Solutions to Exercises 3 - University of...

48 Chapter 3 Complex Integration

Solutions to Exercises 3.11. Given z1 = 1 + i and z2 = −1 − 2i, apply (2), Sec. 3.1, to obtain the parametrizationof the line segment [z1, z2]:

γ(t) = (1 − t)z1 + tz2 = (1− t)(1 + i) + t(−1 − 2i), 0 ≤ t ≤ 1,

orγ(t) = t(−2 − 3i) + 1 + i, 0 ≤ t ≤ 1.

5. Use Example 1(a) with α = −π4 and β = π

4 :

γ(t) = eit, −π4≤ t ≤ π

4.

9. Use Example 1(a) and (c):

γ(t) = z0 +Reit = −3 + 2i+ 5eit, −π2≤ t ≤ 0.

13. Write γ = (γ1, γ2), where γ1 is the circular arc and γ2 the line segment shown in Fig.14, but in reverse direction. To parametrize γ1, use Example 1(a):

γ1(t) = 2e−it, 0 ≤ t ≤ 3π2.

Note the negative sign in the exponent, whose effect is to trace the ar in the negativedirection. If we want to parametrize γ1 using the interval [0, 1], we can change t to 3π

2 t andget

γ1(t) = 2e−i 3π2

t, 0 ≤ t ≤ 1.

To parametrize γ2, use (2):

γ2(t) = (1 − t)2i+ t(5i) = 3it+ 2i, 0 ≤ t ≤ 1.

To parametrize γ2 by the interval [1, 2], we can simply change t to t− 1 and get

γ2(t) = 3i(t− 1) + 2i, 1 ≤ t ≤ 2.

So

γ(t) =

2e−i 3π

2t if 0 ≤ t ≤ 1

3i(t− 1) + 2i if 1 ≤ t ≤ 2.

17. We have x(t) = t and y(t) = sinπt, where 0 ≤ t ≤ 1. Thus y = sin πx, where0 ≤ x ≤ 1, and so the graph is the arch of sin πx for 0 ≤ x ≤ 1.

21. We differentiate using the rules of calculus and treating complex constants as if theywere real. We get

d

dt(2 + i) cos(3it) = (2 + i)

d

dtcos(3it) = −3i(2 + i) sin(3it) = (3 − 6i) sin(3it).

25. To find a particular solution yp of y′′ + y′ + y = cos t, we proceed as follows

Section 3.1 Complex Integration 49

1. Think of cos t as being the real pat of eit.

2. Find a particular solution of y′′+y′+y = eit and let Y denote this particular solution.

3. Compute yp = Re (Y ).

To find Y , try Y = Aeit, Y ′ = Aieit, Y ′′ = −Aeit. Plug into the equation and get

−Aeit + iAeit +Aeit = eit

−A + iA+ A = 1 (Divide by eit.)

iA = 1 ⇒ A = −i.

Hence Y = −ieit and so yp = Re (−ieit) = sin t.

29. Proceed as in Exercise 21: Let Y denote a particular solution of y′′ − 2y′ − 3y = e4it.Then Re (Y ) is a particular solution of y′′ − 2y′ − 3y = cos 4t, and Im (Y ) is a particularsolution of y′′−2y′−3y = sin 4t. So solving the equation with a complex exponential on theright side yields the solutions of two differential equations. Let us now solve y′′−2y′−3y =e4it. For this purpose, we let Y = Ae4it, Y ′ = 4iAe4it, and Y ′′ = −16Ae4it. Plugging intothe equation and solving, we find

−16Ae4it − 8iAe4it − 3Ae4it = e4it

−16A− 8iA− 3A = 1 (Divide by e4it.)

A(−19 − 8i) = 1 ⇒ A = 1−19−8i = − 19

425 + 8425i.

Hence Y = (− 19425 + 8

425i)e4it and so a particular solution of y′′ − 2y′ − 3y = cos 4t is

yp = Re ((− 19425

+8

425i)e4it)) = − 19

425cos 4t − 8

425sin 4t.

Also, a particular solution of y′′ − 2y′ − 3y = sin 4t is

yp = Im((− 19425

+8

425i)e4it)) =

8425

cos 4t− 19425

sin 4t.

33. We have x(t) = a cos t− b cos at2 and y(t) = a sin t − b sin at

2 . So

γ(t) = x(t) + y(t)

= a cos t− b cosat

2+ ai sin t− ib sin

at

2

= a(cos t+ i sin t) − b(cosat

2+ i sin

at

2)

= aeit − beiat2


Solutions to Exercises 3.21. Using (10),

∫ 2π

0e3ix dx =

13ie3ix

∣∣∣∣2π

0

=13i

(=1︷︸︸︷e6iπ −1

)= 0.

5. Write

x+ i

x− i=

x+ i

x− i× x + i

x + i

=x2 + 2ix− 1

x2 + 1=

(x2 + 1) + 2ix− 2x2 + 1

= 1 − 2x2 + 1

+2ixx2 + 1

.

So

∫ 1

−1

x+ i

x− idx =

∫ 1

−11 dx−

∫ 1

−1

2x2 + 1

dx+

=0, odd integrand︷︸︸︷2i

∫ 1

−1

x

x2 + 1dx

= x− 2 tan−1 x∣∣∣1

−1= 2− π.

9. Proceed as in Example 2:

f(x) =

(3 + 2i)x if − 1 ≤ x ≤ 0,

ix2 if 0 ≤ x ≤ 1;

hence an antiderivative

F (x) =

3+2i

2 x2 + C if − 1 ≤ x ≤ 0,i3x

3 if 0 ≤ x ≤ 1.

Setting F (0+) = F (0−), we obtain 0 = 3+2i2 0 + C or C = 0. Hence a continuous antider-

vative of f is

F (x) =

3+2i

2 x2 if − 1 ≤ x ≤ 0,i3x

3 if 0 ≤ x ≤ 1.

13. (a) Let p > 0 be an arbitrary real number, and m and n be arbitrary integers. Writef(x) = ei

mπp

x. Then

f(x+ 2p) = eimπp

(x+2p) = eimπp

x

=1︷︸︸︷ei2mπ = ei

mπp

x = f(x).


Hence f is 2p-periodic.(b) If m 6= n, then

∫ p

−pei mπ

pxei−nπ

pxdx =

∫ p

−pei(m−n)π

pxdx

= − i

p(m− n)ei (m−n)π

px

∣∣∣∣p

−p

= − i

p(m− n)

(ei(m−n)π − e−i(m−n)π

)

= − i

p(m− n)e−i(m−n)π

( =1︷︸︸︷ei2(m−n)π −1

)= 0

If m = n, then∫ p

−p

eimπp

xei−nπ

pxdx =

∫ p

−p

eimπp

xei−mπ

pxdx =

∫ p

−p

dx = 2p.

We have thus established the orthogonlaity relations∫ p

−pei mπ

pxe−i nπ

pxdx =

0 if m 6= n,2p if m = n.

(c) Now suppose m and n are nonnegative integers. If m 6= n, then the equality∫ p

−pei mπ

pxe−i nπ

pxdx = 0

implies that ∫ p

−p

(cos

mπ

px+ i sin

mπ

px)(

cosnπ

px− i sin

nπ

px)dx = 0.

Expanding and then taking real and imaginary parts, we get

(1)∫ p

−p

(cos

mπ

px cos

nπ

px+ sin

mπ

px sin

nπ

px)dx = 0

and

(2)∫ p

−p

(− cos

mπ

px sin

nπ

px+ sin

mπ

px cos

nπ

px)dx = 0.

