c): An antiderivative is (z � 1)4 =4 so that the integral is evalu-ated by evaluating at the endpoints which gives 1/4-1/4 = 0.

Solution (2a): Since 1� e

�z is an antiderivative of e�z, soi=2Ri

e�zdz = 1�

�ei�=2 � ei�

�= (1 + i) =�:

Solution (2b): Since 2 sin (z=2) is an antiderivative of cos (z=2), so�+2iR0

cos�z2

�dz = 2 sin

�z2

�j�+2i0 = 2 sin (�=2 + i) =

�i�ei(�=2+i) � e�i(�=2+i)

�= �i

�ie�1 � (�ie)

�= 1=e+ e:

Solution (2c): Since 14 (z � 2)

4 is an antiderivative of (z � 2)3, so3R1

(z � 2)3 dz = 14 (z � 2)

4 j31 = 14

�(3� 2)� (1� 2)4

�= 0:

3. Use the theorem in Section 42 to show thatRC0

(z � z0)n�1 dz = 0

(n = �1;�2; � � � ) when C0 is any closed contour which does not pass throughthe point z0. [Compare Exercise 10(b), Section 40.]

Solution: The fundamental theorem of calculus implies that the integralsare 0. For n = 0; 1; 2; : : : ; the integrals are 0 even if the contour passes throughz0: For n = �1;�2; : : : ; the integrand is unde�ned at z = z0: Therefore thecontour must not pass through z0:

4. Find an antiderivative F2 (z) of the branch f2 (z) of z1=2 in Example 4,Section 43, to show that the integral (6) there has value 2

p3 (�1 + i). Note

that the value of the integral of the function (5) around the closed contourC2 � C1 in that example is, therefore, �4

p3:

11

Solution: We want to �ndRC2

pzdz with

pz =

prei�=2 for z = rei� with

0 � � < 2�:The branch cut in the de�nition of

pz can be moved to the positive imag-

inary axis in such a way thatpz does not change on the contour C2: De-

�nepz =

prei�=2 for z = rei� with �=2 � � < 5�=2: The antideriva-

tive of this branch ofpz is given by 2

3z3=2 = 2

3r3=2ei3�=2 for z = rei� with

�=2 < � < 5�=2: The integral over the contour C2 can be obtained as follows:23z3=2j3�3 = 2

333=2ei3�=2j�=2��=� = 2

333=2�ei3� � ei3�=2

�= 2

p3 (�1 + i) :

5. Show that1R�1zidz = 1+e��

2 (1� i), where zi denotes the principal branch

zi = exp (i Log z) (jzj > 0;�� < Arg z < �) and where the path of integra-tion is any contour from z = -1 to z = 1 that, except, for its end points, liesabove the real axis.Suggestion: Use an antiderivative of the branch zi = exp (i Log z)�jzj > 0;��

2 < Arg z <3�2

�of the same power function.

Solution: An antiderivative is zi+1

i+1 . Evaluating at the endpoints we get

1i+1= (1 + i) � (�1)i+1 = (i+ 1) : The �rst term is 1= (i+ 1). The second ise(i+1) log(�1)= (1 + i) = e(i+1)i�= (1 + i) = �e��= (1 + i). Putting it together

we get 1�e��

1+i =(1+e��)

2 (1� i) :

Solution: The antiderivative is given by 1(1+i)z

(1+i) = 1(1+i)e

(1+i)Logz; withLog (z) being the principal branch of the logarithm. Of course, Log (1) =0: Since the contour varies on the upper half of the complex plane beforeterminating at -1, we can take Log (�1) = i�: By the fundamental theorem of

calculus, the integral is equal to 1(1+i)

�e0 � ei�(1+i)

�=(1+e��)

2 (1� i) :

Exercises Pages 153-156

1. Apply the Cauchy-Goursat theorem to show thatRC

f (z) dz = 0 when

the contour C is the circle jzj = 1, in either direction, and when(a) f (z) = z2

z�3 ;

(b) f (z) = ze�z;(c) f (z) = 1

z2+2z+2 ;(d) f (z) = sech z;

(e) f (z) = tan z;(f) f (z) = Log (z + 2) :

Solution:

12

According to the Cauchy-Goursat theorem if f(z) is an analytic single-valuedfunction in the convex domain G than for any regular closed curve C containedin G the following integral vanishes:

