Chapter 3

48

Chapter 3

Complex Integration

Solutions to Exercises 3.1 1. Given z1 = 1 + i and z2 = −1 − 2i, apply (2), Sec. 3.1, to obtain the parametrization of the line segment [z1, z2]: γ(t) = (1 − t)z1 + tz2 = (1 − t)(1 + i) + t(−1 − 2i),

0 ≤ t ≤ 1,

or γ(t) = t(−2 − 3i) + 1 + i,

0 ≤ t ≤ 1.

5. Use Example 1(a) with α = − π4 and β = π4 : γ(t) = eit ,

−

π π ≤t≤ . 4 4

9. Use Example 1(a) and (c): γ(t) = z0 + Reit = −3 + 2i + 5eit ,

−

π ≤ t ≤ 0. 2

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≤

3π . 2

Note the negative sign in the exponent, whose effect is to trace the ar in the negative direction. If we want to parametrize γ1 using the interval [0, 1], we can change t to 3π 2 t and get 3π γ1 (t) = 2e−i 2 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, So γ(t) =

(

2e−i

3π t 2

1 ≤ t ≤ 2. if 0 ≤ t ≤ 1

3i(t − 1) + 2i if 1 ≤ t ≤ 2. 17. We have x(t) = t and y(t) = sin πt, where 0 ≤ t ≤ 1. Thus y = sin πx, where 0 ≤ 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 they were real. We get d d (2 + i) cos(3it) = (2 + i) cos(3it) = −3i(2 + i) sin(3it) = (3 − 6i) sin(3it). dt dt 25. To find a particular solution yp of y 00 + y 0 + 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 00 + y 0 + y = eit and let Y denote this particular solution. 3. Compute yp = Re (Y ). To find Y , try Y = Aeit , Y 0 = Aieit , Y 00 = −Aeit . Plug into the equation and get −Aeit + iAeit + Aeit = eit (Divide by eit .)

−A + iA + A = 1 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 00 − 2y 0 − 3y = e4it. Then Re (Y ) is a particular solution of y 00 − 2y 0 − 3y = cos 4t, and Im (Y ) is a particular solution of y 00 − 2y 0 − 3y = sin 4t. So solving the equation with a complex exponential on the right side yields the solutions of two differential equations. Let us now solve y 00 − 2y 0 − 3y = e4it . For this purpose, we let Y = Ae4it , Y 0 = 4iAe4it, and Y 00 = −16Ae4it . Plugging into the equation and solving, we find −16Ae4it − 8iAe4it − 3Ae4it = e4it (Divide by e4it.)

−16A − 8iA − 3A = 1 A(−19 − 8i) = 1 ⇒ A = 19 Hence Y = (− 425 +

8 4it 425 i)e

yp = Re ((−

1 −19−8i

19 = − 425 +

8 425 i.

and so a particular solution of y 00 − 2y 0 − 3y = cos 4t is 19 19 8 8 + i)e4it)) = − cos 4t − sin 4t. 425 425 425 425

Also, a particular solution of y 00 − 2y 0 − 3y = sin 4t is yp = Im ((−

19 8 8 19 + i)e4it)) = cos 4t − sin 4t. 425 425 425 425

at 33. We have x(t) = a cos t − b cos at 2 and y(t) = a sin t − b sin 2 . So

γ(t) = x(t) + y(t) = a cos t − b cos

at at + ai sin t − ib sin 2 2

= a(cos t + i sin t) − b(cos at

= aeit − bei 2

at at + i sin ) 2 2

50

Chapter 3

Complex Integration

Solutions to Exercises 3.2 1. Using (10), Z

2π 0

=1 2π z}|{ 1 1 e6iπ −1 = 0. e3ix dx = e3ix = 3i 3i 0

5. Write x+i x+i × x−i x+i x2 + 2ix − 1 (x2 + 1) + 2ix − 2 = = x2 + 1 x2 + 1 2 2ix = 1− 2 + 2 . x +1 x +1

x+i x−i

=

So odd integrand z Z }| { 1 2 x dx + 2i dx 2 x2 + 1 −1 x + 1 =0,

Z

1

−1

x+i dx = x−i

Z

1

1 dx − −1

Z

1 −1 1

= x − 2 tan−1 x

= 2 − π.

−1

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 2 x i 3 3x

+ C if − 1 ≤ x ≤ 0, if 0 ≤ x ≤ 1.