Replacing n by −n; that is, starting with the identity∫ p

−p

eimπp

xeinπp

x dx = 0,

we get

(3)∫ p

−p

(cos

mπ

px cos

nπ

px− sin

mπ

px sin

nπ

px)dx = 0


and

(4)∫ p

−p

(cos

mπ

px sin

nπ

px+ sin

mπ

px cos

nπ

px)dx = 0.

Yu can now obtain some of the desired integral identities by using linear combinations of(1)–(4). For example, adding (1) and (3) implies that

∫ p

−pcos

mπ

px cos

nπ

px dx = 0.

Adding (2) and (4) implies that∫ p

−pcos

mπ

px sin

nπ

px dx = 0.

All other integral identities with m 6= n follow similarly. Consider the case m = −n 6= 0,∫ p

−pei mπ

pxei mπ

pxdx = 0.

Writing the integrand in terms of cosine and sine and then taking real and imaginary parts,we get

(5)∫ p

−p

(cos2

mπ

px− sin2 mπ

px)dx = 0

Similarly, from the identity ∫ p

−pei mπ

pxei−mπ

pxdx = 2p,

we obtain

(6)∫ p

−p

(cos2

mπ

px+ sin2 mπ

px)dx = 2p.

Adding (5) and (6) shows that∫ p

−pcos2

(mπ

px

)dx = p.

The remaining identities follow similarly.

17. WriteI =

∫

[z1, z2 , z3 ]2z dz =

∫

[z1, z2]2z dz +

∫

[z2, z3 ]2z dz = I1 + I2,

where z1 = 1, z2 = i, z3 = 1+i. To evaluate I1, parametrize [z1, z2] by z = γ1(t) = (1−t)+it,0 ≤ t ≤ 1, dz = (−1 + i) dt. So

∫

[z1 , z2 ]2z dz = 2

∫ 1

0

((1 − t) − it

)(−1 + i) dt = 2(−1 + i)

∫ 1

0

(1 − ((1 + i)t

)dt

= 2(−1 + i)(t − (

1 + i

2t2

)∣∣∣1

0= 2i.


To evaluate I2, parametrize [z2, z3] by z = γ2(t) = (1 − t)i + t(1 + i) = i + t, 0 ≤ t ≤ 1,dz = dt. So ∫

[z2 , z3 ]2z dz = 2

∫ 1

0(t− i) dt = 2(

t

2− it)

∣∣∣1

0= 1 − 2i.

Thus I = 1.

21. Write

I =∫

[z1, z2 , z3 , z4, z1 ]

z dz

=∫

[z1, z2 ]

z dz +∫

[z2 , z3]

z dz +∫

[z3 , z4 ]

z dz +∫

[z4, z1 ]

z dz

= I1 + I2 + I3 + I4,

where z1 = 0, z2 = 1, z3 = 1 + i, and z4 = i. To evaluate I1, parametrize [z1, z2] byz = γ1(t) = t, 0 ≤ t ≤ 1, dz = dt. So

I1 =∫

[z1 , z2 ]z dz =

∫ 1

0t dt =

12.

To evaluate I2, parametrize [z2, z3] by z = γ2(t) = 1 + it, 0 ≤ t ≤ 1, dz = i dt. So

I2 =∫

[z2 , z3 ]z dz =

∫ 1

0(1 + it)i dt = i(t+

i

2t2)

∣∣∣1

0= i(1 +

i

2) = −1

2+ i.

To evaluate I3, parametrize [z3, z4] by z = γ3(t) = (1− t) + i, 0 ≤ t ≤ 1, dz = −dt. So

I3 =∫

[z3 , z4]z dz = −

∫ 1

0((1− t) + i) dt = −(1 + i) +

12

= −12− i.

To evaluate I4, parametrize [z4, z1] by z = γ4(t) = (1− t)i, 0 ≤ t ≤ 1, dz = −idt. So

I4 =∫

[z4 , z1]z dz = −i

∫ 1

0i(1− t) dt = 1 − 1

2=

12.

Finally, adding the four integrals, we obtain I = 0.Note: When you will learn about Cauchy’s Theorem in Section 3.4, or the main result ofthe next section, you will realize that all our work in this solution is unnecessary to concludethat the answer is 0.

25. In our computation, we will suppose that a is an integer 6= 0. Then the curve is tracedin its entirety if we take 0 ≤ t ≤ 4π. We apply the definition of the path integral, withγ(t) = aeit + be−i a

2t, 0 ≤ t ≤ 4π, dz = (aieit − iab

2 e−i a

2t)dt:

∫

γz dz =

∫ 4π

0(aeit + be−i a

2t)(aieit − i

ab

2e−i a

2t) dt

=∫ 4π

0(ia2e2it − i

ab2

2e−iat + iabei(1−

a2)t − i

a2b

2ei(1−

a2)t) dt = 0,


because the integrand is a linear combination of exponential functions with period 4π each.Thus the integal of each exponential fnction is 0.

29. We have f(z) = zz, γ(t) = (1 − t)(2 + i) + t(−1 − i) = (−3 − 2i)t+ 2 + i, 0 ≤ t ≤ 1,dz = (−3 − 2i)dt. So∫

[z1 , z2 ]

(x2 + y2) dz =∫ 1

0

((−3 − 2i)t+ 2 + i)((−3 + 2i)t+ 2 − i)(−3− 2i)dt

= (−3 − 2i)∫ 1

0(13t2 + (−3 − 2i)(2− i)t+ (2 + i)(−3 + 2i)t+ 5)dt

= (−3 − 2i)(133

+(−3 − 2i)(2− i)

2+

(2 + i)(−3 + 2i)2

+ 5) = −4 − 83i.

33. We have γ(t) = (et − t) + 4iet2 , γ ′(t) = (et − 1) + 2ie

t2 ,

|γ ′(t)| =∣∣(et − 1) + 2ie

t2

∣∣ =√

(et − 1)2 + (2et2 )2

=√e2t + 1 + 2et =

√(et + 1)2 = et + 1.

So

L =∫ 1

0|γ ′(t)| dt =

∫ 1

0(et + 1) dt = (et + t)

∣∣∣1

0= e.

37. We apply the ML-inequality (29). We have L = l([z1, z2, z3, z1]) = 2 + 2√

5, as canbe easily verified by plotting the points z1, z2, and z3. Now |z−5| = |z|−5, and for z on thepath [z1, z2, z3, z1], we have |z| ≤ 2

12 . So |z|−5 ≤ 25 for z on the path [z1, z2, z3, z1]. Thus,

by the ML-inequality,∣∣∣∣∣

∫

[z1, z2 , z3 , z1 ]z−5 dz

∣∣∣∣∣ ≤ 2−52 (2 + 2

√5).

41. We have∫

Γf(z) dz =

∫

(γ,−γ)f(z) dz

=∫

γf(z) dz +

∫

−γf(z) dz

=∫

γf(z) dz −

∫

γf(z) dz = 0.

Section 3.3 Independence of Path 55

Solutions to Exercises 3.3

1. An antiderivative of f(z) = z2+z−1 is simply F (z) = 13z

3+ 12z

2−z+C where C is an arbitrarycomplex constant. The function F is entire and so we can take Ω = C.

5. To find an antiderivative of 1(z−1)(z+1)

, we proceed as we would have done in calculus. Usingpartial fractions, write

1(z − 1)(z + 1)

=A

z − 1+

B

z + 11

(z − 1)(z + 1)=

A(z + 1) + B(z − 1)(z − 1)(z + 1)

1 = A(z + 1) + B(z − 1).

Taking z = −1, it follows that B = −12 . Taking z = 1, it follows that A = 1

2 . Hence

1(z − 1)(z + 1)

=1

2(z − 1)− 1

2(z + 1).