RC

f (z) dz = 0:

(a) In our case f (z) = z2= (z � 3) : This function has the only singularityat point z = 3. Throughout the disk jzj � 2 the function f(z) is analytic andsingle-valued. Hence, according to the Cauchy-Goursat theorem, its integralover the curve C : jzj = 1 is equal to zero:

Rjzj=1

z2dz(z�3) = 0:

(b) As before, the function f (z) = ze�z is analytic and single-valuedthroughout the disk jzj � 2: Hence, by the Cauchy-Goursat theoremR

C

ze�zdz = 0:

(c) The function f (z) = 1=�z2 + 2z + 2

�= (z + 1 + i) (z + 1� i) : Hence,

the function f(z) has singularities only at points z = -1-i and z = -1+i. Bothpoints lie outside of the disk jzj � 1:2 since j�1� ij =

p2 ' 1:4: Hence, the

function is analytic throughout the disk jzj � 1:2 and by the Cauchy-Goursattheorem

RC

dz(z2+2z+2) = 0:

(d) The function f (z) = sechz does not have singularities and is analyticin any �nite disk jzj � R: Hence, by the Cauchy-Goursat theoremR

jzj=1sechzdz = 0:

(e) The function f (z) = tan z = sin zcos z is not de�ned only if the denominator

is zero; that is cos z = 0. The general solution is z = (2n+ 1) �2 ; where n is aninteger. Whatever the values of n may be, the value of z will not lie inside theunit circle. So tan z is analytic throughout the region. Since f(z) is analyticinside and on C, the integral of f(z) over C is zero by the Cauchy-Goursattheorem.(f) The function f (z) = Log (z + 2) has a singular point z = -2. It is

analytic throughout the disk jzj � 1:5; and by the Cauchy-Goursat theoremRjzj=1

Log (z + 2) dz = 0:

2. Let C1 denote the positively oriented circle jzj = 4 and C2 the positivelyoriented boundary of the square whose sides lie along the lines z = �1; y =�1 (Figure 61). With the aid of Corollary 2 in Section 46, point out whyRC1

f (z) dz =RC2

f (z) dz when

(a) f (z) = 13z2+1 ;

(b) f (z) = z+2sin(z=2) ;

(c) f (z) = z1�ez :

Solution (2a): The singularities of this function are atp�1=3 which are

inside C2. Therefore by Cauchy for doubly connected domains,RC1

f (z) dz =RC2

f (z) dz:

13

Solution (2a): Note that f is analytic in Cnn� ip

3

o, and � ip

3are all interior

to C2. Both C1 and C2 are positively oriented, soRC1

fdz =RC2

fdz:

Solution (2b): Note that f is analytic in Cn f2n� : n 2 Ng. When n = 0,2n� = 0 lies inside C2. When n 6= 0; j2n�j � 2� > 4, so 2n� lies outside C1.Thus

RC1

fdz =RC2

fdz:

Solution (2c): Note that f is analytic in Cn f2n�i : n 2 Ng. When n = 0,2n�i = 0 lies inside C2. When n 6= 0; j2n�ij � 2� > 4, so 2n�i lies outside C1.Thus

RC1

fdz =RC2

fdz:

3. If C0 denotes a positively oriented circle jz � z0j = R, thenRC0

(z � z0)n�1 dz =�0 when n = � 1;�2; � � � ;

2�i when n = 0.

�according the Exercise 10, Section 40. Use that result and Corollary 2 in

Section 46 to show that if C is the boundary of the rectangle0 � x � 3; 0 � y � 2, described in the positive sense, thenRC

(z � 2� i)n�1 dz =�0 when n = � 1;�2; � � � ;

2�i when n = 0.

�Solution (3): Note that 2+i lies inside C. Let C0 be a positively ori-

ented circle fjz � (2 + i)j = 1=2g. Then C0 lies inside C and (z � (2 + i))n�1

is analytic between C0 and C for any n 2 N . ThusRC

(z � 2� i)n�1 dz =RC0

(z � 2� i)n�1 dz =�

0 when n 6= 0;2�i when n = 0.