An antiderivative of this function is

F (z) =12(Log (z − 1) − Log (z + 1)

)+ C,

where C is an arbitrary complex constant. (You could also use a different branch of the logarithm.)The function Log (z − 1) is analytic in C \ (−∞, 1], while the function Log (z + 1) is analytic inC\(−∞,−1]. So the function F (z) = 1

2

(Log (z−1)− Log (z+1)

)+C, is analytic in C\(−∞, 1] and

we may take Ω = C\(−∞, 1]. In fact, the function Log (z−1)− Log (z+1) = Log z−1z+1 is analytic in

a larger region C \ [−1, 1]. There are at least two possible ways to see this. One way is to note thatthe linear fractional transformation w = z−1

z+1takes the interval [−1, 1] onto the half-line (−∞, 0].

All other values of z outside the interval [−1, 1] are mapped into C\ (−∞, 0] and so the compositionLog z−1

z+1 is analytic everywhere on C \ [−1, 1]. Another way to show that Log (z − 1)− Log (z + 1)is analytic in C \ [−1, 1] is to use Theorem 4, Sec. 2.3. Let g(z) = Log (z − 1) − Log (z + 1) andf(z) = ez. The function g(z) is continuous on C \ [−1, 1], because the discontinuities of Log (z − 1)and Log (z + 1) cancel on (−∞, −1). The function f(z) = ez is obviously entire. The compositionf(g(z)) is equal to the function z−1

z+1, which is analytic except at the points z = ±1. According to

Theorem 4, Sec. 2.3, the function g(z) is analytic in C \ [−1, 1].

9. An antiderivative of z sinh z2 is 12 cosh z2 + C, as you can verify by differentiation. The an-

tiderivative is valid for all z.

13. All branches of the logarithm, logα z, have the same derivative 1z . The only difference is that

the domain of analyticity differs in each case, since the branch cut is different. Again, taking a hintfrom calculus and the fact that an antiderivative of lnx is −x+x lnx+C (as you can verify directly orintegrate lnx by parts), we find that an antiderivative of f1(z) = log0 z is F1(z) = −z+ z log0 z+C.The functions f1 and F1 are analytic in the region Ω1 = C \ [0, ∞). Similarly, an antiderivative off2(z) = logπ

2z is F2(z) = −z + z logπ

2z + C. The functions f2 and F2 are analytic in the region

Ω2 = C minus the ray at angle π2. So, an antiderivative of the function log0 z + logπ

2z + 1

zis

−2z + z log0 z + z logπ2z + Log z +C, and the antiderivative is good in the region C minus the rays

at angle 0, π2 , and π, or, simply, C minus the real line and the ray at angle π

2 .

17. The integrand is entire and has an antiderivative in a region containing the path. So, by


Theorem 1,∫

γ

z2 dz =13z3

∣∣∣∣γ( π

4 )

γ(0)

=13((ei π

4 + 3e2iπ4 )3 − (ei·0 + 3e2i·0)3

)

=13

(ei 3π

4 + 9ei 2π4 ei π

2 + 27eiπ4 e2i π

2 + 27ei3π2 − (4)3

)

=13

(−√

22

+ i

√2

2− 9 − 27(

√2

2+ i

√2

2) − 27i− (4)3

)

=13

((−14

√2 − 73 + i(−13

√2 − 27)

)

21. Evaluate using Theorem 1:∫

γ

sin z dz = − cos z∣∣∣γ( π

2 )

γ(0)

= cos(2ei·0) − cos(2ei π2

= cos 2 − cos(2i) = cos 2 − cosh 2.

(Use (25), Sec. 1.6.)

25. The function Log z has an analytic antiderivative in a region that contains the path [z1, z2, z3, z1];for example, the function F (z) = z Log z − z. So by Theorem 1,

∫

[z1, z2, z3, z1]

z Log z dz = 0.

29. Part (a) is clear. For (b), the path γ1 lies completely in a region on which the function Log zis analytic. Since Log z is an antiderivative of 1

z , it follows from Theorem 1 that∫

γ1

1zdz = Log (z1) − Log (z2).

Similarly, the path γ2 lies completely in a region on which the function log0 z is analytic. Sincelog0 z is an antiderivative of 1

z, it follows from Theorem 1 that and

∫

γ2

1zdz = log0(z2) − log0(z1).

Now ∫

CR(z0)

1zdz =

∫

γ1

1zdz +

∫

γ2

1zdz = Log z1 − Log z2 + log0 z2 − log0 z1.

Recall that z1 is on the positive y-axis, so Log z1 = log0 z1. Also z2 is on the negative y-axis, soarg 0z2 = 3π

2 and Arg z2 = −π2 . Hence log0 z2 − Log z2 = 2πi. Putting this information together,

we get ∫

CR(z0)

1zdz = 2πi.

33. If f is analytic in a region Ω, then obviously f is an antiderivative of f ′ in Ω. Let z0 and z1 beany two points in Ω. Since Ω is connected, we can join z0 to z1 by a polygonal path γ, with initialpoint z0 and terminal point z1. By Theorem 1,

∫

γ

f ′(z) dz = f(z1) − f(z0).

But f ′(z) = 0, so f(z1) − f(z0) = 0 implying that f(z0) = f(z1). Since z1 is arbitrary, it followsthat f is constant equal to f(z0).

Section 3.4 Cauchy’s Integral Theorem 57

Solutions to Exercises 3.4

1. Each path is continuously deformable to a point in Ω and Ω is connected. So the two paths aremutually continuously deformable.

5. Same reasoning as in Exercise 1.

9. The path in Figure 35 is an ellipse centered at 3i with length of major axis 6 and length ofminor axis 3. The ellipse can be parametrized by

γ0(t) = 3i+ 3 cos t+ i sin t, 0 ≤ t ≤ 2π.

To parametrize γ0 by the interval [0, 1], we can use

γ0(t) = 3i+ 3 cos(2πt) + i sin(2πt) = 3 cos(2πt) + i(3 + sin(2πt)

), 0 ≤ t ≤ 1.

It is clear from the figure that we can deform γ0 continuously to 3i, its center (there are, of course,many other possible points). Since the interior of the ellipse is convex, we can use (8) to constructthe desired mapping of the unit square. We take γ0 as described above and γ1(t) = 3i. Then, for0 ≤ t ≤ 1 and 0 ≤ s ≤ 1:

H(t, s) = (1 − s)γ0(t) + sγ1(t)= 3(1 − s) cos(2πt) + i(1 − s)

(3 + sin(2πt)

)+ 3is

= 3(1 − s) cos(2πt) + i(3 + (1 − s) sin(2πt)

).

13. The function f(z) = ez

z+2 is analytic inside an on the simple path C1(0). By Theorem 5,

∫

C1(0)

ez

z + 2dz = 0.

17. The function f(z) = ez

z+i has a problem at z = −i. Hence it is analytic inside an on the simplepath γ(t) = i + eit, 0 ≤ t ≤ 2π (circle, centered at i with radius 1). By Theorem 5,

∫

γ

ez

z + idz = 0.

21. The path [z1, z2, z3, z1] is contained in a region that does not intersect the branch cut of Log z.Hence the function f(z) = z2 Log z is analytic inside an on the simple path [z1, z2, z3, z1], and soby Theorem 5, ∫

[z1, z2, z3, z1]

z2 Log z dz = 0.

25. The path C2(0) contains both roots of the polynomial z2 − 1. We will evaluate the integral byusing the method of Example 5. We have

z

z2 − 1=

A

z − 1+

B

z + 1⇒ z = A(z + 1) + B(z − 1).

Setting z = 1, we get 1 = 2A or A = 12 . Setting z = −1, we get −1 = −2B or B = 1

2 . Hence

z

z2 − 1=

12(z − 1)

+1

2(z + 1),


and so ∫

C2(0)

z

z2 − 1dz =

12

∫

C2(0)

1z − 1

dz +12

∫

C2(0)

1z + 1

dz =122πi +

122πi = 2πi,

where we have applied the result of Example 4 in evaluating the integrals.

29. Write γ = (γ1, γ2), where γ1 is the circle centered at i and γ2 is the circle centered at −1. Then∫

γ

1(z + 1)2(z2 + 1)

dz =∫

γ1

1(z + 1)2(z2 + 1)

dz +∫

γ2

1(z + 1)2(z2 + 1)

dz = I1 + I2.

The partial fraction decomposition of the integrand is

1(z + 1)2(z2 + 1)

=1

2(z + 1)+

12(z + 1)2

− 14(z + i)

− 14(z − i)

.

We have

I1 =12

∫

γ1

dz

z + 1+

12

∫

γ1

dz

(z + 1)2− 1

4

∫

γ1

dz

z + i− 1

4

∫

γ1

dz

z − i

=12· 0 +

12· 0 − 1

4· 0 − 1

4· 2πi = −π

2i,

where the first 3 integrals are 0 because the integrands are analytic inside an on γ1, and the fourthintegral follows from Example 4. Similarly,

I2 =12

∫

γ2

dz

z + 1+

12

∫

γ2

dz

(z + 1)2− 1

4

∫

γ2

dz

z + i− 1

4

∫

γ2

dz

z − i

=12· 2πi+ 1

2· 0 − 1

4· 0 − 1

4· 0 = πi,

where the first, third, and fourth integrals follow from Example 4, and the second integral followsfrom Example 4, Sec. 3.2. Thus, the desired integral is equal to

I1 + I2 =π

2i.

33. (a) Let z1, z2, . . . , zn be distinct complex numbers (n ≥ 2). Reducing to the common denomi-nator in the partial fraction decomposition

1(z − z1)(z − z2) · · · (z − zn)

=A1

z − z1+

A2

z − z2+ · · ·+ An

z − zn,

we obtain1

p(z)=

A1(z − z2) · · · (z − zn)p(z)

+A2(z − z1)(z − z3) · · · (z − zn)

p(z)+

· · ·+ An(z − z1)(z − z2) · · · (z − zn1)p(z)

,

where p(z) = (z − z1)(z − z2) · · · (z − zn).. From this expression, we see clearly that the coefficientof zn−1 in the numerator on the right side is A1 + A2 + · · ·+ An. Comparing with the left side, wesee that A1 + A2 + · · ·+An = 0.(b) Suppose that C is a simple closed path that contains the points z1, z2, . . . , zn in its interior.Using the partial fraction decomposition, we have

∫

C

1(z − z1)(z − z2) · · · (z − zn)

dz = A1

∫

C

dz

z − z1+ A2

∫

C

dz

z − z2+ · · ·+An

∫

C

dz

z − zn

= 2πi(A1 + A2 + · · ·+An) = 0,

where we have used Example 4 to evaluate the integrals and part (a) to sum the Aj’s.

Section 3.5 Proof of Cauchy’s Theorem 59

Solutions to Exercises 3.51. An example of a closed and bounded subset K of the real line and a bounded subsetS of the real line such that K and S are disjoint but the distance from K to S is 0: LetK = [0, 1] and S = (1, 2). Then K ∩ S = ∅ but the distance from K to S is 0.

5. The distance formula: for z1 = a1 + ib1, z2 = a2 + ib2, w1 = c1 + id1, and w2 = c2 + id2:∣∣(z1, w1) − (z2, w2)

∣∣ =√

(a1 − a2)2 + (b1 − b2)2 + (c1 − c2)2 + (d1 − d2)2.

(a) If z = a + ib and w = c+ id, then by definition

|(z, w)| =∣∣(z, w)− (0, 0)

∣∣ =√a2 + b2 + c2 + d2.

It is clear then that |(z, w)| ≥ 0 and

(z, w) = (0, 0) ⇔ a2 + b2 + c2 + d2 = 0 ⇔ a = b = c = d = 0.

(b) For any complex number α, write α = A+ iB, where A and B are real. Then

|α(z, w)| = |(αz, αw)| =∣∣(Aa−Bb + i(Ab+Ba), Ac−Bd + i(Ad+ Bc)

)∣∣

=√

(Aa− Bb)2 + (Ab+Ba)2 + (Ac− Bd)2 + (Ad+Bc)2

=√

(A2 + B2)(a2 + b2 + c2 + d2)

=√A2 +B2

√a2 + b2 + c2 + d2

= |α| |(z, w)|.

(c) We will prove the triangle inequality: |(z1, w1) + (z2, w2)| ≤ |(z1, w1)| + |(z2, w2)| byfollowing the proof of the triangle inequality for the usual absolute value (see Sec. 1.2).First note that from the definition of the absolute value on C × C, we have

|(z, w)|2 = |z|2 + |w|2,

where on the right side we are using the usual absolute value on C. Also note that forcomplex numbers z1 = a1 + ib1, z2 = a2 + ib2, w1 = c1 + id1, and w2 = c2 + id2, we have

|z1| |z2|+ |w1| |w2| ≤√

(|z1|2 + |w1|2)2 · (|z2|2 + |w2|2)2,

as can be verified by squaring both sides. We can now prove the desired identity:

|(z1, w1) + (z2, w2)|2 = |(z1 + z2, w1 + w2)|2 = |z1 + z2|2 + |w1 + w2|2

≤ (|z1|+ |z2|)2 + (|w1| + |w2|)2

= |z1|2 + |z2|2 + 2|z1| |z2|2 + |w1|2 + |w2|2 + 2|w1| |w2|2

≤ |z1|2 + |w1|2 + |z2|2 + |w2|2 + 2√

(|z1|2 + |w1|2)2 · (|z2|2 + |w2|2)2

=(

|z1|2 + |w1|21

2 +|z2|2 + |w2|2

12

)2

.

The desired inequality follows upon taking square roots on both sides.


9. Pick y0 in the interval (c, d). We have to sow that

limh→0

1h

∫ b

a

(u(x, y0 + h) − u(x, y0)

)dx =

∫ b

a

∂

∂yu(x, y0) dx.

Equivalently, we have to show that the difference∣∣∣∣1h

∫ b

a

(u(x, y0 + h) − u(x, y0)

)dx−

∫ b

a

∂

∂yu(x, y0) dx

∣∣∣∣

can be made arbitrarily small for all small values of h. Since ∂∂yu is continuous in R, it is

uniformly continuous. So, given ε > 0, we can find a δ > 0 such that if 0 < h < δ, then∣∣∣∣∂

∂yu(x, s) − ∂

∂yu(x, y0)

∣∣∣∣ < ε for all y0 < s < y0 + h.

Now using the fact that

u(x, y0 + h) − u(x, y0) =∫ y0+h

y0

∂

∂yu(x, s) ds,

we find that for 0 < h < δ

∣∣∣∣1h

∫ b

a

(u(x, y0 + h) − u(x, y0)

)dx−

∫ b

a

∂

∂yu(x, y0) dx

∣∣∣∣

=∣∣∣∣1h

∫ b

a

∫ y0+h

y0

∂

∂yu(x, s) ds dx−

∫ b

a

∂

∂yu(x, y0) dx

∣∣∣∣

≤∫ b

a

1h

∫ y0+h

y0

<ε︷︸︸︷∣∣∣∣∂

∂yu(x, s) − ∂

∂yu(x, y0)

∣∣∣∣ ds dx

≤ 1h

∫ b

a

∫ y0+h

y0

ε ds dx

= ε(b− a).