�Solution: Using the statement of Exercise 40.20 and Corollary 2 in section

46 we only have to show that we can take a positively circle centered at z0 = 2+isuch that f (z) = (z � z0)n�1 is closed on this circle, on the contour C and inthe region between them. If the exponent n-1 is positive or equal to 0, then fis entire and we can take this circle to be any circle that is large enough. If theexponent is negative, then f has a singular point in 2+i. This point is containedin the interior of the rectangle given by C, so again we may take our circle to bea huge circle enclosing this rectangle, then f has the desired property of beinganalytic outside the interior of the rectangle and we may apply the corollary.

4. Use the method described below to derive the integration formula1R0

e�x2

cos 2bxdx =p�2 e

�b2 (b > 0) :

(a) Show that the sum of the integrals of exp��x2

�along the lower and

upper horizontal legs of the rectangular path in Figure 62 can be written

2aR0

e�x2

dx � 2eb2aR0

e�x2

cos 2bxdx and that the sum of the integrals along the

vertical legs on the right and left can be written

ie�a2bR0

ey2

e�i2aydy � ie�a2bR0

ey2

ei2aydy:

14

Thus, with the aid of the Cauchy-Goursat theorem, show thataR0

e�x2

cos 2bxdx = e�b2aR0

e�x2

dx+ e�(a2+b2)

bR0

ey2

sin 2aydy:

(b) By accepting the fact that1R0

e�x2

dx =p�2 and observing that����� bR0 ey2 sin 2aydy

����� < bR0

ey2

d y, obtain the desired integration formula by letting

a tend to in�nity in the equation at the end of part (a).

Solution: For part (a), we �rst considerRC1

e�x2

dz +RC2

e�z2

dz;

where C1 and C2 are the lower and upper legs of the rectangular contourshown in Figure 62. The lower leg can be parameterized as z = x;�a � x < a;and the upper leg as z = x + ib; with x decreasing from a to -a. For bothparameterizations dz = dx. Along the lower leg e�z

2

= e�x2

; and along theupper leg e�z

2

= e�(x+ib)2

= e�x2+b2�2ibx = e�x

2+b2 [cos (2bx) + i sin (2bx)] :Therefore the sum of the integrals along the lower and upper legs is equal toaR�ae�x

2

dx+ eb2�aRa

e�x2

[cos (2bx) + i sin (2bx)] dx:

Notice that e�x2

sin (2bx) is an odd function of x while e�x2

and e�x2

cos (2bx)are even functions of x. Therefore, the sum of the two integrals can be expressedas

2aR0

e�x2

dx� 2eb2aR0

e�x2

cos (2bx) dx:

The left and right legs of the contour in Figure 62 can be parameterized asz = a+ iy with y increasing from 0 to b and as z = �a+ iy with y decreasingfrom b to 0. Use of this parameterization gives the sum of the integrals alongthe right and left legs to be

ie�a2bR0

ey2

e�i2aydy � ie�a2bR0

ey2

ei2aydy:

If Euler�s formula is used for e�i2ay and ei2ay, we get

2e�a2bR0

ey2

sin (2ay) dy

for the sum of the integrals along the right and left legs.Since e�z

2

is analytic on the contour shown in Figure 62 and everywhereinside it, Cauchy�s theorem implies that the sum of the integrals

2aR0

e�x2

dx�2eb2aR0

e�x2

cos (2bx) dx and 2e�a2bR0

ey2

sin (2ay) dy is 0. Therefore

aR0

e�x2

cos (2bx) dx = e�b2aR0

e�x2

dx+ e�(a2+b2) +

bR0

ey2

sin (2ay) dy:

For part (b), we merely have to note that the third term inaR0

e�x2

cos (2bx) dx =

15

e�b2aR0

e�x2

dx+ e�(a2+b2) +

bR0

ey2

sin (2ay) dy above tends to 0 as a!1 for any

b > 0:6. Let C denote the positively oriented boundary of the half disk 0 � r �

1; 0 � � � �, and let f(z) be a continuous function de�ned on that half disk bywriting f(0) = 0 and using the branch f (z) =

prei�=2

�r > 0;��

2 < � <3�2

�of the multiple-valued function z1=2. Show that

RC

f (z) dz = 0 by evaluating

separately the integrals of f(z) over the semicircle and the two radii which makeup C. Why does the Cauchy-Goursat theorem not apply here?