Since ε is arbitrary, the proof is complee.

Section 3.6 Cauchy’s Integral Formula 61

Solutions to Exercises 3.61. Apply Cauchy’s formula with f(z) = cos z at z = 0. Then

∫

C1(0)

cos zz

dz =∫

C1(0)

cos zz − 0

dz = 2πif(0) = 2πi.

5. Apply Cauchy’s formula with f(z) = −Log z at z = i. Then∫

C 12(i)

Log z−z + i

dz =∫

C 12(i)

−Log zz − i

dz = −2πiLog i = −2πi(ln 1 + iπ

2) = π2.

9. Apply the generalized Cauchy formula (6), with f(z) = sin z at z = π, with n = 2.Then ∫

γ

sin z(z − π)3

dz =2πi2!f (2)(π) = πi(− sinπ) = 0.

13. Follow the solution in Example 2. Draw small nointersecting negatively oriented circlesinside γ, γ1 centered at 0 and γ2 centered at i. Then

∫

γ

z + cos(πz)z(z2 + 1)

dz =∫

γ1


dz +∫

γ2


dz = I1 + I2.

Apply Cauchy’s formula with f(z) = z+cos(πz)z2+1

at z = 0. Then (recall γ1 is negativelyoriented)

I1 =∫

γ1


dz = −2πif(0) = −2πi0 + cos 002 + 1

= −2πi.

Apply Cauchy’s formula with f(z) = z+cos(πz)z(z+i) at z = i. Then

I2 =∫

γ2


dz =∫

γ2

z + cos(πz)z(z + i)(z − i)

dz

= −2πif(i) = −2πii+ cosπii(2i)

= +πi(i+ cosh π).

So I1 + I2 = −π − iπ(2− cosh π).

17. Factor the denominator as z3−3z+2 = (z+2)(z−1)2. Apply the generalized Cauchyformula (6), with f(z) = 1

z+2 at z = 1, with n = 1. Then∫

C 32(0)

dz

(z + 2)(z − 1)2= 2πif ′(1) = 2πi

−132

= −2πi9.

21. (a) For |z| < 1, let F (z) = 12π

∫ 2π0

eit

eit−zdt (we changed the notation of the exercise).

We will apply Theorem 4, Sec. 3.5, with f(z, t) = eit

eit−z, where t is on the path [0, 2π] and

z in the region |z| < 1. Because |eit| = 1 and |z| < 1, it follows that the function eit

eit−zis

continuous for all z and t in their respective domains. Moreover, we have

d

dzf(z, t) =

eit

(eit − z)2,


which is also continuous for all z and t, in their respective domains, for the same reason.By Theorem 4, Sec. 3.5, the function F (z) is analytic for all |z| < 1.(b) Let eit = ζ, ieitdt = dζ or dt = dζ

iζ . As t runs through [0, 2π], ζ runs through the unitcircle in the positive direction. Hence

F (z) =12π

∫

C1(0)

ζ

ζ − z

dζ

iζ=

12πi

∫

C1(0)

dζ

ζ − z= 1.

25. Define F (z) =∫ 10 cos(zt) dt. Let f(z, t) = cos(zt). Repeat the argument in Exercise

21(a) to see that F (z) is entire. If z = 0, F (z) =∫ 10 dt = 1. For z 6= 0, we have

F (z) =∫ 1

0cos zt dt =

1z

sin zt∣∣∣1

t=0=

sin zz.

Thus

F (z) =

1 if z = 0,sin z

z if z 6= 0,

is an entire function.

29. For z inside C, by Cauchy’s formula,

f(z) =1

2πi

∫

C

f(ζ)ζ − z

dζ.

But f(ζ) = g(ζ) for ζ on C. So, for all z inside C,

f(z) =1

2πi

∫

C

f(ζ)ζ − z

dζ =1

2πi

∫

C

g(ζ)ζ − z

dζ = g(z),

by Cauchy’s formula applied to g.

33. (a) Since our proof depends heavily on Exercise 32, let us outline the proof of part(a) of that exercise. To prove that

f(z) − f(z0)z − z0

− f ′(z0) =1

2πi

∫

CR(z0)

(z − z0)(ζ − z)(ζ − z0)2

f(ζ) dζ,

note that the integrand on the right side is equal to

f(ζ)[

1(z − z0)(ζ − z)

− 1(ζ − z0)2

− 1(z − z0)(ζ − z0)

]

Integrating the partial fraction and using Cauchy’s formula, we obatin

(z − z0)(ζ − z)(ζ − z0)2

f(ζ)

=1

2πi

∫

CR(z0)

f(ζ)1

(z − z0)(ζ − z)dζ +

12πi

∫

CR(z0)

f(ζ) − 1(ζ − z0)2

dζ

− 12πi

∫

CR(z0)

f(ζ)1

(z − z0)(ζ − z0)dζ

=f(z)z − z0

− f ′(z0) −f(z0)z − z0

,

Section 3.6 Cauchy’s Integral Formula 63

which yields the desired identity.In the preceding identity, replace f(z) by f(z, w), and get

f(z, w)− f(z0, w)z − z0

− d

dz(z0, w) =

12πi

∫

CR(z0)

(z − z0)(ζ − z)(ζ − z0)2

f(ζ, w) dζ.

Integrate the identity over γ and get∫γ f(z, w) dw−

∫γ f(z0, w) dw

z − z0−

∫

γ

d

dz(z0, w) dw

=1

2πi

∫

γ

∫

CR(z0)

(z − z0)(ζ − z)(ζ − z0)2

f(ζ, w) dζdw,

or

F (z) − F (z0)z − z0

−∫

γ

d

dz(z0, w) dw =

12πi

∫

γ

∫

CR(z0)

(z − z0)(ζ − z)(ζ − z0)2

f(ζ, w) dζdw,

So∣∣∣∣F (z) − F (z0)

z − z0−

∫

γ

d

dz(z0, w) dw

∣∣∣∣ =

∣∣∣∣∣1

2πi

∫

γ

∫

CR(z0)

(z − z0)(ζ − z)(ζ − z0)2

f(ζ, w) dζdw

∣∣∣∣∣≤ A · l(γ),

where A is the maximum value of∣∣∣∣∣

12πi

∫

CR(z0)

(z − z0)(ζ − z)(ζ − z0)2

f(ζ, w) dζ

∣∣∣∣∣ .

According to Exercise 32(c), A ≤ 2MR2 |z − z0|, which implies

∣∣∣∣F (z) − F (z0)

z − z0−

∫

γ

d

dzf(z0, ζ) dζ

∣∣∣∣ ≤2MR2

|z − z0| l(γ).

(b) Letting z → z0 in (a), the right side tends to 0, and the left side tends to F ′(z0) −∫γ

ddz f(z0, ζ) dζ. Hence

F ′(z0) =∫

γ

d

dzf(z0, ζ) dζ.

37. We will show the following version of Morera’s theorem. Suppose that f is continuousand complex-valued in Ω such that for every point z0 in Ω, there is a disk BR(z0) containedin Ω such that ∫

γf(z) dz = 0

for every closed triangular path contained in BR(z0). This is a stronger version than theone in the text, since we do assume that

∫γ f(z) dz = 0 for all closed paths γ in Ω.