Solution: The branch of z1=2 which is taken to be f (z) is analytic everywhereexcept on the negative imaginary axis and at z = 0: The contour C passesthrough z = 0, a point where f (z) is not analytic. Therefore Cauchy�s theoremdoes not apply. C can be deformed very slightly by bumping it up a bitnear the origin. Then

Rf (z) changes only slightly as f (z) is continuous at

z = 0: Cauchy�s theorem applies over the deformed contour and implies thatRf (z) dz = 0: By making the bump smaller and smaller and taking the limit,

we can conclude thatRC

f (z) dz = 0:

Exercises Pages 162-1641. Let C denote the positively oriented boundary of the square whose sides

lie along the lines x = �2 and y = �2. Evaluate each of these integrals:(a)

RC

e�zdzz�(�i=2) ;

(b)RC

cos zz(z2+8)dz;

(c)RC

zdz2z+1 ;

(d)RC

cosh zz4 dz;

(e)RC

tan(z=2)

(z�z0)2dz (�2 < x0 < 2)

Solution (1a): The function e�z is analytic so by CIF the integral is2�ie��i=2 = 2�i (�i) = 2�:Solution (1b): The function cos (z) =

�z2 + 8

�is analytic inside the contour

so the integral is 2�i cos (0) =8 = �i=4:

Solution (1a): Since �i=2 lies inside C, and cos zz2+8 is analytic inside and on

C (it is analytic throughout C except at �p8i, which both lie inside C), soR

C

e�zz�(�i=2) = 2�ie

��i=2 = 2�:

Solution (1b): Since 0 lies inside C, and cos zz2+8 is analytic inside and on

C (it is analytic throughout C except at �p8i, which both lie outside C), soR

C

cos zz(z2+8)dz = 2�i

cos(0)02+8 =

�i4 :

16

Solution (1c): Since -1/2 is inside C and z/2 is analytic in C, soRC

z2z+1dz =R

C

z=2z�(�1=2)dz = 2�i ((�1=2) =2) = ��i=2:

Solution (1d): Since 0 lies inside C, and cosh is analytic in C, soRC

cosh zz4 dz =

2�i3! � cosh

(3) (0) = �i3 sinh (0) = 0:

Solution (1e): Since x0 lies inside C, and tan (z=2) is analytic in C, soRC

tan(z=2)

(z�x0)2dz = 2�i ddz tan (z=2) jz=z0 = �i sec

2 (x0=2) :

2. Find the value of the integral of g(z) around the circle jz � ij = 2 in theportion sense when(a) g (z) = 1

z2+4 ;

(b) g (z) = 1(z2+4)2

:

Solution (2a): We have 1z2+4 =

1z+2i

1z�2i : The �rst factor is analytic inside

the contour so the integral is 2�i 14i =�2 :

Solution (2b): For the second g (z) = f (z) 1(z�2i)2 where f (z) =

1(z+2i)2

so

that by CIF for derivatives the integral is 2�if 0 (2i) = �4�i(4i)3

= �16 :

Solution (2a): Let C denote the positively oriented circle fjz � ij = 2g.Note that 2i lies inside C, and -2i does not. So 1

(z+2i) and1

(z+2i)2are analytic

inside and on C.RC

1z2+4dz =

RC

1=(z+2i)z�2i dz = 2�i 1

2i+2i = �=2:

Solution (2b): Let C denote the positively oriented circle fjz � ij = 2g.Note that 2i lies inside C, and -2i does not. So 1

(z+2i) and1

(z+2i)2are an-

alytic inside and on C.RC

1(z2+4)2

dz =RC

1=(z+2i)2

(z�2i)2 dz = 2�i ddz

�1

z+2i

�jz=2i =

2�i �2(2i+2i)3

= 2�i �2�64i = �=16:

3. Let C be the circle jzj = 3, described in the positive sense. Show that ifg (w) =

RC

2z2�z�2z�w dz (jwj 6= 3), then g (2) = 8�i. What is the value of g(z)

when jwj > 3?Solution (3): Let f (z) = 2z2 � z � 2. By CIF g (2) = 2�if (2) = 8�i: For

the second part since the function is analytic inside the contour when jwj > 3the integral is 0 by Cauchy Theorem.