(a) It is enough to show that, given any point z0 in Ω, there is is a disk BR(z0) containedin Ω such that f has a analytic antiderivative F (z) in BR(z0). Indeed, if F is analytic and


F”(z) = f(z) for all z in BR(z0), it follows from Corollary 1 that f is analytic in BR(z0),and since this holds for all z0 in Ω, it follows that f is analytic in Ω.(b) Given z0, pick a disk BR(z0) such that integrals over triangular paths in BR(z0) are 0.To show that f has an analytic antiderivative in BR(z0), for z in BR(z0), define F (z) =∫[z0 , z] f(ζ) dζ. Then F is analytc in BR(z0) by There 4, Sec. 3.5. We no show thatF ′(z) = f(z). We have

F (z + ∆z) − F (z)∆z

− f(z) =1

∆z

[∫

[z0, z+∆z]

f(ζ) dζ −∫

[z0, z]

f(ζ) dζ

]

=1

∆z

[∫

[z0, z+∆z]

f(ζ) dζ +∫

[z, z0 ]

f(ζ) dζ

]

=1

∆z

∫

[z, z+∆z]f(ζ) dζ.

In the last equality, we used the fact that the integral over a triangular path in BR(z0) is 0.(c) Using Lemma 1, Section 3.3, we find that the limit in (b) as ∆z → 0 is f(z). HenceF ′(z) = f(z), as desired.

41. If f and g are analytic at z0, such that f(z0) = f ′(z0) = · · · = f (m−1)(z0) = 0 andg(z0) = g′(z0) = · · · = g(m−1)(z0) = 0, but g(m)(z0) 6= 0, then according to Exercise 40, wecan write f(z) = (z− z0)mφ(z) and g(z) = (z− z0)mψ(z) where φ and ψ are analytic at z0,φ(z0) = f (m)(z0)

m! , and ψ(z0) = g(m)(z0)m! . Then

limz→z0

f(z)g(z)

= limz→z0

(z − z0)mφ(z)(z − z0)mψ(z)

= limz→z0

φ(z)ψ(z)

=f (m)(z0)g(m)(z0)

.

Section 3.7 Bounds for Moduli of Analytic Functions 65

Solutions to Exercises 3.71. We have |f(z)| = |z|. Obviously, if z belongs to the unit disk, |z| ≤ 1, then the largest valueof |f(z)| is 1 and its smallest value is 0. Hence |f(z)| attains its maximum value at all points ofthe boundary, and it attains its minimum value at the point z = 0, inside the region. The fact thatthe minimum is attained inside the region does not contradict Corollary 3, because f(z) vanishes atz = 0 inside of Ω¿

5. The function f(z) = zz2+2 is continuous for all z such that 2 ≤ |z| ≤ 3 and does not vanish inside

this annular region. Thus according to Corollary 3, |f | attains its maximum and minimum at theboundary; that it at points where |z| = 2 or |z| = 3. Using the triangle inequality, we have

|z2 + 2| ≤ |z|2 + 2 ⇒ 1|z|2 + 2

≤ 1∣∣z2 + 2∣∣ ;

|z2 + 2| ≥∣∣|z|2 − 2

∣∣ ⇒ 1∣∣z2 + 2∣∣ ≤

1∣∣|z|2 − 2∣∣ .

On the part of the boundary |z| = 2, we have

13

=2

|2|2 + 2≤ |f(z)| =

∣∣∣∣z

z2 + 2

∣∣∣∣ ≤2

|2|2 − 2|= 1.

On the part of the boundary |z| = 3, we have

311

=3

|3|2 + 2≤ |f(z)| =

∣∣∣∣z

z2 + 2

∣∣∣∣ ≤3

|3|2 − 2|=

37.

Thus the smallest value of |f(z)| is 311 . It is attained at a point z with |z| = 3. For this value of z,

we must have |z2 + 2| = 11. The only possibilities are z = ±3.The largest value of |f(z)| is 1. It is attained at a point z with |z| = 2. For this value of z, we

must have |z2 + 2| = 2 or z2 = −4. The only possibilities are z = ±2i.

9. We have

|f(z)| = | ln |z|+ iArg z|;|f(z)|2 = (ln |z|)2 + ( Arg z)2.

The largest value (respectively, minimum value) of |f(z)| is attained when |f(z)|2 attains its largestvalue (respectively, minimum value). The largest value of |f(z)|2 = (ln |z|)2 + ( Arg z)2 is clearlyattained when |z| = 2 and Arg z = π

4. So |f(z)| attains its maximum value

√(ln 2)2 + (π

4)2 when

z = 2ei π4 .

The smallest value of |f(z)|2 = (ln |z|)2+( Arg z)2 is clearly attained when |z| = 1 and Arg z = 0.So |f(z)| attains its smallest value 0 when z = 1.

13. (a) To verify the identity

zn − wn = (z − w)(zn−1 + zn−2w + zn−3w2 + · · ·+ z wn−2 + wn−1),

expand the right side. All terms cance except for zn −wn.(b) If p(z) = pnz

n + pn−1zn−1 + · · · + p1z + p0 is a polynomial of degree n ≥ 2, and if p(z0) = 0,

then pnzn0 + pn−1z

n−10 + · · ·+ p1z0 + p0 = 0. From (a),

p(z) = p(z) − p(z0) = pn(zn − zn0 ) + pn−1(zn−1 − zn−1

0 ) + · · ·+ p1(z − z0)= (z − z0)q(z),

where q(z) = (zn−1 + zn−2z0 + zn−3z20 + · · ·+ z zn−2

0 + zn−10 ) is a polynomial of degree n− 1 in z.


17. Suppose that f is entire and that it omits an open nonempty set, say, there is an open diskBR(w0) with R > 0 in the w-plane such that f(z) is not in BR(w0) for all z. Let g(z) = 1

f(z)−w0.

Then g(z) is also entire, because f(z) 6= w0 for all z. In fact, since f(z) is not in BR(w0), its distanceto w0 is always greater than R. That is, |f(z) − w0| ≥ R. But this implies that |g(z)| ≤ 1

R , whichin turn implies that g is constant, by Liouville’s theorem. Since g(z) 6= 0, this constant is obviouslynot 0. So C = 1

f(z)−w0, hence f(z) = 1

C+ w0 is constant.

21. Suppose that f(z) = f(x + i y) is periodic in x and y, and let T1 > 0 and T2 > 0 be such thatf((x+T1)+i (y+T2)) = f(x+i y) for all z = x+i y. Because f is periodic in both x and y, its valuesrepeat on every T1 × T2-rectangle. This means that, if we take the rectangle R = [0, T1] × [0, T2]and consider the values of f on this rectangle, then these are all the values taken by f(z), for z inC. The reason is that the complex plane can be tiled by translates of R in the x and y direction,where in the x drection we translate by T1 units at a time, and in the y direction we translate by T2

units at a time. Now sice f is continuous, it is bounded on R; that is |f(z)| ≤ M for some constantM and all z in R. But since f takes on all its values in R, we concude that |f(z)| ≤ M for all z inC, and thus f must be constant by Liouville’s theorem.

If f is constant in x or y alone, f need not be constant. As an example, take f(z) = sinz then fis 2π-periodic in the x variable. By considering f(iz) = sin(iz), we obtain a function that is constantin y alone. Both functions are entire and clearly not constant.

25. Suppose that f is analytic on |z| < 1 and continuous on |z| ≤ 1. Suppose that f(z) is real-valued for all |z| = 1. Consider g(z) = eif(z). Then g is analytic and nonvanishing in |z| ≤ 1, becausethe complex exponential function is never 0. Moreover g is continuous on |z| = 1. By Corollary 3, gattains its maximum and minimum on the boundary |z| = 1. But since f(z) is real-valued if |z| = 1,it follows that |g(z)| = |eif(z)| = 1 for all |z| = 1. Consequently, by Corrolary 3, g is constant insidethe unit disk. Hence, for all z inside the unit disk, f(z) = c + 2kπ, where c is a constant and k isan integer. Now it is easy to see that since f is continuous, its values cannot hop around a discreteset. (A proof of this fact may be constructed in a way similar to the proof of Lemma 2, Sec. 5.7.)So f(z) = c+ i2k0π for all z, and so f is constant.

29. Suppose that f is analytic with |f(z)| ≤ 1 for all |z| ≤ 1. By the maximum principle, we havethat |f(z)| ≤ 1 for all |z| ≤ 1 Given z in the open unit disk, we can form a circle centerd at z withradius 1 − |z|. This circle is contained in the unit disk. Since f is analytic on C1−|z|(z), we have byCauchy’s generalized formula,

f (n)(z) =n!2πi

∫

C1−|z|(z)

f(ζ)(ζ − z)n+1

dζ.

For ζ on C1−|z|(z), we have |ζ − z| = 1 − |z|. So, since |f(ζ| ≤ 1, using the familiar inequality forpath integrals, we obtain

|f (n)(z)| =

∣∣∣∣∣n!2πi

∫

C1−|z|(z)

f(ζ)(ζ − z)n+1

dζ

∣∣∣∣∣ ≤n!2π

2π(1 − |z|)(1 − |z|)n+1

=n!

(1 − |z|)n.

Section 3.8 Applications to Harmonic Functions 67

Solutions to Exercises 3.81. (a) The function u(x, y) = xy is harmonic on Ω = C, because

uxx = 0, uyy = 0, and so ∆u = 0.

(b) To find the conjugate gradient of u, apply Lemma 1: for z = x+ iy,

φ(z) = ux − iuy = y − ix = iz.

Clearly, φ is analytic in Ω as asserted by Lemma 1.

5. In this problem we are told that u is the real part of an analytic function f ; that is f = u + i von a region Ω. To find the conjugate gradient of u, we apply Lemma 1 and find φ(z) = ux − iuy.But uy = −vx, by the Cauchy-Riemann equations, so φ(z) = ux + iuy = f ′(z), by (8), Sec. 2.4.

9. Suppose that f is analytic on a region Ω. We want to prove that |f | is harmonic on Ω if andonly if f is constant. Note that one implication is immediate: If f is constant, then |f | is constantand hence it is harmonic. It is the other implication that is more interesting. We will offer twoproofs! The following results are needed in the proofs.Lemma 1 Suppose f is analytic in a region Ω and |f | is harmonic in Ω. If f(z0) = 0 for some z0 inΩ and BR(z0) is any disk centered at z0 and contained in Ω, then f is identically 0 in BR(z0).Proof Suppose that f(z0) = 0. Let BR(z0) be an open disk contained in Ω. Since |f | is harmonicon BR(z0), applying the mean value property, we find that for any 0 < r < R,

0 = |f(z0)| =12π

∫ 2π

0

|f(z0 + reit)| dt.

Since the integrand is continuous and nonnegative, the only way for the integral to equal zero is for|f(z0 + reit)| to be identically zero for all t in [0, 2π], which implies that |f | and hence f is zero onthe circle of radius r and center at z0. Since this is true for all 0 < r < R, it follows that f = 0 onBR(z0).

Lemma 1 Suppose f is analytic in a region Ω and |f | is harmonic in Ω. Either |f | is identically0 in Ω or |f | does not vanish in Ω.Proof Let Ω1 = z : z ∈ Ω, |f(z)| = 0 and Ω2 = z : z ∈ Ω, |f(z)| 6= 0. Then Ω = Ω1 ∪ Ω2 andΩ1 ∩ Ω2 = ∅. If we can show that both Ω1 and Ω2 are open, then by the connectedness of Ω, eitherΩ1 = Ω or Ω2 = Ω. Since |f | is continuous, it follows that Ω2 is open, because Ω2 is the inverseimage of the open set (0, ∞). The fact that Ω1 is open follows from Lemma 1, because every z0 inΩ1 is contained in a disk on which |f | vanishes, i.e., the disk is contained in Ω1. This proves Lemma2.

We now return to the proof of Exercise 9.First Proof. This one is more direct, but requires more computations.

In view of Lemma 2, we can suppose that |f | 6= 0 in Ω, otherwise |f | is identically 0 in Ω.Write f = u+ iv, then φ = |f | = (u2 + v2)1/2. We are given that φxx + φyy = 0. Compute the

partial derivatives explicitely:

φx = (u2 + v2)−1/2(uux + vvx);φxx = −(u2 + v2)−3/2(uux + vvx)2 + (u2 + v2)−1/2

((ux)2 + uuxx + (vx)2 + vvxx

)

= (u2 + v2)−3/2[− (uux + vvx)2 + (u2 + v2)

((ux)2 + uuxx + (vx)2 + vvxx

)]

= (u2 + v2)−3/2[(uvx − vux)2 + (u2 + v2)

(uuxx + vvxx

)].

Changing x to y, we obtain

φyy = (u2 + v2)−3/2[(uvy − vuy)2 + (u2 + v2)

(uuyy + vvyy

)].


Since φ is harmonic, we obtain

0 = φxx + φyy

= (u2 + v2)−3/2[(uvx − vux)2 + (u2 + v2)

(uuxx + vvxx

)]

+(u2 + v2)−3/2[(uvy − vuy)2 + (u2 + v2)

(uuyy + vvyy

)]

= (u2 + v2)−3/2[(uvx − vux)2 + (uvy − vuy)2 + (u2 + v2)

(uuxx + vvxx + uuyy + vvyy

)]

= (u2 + v2)−3/2[(uvx − vux)2 + (uvy − vuy)2 + (u2 + v2)

(u

=0︷︸︸︷(uxx + uyy) +v

=0︷︸︸︷(vxx + vyy)

)]

= (u2 + v2)−3/2[(uvx − vux)2 + (uvy − vuy)2

];

⇒ (uvx − vux)2 = 0 and (uvy − vuy)2 = 0⇒ u2(vx)2 + v2(ux)2 − 2uvuxvx = 0 and u2(vy)2 + v2(uy)2 − 2uvvyuy = 0.

Since f = u+ iv is analytic, the Cauchy-Riemann equations hold: ux = vy and uy = −vx and so

u2(uy)2 + v2(ux)2 − 2uvuxvx = 0 and u2(ux)2 + v2(uy)2 + 2uvuxvx = 0.

Adding the two equations, we get

u2[(ux)2 + (uy)2] + v2[(ux)2 + (uy)2] = 0 ⇒ (u2 + v2)[(ux)2 + (uy)2] = 0

By assumption, |f | 6= 0, hence u2 + v2 6= 0, and so (ux)2 + (uy)2 = 0, implying that ux = 0 anduy = 0, which in turn implies that u is identically constant in Ω. Consequently, v is identicallyconstant in Ω and so f is identically constant in Ω.

For the second proof, we need two lemmas that are interesting in their own right.Lemma 3 If u and eu are harmonic in Ω then u is identically constant.Proof Let φ = eu. Then φx = euux and φxx = eu(ux)2 + euuxx. Similarly, φy = euuy andφyy = eu(uy)2 + euuyy. Using ∆φ = 0 and ∆u = 0, we get

eu[(ux)2 + (uy)2] = 0 ⇒ (ux)2 + (uy)2 = 0 ⇒ ux = 0 and uy = 0.

Hence u is constant.The following lemma is extremely useful.