Solution (3): Let f (z) = 2z2 � z � 2. Then f is an entire function. Since2 lies inside C, so by Cauchy�s integral formula, g (2) =

RC

f(z)z�2dz = 2�if (2) =

2�i4 = 8�i: If w is outside C, let hw (z) = f (z) = (z � w). Then h is analyticon and inside C. Thus by Cauchy-Goursat theorem, g (w) =

RC

h2 (z) dz = 0:

4. Let C be any simple closed contour, described in the positive sense inthe z plane, and write g (w) =

RC

z3+2z(z�w)3 dz. Show that g (w) = 6�iw when w is

inside C and that g(w) = 0 when w is outside C.

Solution (4): Since z3 + 2z is analytic in C, so if w is inside C, we have

17

g (w) =RC

z3+2z(z�w)3 dz =

2�i2

d2

dz2

�z3 + 2z

�jz=w = 6�iw: If w is outside C, then

(z3+2z)(z�w)3 is analytic inside and on C. Thus g (w) =

RC

z3+2z(z�w)3 dz = 0:

5. Show that if f is analytic within and on a simple closed contour C andz0 is not on C, then

RC

f 0(z)dzz�z0 =

RC

f(z)dz

(z�z0)2:

Solution: The Cauchy Integral Formula applied to f�gives f 0 (z0) =RC

f 0(z)(z�z0)dz:

The formula for higher derivatives (p.161, (4)) gives f 0 (z0) = 1!2�i

RC

f(z)

(z�z0)2dz:

This implies the claim.

6. Let f denote a function that is continuous on a simple closed contour C.

Prove that the function g (z) = 12�i

RCf(s)dss�z is analytic at each point z interior

to C and that g0 (z) = 12�i

RCf(s)ds

(s�z)2 at such a point.

Solution:In order to prove the analyticity of the function g(z) we need to demonstrate

that it is di¤erentiable. To this end we need to �nd the linear part of theincrement of the function g (z) : g (z + h)� g (z) for every two points z and z+hinterior to C. We have

1[s�(z+h)] �

1(s�z) =

h(s�z)[s�(z+h)]

= h(s�z)2 +

h2

(s�z)2[s�(z+h)] :

From the de�nition of function g(z) and from the formula above it followsthatg (z + h)� g (z) = 1

2�i

RC

f(s)ds[s�(z+h)] �

12�i

RC

f(s)ds(s�z)

= h

�12�i

RC

f(s)ds

(s�z)2

�+ h2

�12�i

RC

f(s)ds

(s�z)2[s�(z+h)]

�:

Now we have to prove that the second term is of order O�h2�: In other

words, we need to limit the second integral in the right-hand side by a constant.Since both points z and z+h are interior to C, there exists a constant � suchthat js� zj > �; js� (z + h)j > �: Besides, since f(z) is continuous on C it isbounded throughout C, i.e., there exists a constant M such that jf (z)j �M foreach point z 2 C: Hence, by the Mean Value Theorem�� 1

2�i

�� RC

f(s)ds

(s�z)2[s�(z+h)] �ML2��3 =constant.

Here L is the length of C. Thus, we obtain that the linear part of theincrement of the function g(z) exists and is given by g0 (z) = 1

2�i

RC

f(s)ds

(s�z)2 : Hence,

the proof that the function g(z) is analytical is complete, and its derivative isgiven by the required formula.

18

7. Let C be the unit circle z = ei� (�z � � � �). First show that, forany real constant a,

RC

eaz

z dz = 2�i: Then write this integral in terms of � to

derive the integration formula�R0

ea cos � cos (a sin �) d� = �:

Solution (7): Since eaz is analytic in C and 0 lies inside C, so we haveRC

eaz

z dz = 2�iea�0 = 2�i: From the de�nition of a line integral, we haveRC

eaz

z dz =�R��

exp(aei�)ei�

iei�d� = i�R��exp

�aei�

�d�: Thus

�R��exp

�aei�

�d� = 2�.

Since exp�aei�

�= exp (a cos � + ia sin �) =

ea cos � cos (a sin �) + iea cos � sin (a sin �) ; so�R��ea cos � cos (a sin �) d� =

Re

�R��exp

�aei�

�d�

!= 2�: Since ea cos(��) cos (a sin (��)) =

ea cos � cos (�a sin �) = ea cos � cos (a sin �), so0R

��ea cos � cos (a sin �) d� =

�R0

ea cos � cos (a sin �) d�: Thus�R0

ea cos � cos (a sin �) d� =

12

�R��ea cos � cos (a sin �) d� = �:

9. Follow the steps below to verify the expression f 00 (z) = 1�i

RC

f(s)ds

(s�z)3 in

the lemma in Section 48.(a) Use the expression for f�(z) in the lemma to show thatf 0(z+�z)�f 0(z)

�z � 1�i

RC

f(s)ds

(s�z)3 =12�i

RC

3(s�z)�z�2(�z)2(s�z��z)2(s�z)3 f (s) ds:

(b) Let D and d denote the largest and smallest distances, respectively,from z to points on C. Also, let M be the maximum value of jf (s)j on C and Lthe length of C. With the aid of the triangle inequality and by referring to thederivation of the expression for f�(z) in the lemma, show that when 0 < j�zj < d,the value of the integral on the right-hand side in part (a) is bounded from above

by (3Dj�zj+2j�zj2)M(d�j�zj)2d3 L:

(c) Use the results in parts (a) and (b) to obtain the desired expression forf�(z).

Solution: In view of the expression for f�in this lemma.f 0(z+�z)�f 0(z)

�z = 12�i

RC

h1

(s�z��z)2 �1

(s�z)2

if(s)ds�z

= 12�i

RC

h2(s�z)��z

(s�z��z)2(s�z)2

if (s) ds:

Thenf 0(z+�z)�f 0(z)

�z � 1�i

RC

f(z)

(s�z)3 =12�i

RC

h2(s�z)��z

(s�z��z)2(s�z)2 �2

(s�z)3

if (s) ds

= 12�i

RC

h3(s�z)�z�2(�z)2(s�z��z)2(s�z)3

if (s) ds:

19

b) The triangle inequality tells us that���3 (s� z)�z � 2 (�z)2��� � 3 js� zj j�zj+ 2 j�zj2 � 3D j�zj+ 2 j�zj2 : Wealso can see that js� z ��zj � d� j�zj > 0 (p.160, line 3). Hence,���(s� z ��z)2 (s� z)3��� � (d� j�zj)2 d3 > 0: Together we obtain����R

C

h3(s�z)�z�2(�z)2(s�z��z)2(s�z)3 f (s) ds

i���� � (3Dj�zj+2j�zj2)M(d�j�zj)2d3 :

c) If we let �z tend to 0 in this inequality, we �nd

lim�z!0

12�i

RC

h3(s�z)�z�2(�z)2(s�z��z)2(s�z)3

if (s) ds = 0:

This, together with the result in part a), yields the desired expression for f�.

Exercises Pages 171-1731. Let f be an entire function such that jf (z)j � A jzj for all z, where A is a

�xed positive number. Show that f (z) = a1z, where a1 is a complex constant.Suggestion: Use Cauchy�s inequality (Section 49) to show that the second

derivative f�(z) is zero everywhere in the plane. Note that the constant MR inCauchy�s inequality is less than or equal to A (jz9j+R) :

Solution: By Cauchy inequality we have jf 00 (z0)j � 2MR

R2 where MR is themaximum of jf (z)j on the circle of radius R centered at z0. In this case jzj �jz0j + R so MR � A (jz0j+R). Plugging that in we get jf 00 (z0)j � A(jz0j+R)

R2 .As R!1 the right hand side goes to 0 and this says that f 00 (z0) = 0. Sincethis is true for any z0 we have f 00 (z) = 0 for all z and this means f�is a constanta1. But this means that f (z) = a1z + b. Since jf (z)j � A jzj we have b1 = 0:

Solution: For R > 0, let CR denote the positively oriented circlefz 2 C : jzj = Rg : Fix z0 2 C: Then z0 lies inside CR if R > jzj : From

Cauchy�s formula, f 00 (z0) = 22�i

RCR

f(z)

(z�z0)3dz: For z 2 CR;

��� f(z)

(z�z0)3

��� � AR(R�jz0j)3

:

Thus jf 00 (z0)j � L(CR)2�

AR(R�jz0j)3

= AR2

(R�jz0j)3: This inequality holds for all R >

jz0j. Since AR2

(R�jz0j)3! 0 asR!1, so jf 00 (z0)j = 0, i.e., f 00 (z0) = 0: Since z0

is chosen from C arbitrarily, so f�(z) = 0 for all z 2 C: Thus f�(z) is constant inC. Let a1 be the constant and g (z) = f (z)�a1z. Then g�(z) = 0 for all z 2 C:Thus g(z) is a constant in C. Let the constant be C. Then f (z) = a1z+C forall z 2 C: From jf (0)j � A0 = 0 we get f(0) = 0. Thus C = 0 and f (z) = a1z:2. Suppose that f(z) is entire and that the harmonic function u (x; y) =