Lemma 4 If f is analytic an nonvanishing on a region Ω, then ln |f(z)| is harmonic on Ω.Proof One way to proceed is to realize that if f(z) is analytic and nonvanishing, then ln |f(z)| isthe real part of an analytic branch of the logarithm of f(z), and so it is harmonic (see Exercise 36,Sec. 3.6). Another way is to simply verify Laplace’s equation. Let φ = 2 ln |f(z)| = ln(u2 + v2) andcheck that ∆φ = 0. This will imply that φ is harmonic and hence that ln |f(z)| is harmonic. Tosimplify notation, let ψ(x, y) = u2 + v2 so φ = lnψ. Let’s compute:

φx =ψx

ψ;

φxx =ψxxψ − ψ2

x

ψ2;

φyy =ψyyψ − ψ2

y

ψ2;

φxx + φyy =(ψxx + ψyy)ψ − (ψ2

x + ψ2y)

ψ2.


But

(ψxx + ψyy)ψ =(2u2

x + 2uuxx + 2v2x + 2vvxx + 2u2

y + 2uuyy + 2v2y + 2vvyy

)ψ

=(2u2

x + 2u2y + 2u(

=0︷︸︸︷uxx + uyy) + 2v(

=0︷︸︸︷vxx + vyy)

)ψ

= 2(u2x + u2

y + v2x + v2

y)(u2 + v2)

= 4(u2x + u2

y)(u2 + v2),

because by the Cauchy-Riemann equation ux = vy and uy = −vx. Also

ψ2x + ψ2

y = (2uux + 2vvx)2 + (2uuy + 2vvy)2

= 4((uux)2 + (vvx)2 + (uuy)2 + (vvy)2

)+

=0 by Cauchy-Riemann eq.︷︸︸︷8uuxvvx + 8uuyvvy

= 4(u2x + u2

y)(u2 + v2).

Thus ∆φ = 0, as desired.Second Proof. In view of Lemma 2, we can suppose that f is nonvanishing on Ω. By Lemma 3,ln |f | is harmonic. Also, eln |f| = |f | is harmonic, by assumption. By Lemma 3, ln |f | is constant,but this implies that |f | is constant.

13. Apply (7) and you get the solution

u(r, θ) = 1 − r cos θ + r2 sin 2θ (0 ≤ r ≤ 1, all θ.

17. To find the isotherms in Exercise 13, it is easier to use Cartesian coordinates. With x = r cos θand y = r sin θ, we have

u(r, θ) = 1 − r cos θ + r2 sin 2θ= 1 − r cos θ + 2r sin θr cos θ= 1 − x+ 2xy.

If T is a real number, the isotherm corresponding to T is determined by the equation

1 − x+ 2xy = T ⇒ y =T − 1 + x

2x.

The isotherm is the portion of this hyperbola that lies in the unit disk. Note that the values of theboundary temperature vary between −2 and 3. So if T < −2 or T > 3, the isotherm correspondingto T is empty.

21. The solution of the Dirichlet problem on the unit disk with boundary data

f(θ) =

100 if 0 ≤ θ ≤ π,0 if π < θ < 2π

is given by the Poisson formula (12):

u(r, θ) =1 − r2

2π

∫ 2π

0

f(φ)1 − 2r cos(θ − φ) + r2

dφ

=1 − r2

2π

∫ π

0

1001 − 2r cos(φ − θ) + r2

dφ

= 1001− r2

2π

∫ π−θ

−θ

dt

1 − 2r cos t+ r2dt,


where in the last integral we have changed variables: t = φ− θ, dt = dφ. By Exercise 20, we have∫

dt

1 − 2r cos t+ r2dt =

21 − r2

tan−1

(1 + r

1 − rtan(

t

2))

+ C.

This formula gives an antiderivative up to a constant C. To evaluate a definite integral using it,we need to choose C in such a way that the antiderivative becomes continuous. One way to get acontinuous antiderivative is to use the following formula:

F (t) =

2

1−r2 tan−1(

1+r1−r tan( t

2 ))

if− π < t < π,

F (t− 2π) + π otherwise.

(A good way to see that this formula yields a continuous function, just plot it.) This formula saysthat F (t ± 2π) = F (t) + π. So between two consecutive intervals of length 2π, the values of Fincrease by π. Note that F is also odd: F (−θ) = −F (θ). If 0 < θ < 2π, then 0 < 2π − θ < 2π, and−π < π − θ < π. We have

u(r, θ) =100πF (t)

∣∣∣π−θ

−θ=

100π

(F (π − θ) − F (−θ)

)=

100π

(F (π − θ) + F (θ)

).

To write down the formula explicitly, we distinguish two cases: If 0 < θ < π, then 0 < π − θ < π,and

u(r, θ) =100π

[tan−1

(1 + r

1 − rtan(

π

2− θ

2))

+ tan−1

(1 + r

1 − rtan(

θ

2))]

=100π

[tan−1

(1 + r

1 − rcot(

θ

2))

+ tan−1

(1 + r

1 − rtan(

θ

2))]

,

where we have used the identities tan(φ2− α) = cotα and tan−1(−α) = − tan−1 α. If π < θ < 2π,

then −π < π − θ < 0, and −π < θ − 2π < 0, F (θ) = F (θ − 2π) + π. So

u(r, θ) =100π

[tan−1

(1 + r

1 − rcot(

θ

2))

+ tan−1

(1 + r

1 − rtan(

θ

2))

+ π].

It is useful at this point to check that u verifies the boundary conditions. If 0 < θ < π, then0 < θ

2< π

2, hence tan θ

2and cot θ

2are both positive, and so

limr→1

1 + r

1 − rtan(

θ

2) = +∞ and lim

r→1

1 + r

1 − rcot(

θ

2) = +∞.

Consequently, if 0 < θ < π,

limr→1

u(r, θ) =100π

(tan−1(+∞) + tan−1(+∞)

)=

100π

(π2

+π

2)

= 100,

which is the correct boundary value for u. If π < θ < 2π, then by a similar reasoning, we find thatlimr→1 u(r, θ) = 0.

25. All the stated formulas are different ways to express a convolution of f with the Poisson kernel(see Sec. 7.5 for details about convolutions). Let

Pr(θ) = P (r, θ) =R2 − r2

R2 − 2rR cos θ + r2.

Then according to (12), the Poisson integral formula is

12π

∫ 2π

0

f(φ)Pr(θ − φ) dφ.


This is precisely the definition of the convolution of the two 2π-periodic functions f(φ) and Pr(φ).Other equivalent are derived by using properties of integrals of periodic functions. The statedformulas are special cases of formulas given in Sec. 7.5. In particular, see (13) and Corollary 2 ofSec. 7.5.


Solutions to Exercises 3.91. Consider calculus of a real variable and take the function f(x) = x2 sin(1/x) if x 6= 0 andf(0) = 0.(a) For x 6= 0,

f ′(x) = 2x sin1x− x2 1

x2cos

1x

= 2x sin1x− cos

1x.

(b) If x = 0, then

f ′(0) = limh→0

f(0 + h) − f(0)h

= limh→0

h sin1h

= 0,

because h → 0 and sin 1h is bounded between −1 and 1, so the limit is 0 by the squeeze theorem.

(c) Even though f ′(x) exists for all x, it is not continuous at 0. First note that limx→0 cos 1x does

not exist. To see this, take x = xn = 2(2n+1)π , then

cos1xn

= cos(2n+ 1)π

2= (−1)n,

and clearly the limit as n → ∞ does not exist. Now if limx→0 f′(x) does exist, then because

f ′(x) = 2x sin 1x − cos 1

x and limx→0 2x sin 1x = 0 by the squeeze theorem, this would imply that

limx→0 cos 1x

exists, which is a contradiction. So limx→0 f′(x) does not exist and hence f ′ is not

continuous at x = 0.

Solutions to Exercises 3 - University of...

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