Re [f (z)] has an upper bound u0; that is, u (x; y) � u0 for all points (x; y) inthe xy plane. Show that u (x; y) must be constant throughout the plane.Suggestion: Apply Liouville�s theorem (Section 49) to the function g (z) =

exp [f (z)] :

Solution: Let g (z) = ef(z) = eu+iv. Since��ef(z)�� = eu and u is assumed

bounded, we have ef(z) a constant. But this means eu constant which meansthat u is a constant.

20

Solution: Let g(z) = exp(f(z)). Then g is entire and jg (z)j = exp (u (z)) �u0. From Liouville�s Theorem, g must be a constant. Thus u (z) = ln (jg (z)j)is also a constant throughout C.4. Let a function f be continuous in a closed bounded region R, and let

it be analytic and not constant throughout the interior of R. Assuming thatf (z) 6= 0 anywhere in R, prove that jf (z)j has a minimum value m in R whichoccurs on the boundary of R and never in the interior. Do this by applying thecorresponding result for maximum values (Section 50) to the function g (z) =1=f (z) :

Solution: Let g(z) = 1/f(z). Since f is continuous on R, analytic in theinterior of R and f (z) 6= 0 for all z 2 R, so g is continuous in R and analytic inthe interior of R. Since R is closed and bounded, and jgj is continuous, so jgjattain its maximum on R. That means there is z0 2 R such that jg (z)j � jg (z0)jfor all z 2 R: Thus jf (z)j � jf (z0)j for all z 2 R: If z0 2 int R then frommaximum modulus principle we have g is a constant in int R. Because g iscontinuous, so g is also constant in R. From f(z) = 1/g(z) we then have that fis constant in R which is a contradiction. Thus z0 must lie on the boundary ofR.6. Consider the function f (z) = (z + 1)2 and the closed triangular region

R with vertices at the points z = 0, z = 2, and z = i. Find points in R wherejf (z)j has its maximum and minimum values, thus illustrating results in Section50 and Exercise 4.Suggestion: Interpret jf (z)j as the square of the distance between z and -1.Solution: We must have jf (z)j taking its maximum on the boundary and

since it is never 0 also its minimum on the boundary. Since jf (z)j is the squareof the distance between z and -1 the maximum is taken on at the furthest pointfrom -1 which is at z = 2. The closest point is at z = 0.7. Let f (z) = u (x; y)+ iv (x; y) be a function that is continuous on a closed

bounded region R and analytic and not constant throughout the interior of R.Prove that the component function u (x; y) has a minimum value in R whichoccurs on the boundary of R and never in the interior. (See Exercise 4.)Solution???: Suppose u has a minimum at a point in the interior say at

u (z0). Let f (z) = ef(z) which is never 0. We have jf (z)j = eu. Theminimum of this is at z0. But this violates (4).10. Let z0 be a zero of the polynomial P (z) = a0 + a1z + a2z

2 + � � � +anz

n (an 6= 0) of degree n (n � 1). Show in the following way that P (z) =(z � z0)Q (z), where Q (z) is a polynomial of degree n-1.(a) Verify that zk�zk0 = (z � z0)

�zk�1 + zk�2z0 + � � �+ zzk�20 + zk�10

�(k = 2; 3; � � � :) :(b) Use the factorization in part (a) to show thatP (z) � P (z0) = (z � z0)Q (z), where Q(z) is a polynomial of degree n-1,

and deduct the desired results from this.

Solution (10a): We calculate the right-hand side of the equation:

21

(z � z0)�zk�1 + zk�2z0 + � � �+ zzk�20 + zk�10

�= zk+zk�1z0+� � �+z2zk�20 +

zzk�10 ��zk�1z0 + � � �+ zk0

�= zk � zk0 :

(10b) Now P (z)� P (z0) = a0 + � � �+ anzn � (a0 + � � �+ anzn0 )= (a� a0) + a1 (z � z0) + a2 (z � z0)2 + � � �+ an (z � z0)n

= (z � z0)ha1 + a2 (z � z0) + � � �+ an (z � z0)n�1

i= (z � z0)Q (z) :

22