Setting F (0+) = F (0−), we obtain 0 = 3+2i 2 0 + C or C = 0. Hence a continuous antidervative of f is ( 3+2i 2 if − 1 ≤ x ≤ 0, 2 x F (x) = i 3 if 0 ≤ x ≤ 1. 3x 13. (a) Let p > 0 be an arbitrary real number, and m and n be arbitrary integers. Write mπ f (x) = ei p x . Then =1

f (x + 2p) = e

i mπ (x+2p) p

=e

z }| { mπ e = ei p x = f (x).

i mπ x i2mπ p

Section 3.2

Hence f is 2p-periodic. (b) If m 6= n, then Z p Z i mπ x i −nπ x e p e p dx = −p

p

e

i

(m−n)π x p

Complex Integration

dx

−p

p i i i (m−n)π x i(m−n)π −i(m−n)π =− = − − e e p e p(m − n) p(m − n) −p =1

z }| { i e−i(m−n)π ei2(m−n)π −1 = 0 = − p(m − n) If m = n, then Z

p

e

i mπ x i −nπ x p p

e

dx =

−p

Z

p

e

i mπ x i −mπ x p p

e

dx =

−p

Z

p

dx = 2p. −p

We have thus established the orthogonlaity relations Z p 0 if m 6= n, i mπ x −i nπ x p p e e dx = 2p if m = n. −p (c) Now suppose m and n are nonnegative integers. If m 6= n, then the equality Z p i mπ x −i nπ x e p e p dx = 0 −p

implies that

Z

p

cos −p

mπ mπ nπ nπ x + i sin x cos x − i sin x dx = 0. p p p p

Expanding and then taking real and imaginary parts, we get Z p mπ nπ mπ nπ (1) cos x cos x + sin x sin x dx = 0 p p p p −p and (2)

Z

p

− cos −p

mπ nπ mπ nπ x sin x + sin x cos x dx = 0. p p p p

Replacing n by −n; that is, starting with the identity Z p mπ nπ ei p x ei p x dx = 0, −p

we get (3)

51

Z

p

−p

cos

mπ nπ mπ nπ x cos x − sin x sin x dx = 0 p p p p

52

Chapter 3

Complex Integration

and Z

(4)

p

cos

−p

mπ nπ mπ nπ x sin x + sin x cos x dx = 0. p p p p

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 Z p mπ nπ cos x cos x dx = 0. p p −p Adding (2) and (4) implies that Z

p

cos −p

mπ nπ x sin x dx = 0. p p

All other integral identities with m 6= n follow similarly. Consider the case m = −n 6= 0, Z p i mπ x i mπ x e p e p dx = 0. −p

Writing the integrand in terms of cosine and sine and then taking real and imaginary parts, we get Z p mπ mπ (5) cos2 x − sin2 x dx = 0 p p −p Similarly, from the identity

Z

p

e

i mπ x i −mπ x p p

e

dx = 2p,

−p

we obtain Z

(6)

p

cos2 −p

mπ mπ x + sin2 x dx = 2p. p p

Adding (5) and (6) shows that Z

p

cos

2

−p

mπ x p

dx = p.

The remaining identities follow similarly. 17. Write I=

Z

2z dz = [z1 , z2 , z3 ]

Z

2z dz + [z1 , z2 ]

Z

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

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 Z Z 1 Z 1 (1 − t) − it (−1 + i) dt = 2(−1 + i) 1 − ((1 + i)t dt 2z dz = 2 [z1 , z2 ]

0

1 + i 2 1 = 2(−1 + i) t − ( t = 2i. 0 2

0

Section 3.2

Complex Integration

53

To evaluate I2, parametrize [z2, z3] by z = γ2 (t) = (1 − t)i + t(1 + i) = i + t, 0 ≤ t ≤ 1, dz = dt. So Z Z 1 1 t 2z dz = 2 (t − i) dt = 2( − it) = 1 − 2i. 2 0 [z2 , z3 ] 0 Thus I = 1. 21. Write

I

=

Z

z dz [z1 , z2 , z3 , z4 , z1 ]

=

Z

z dz + [z1 , z2 ]

Z

z dz + [z2 , z3 ]

Z

z dz + [z3 , z4 ]

Z

z dz [z4 , z1 ]

= I1 + I2 + I3 + I4 , where z1 = 0, z2 = 1, z3 = 1 + i, and z4 = i. To evaluate I1 , parametrize [z1 , z2 ] by z = γ1(t) = t, 0 ≤ t ≤ 1, dz = dt. So Z Z 1 1 I1 = z dz = t dt = . 2 [z1 , z2 ] 0 To evaluate I2 , parametrize [z2, z3] by z = γ2(t) = 1 + it, 0 ≤ t ≤ 1, dz = i dt. So Z Z 1 i 1 i 1 I2 = z dz = (1 + it)i dt = i(t + t2 ) = i(1 + ) = − + i. 2 2 2 0 [z2 , z3 ] 0 To evaluate I3 , parametrize [z3, z4] by z = γ3(t) = (1 − t) + i, 0 ≤ t ≤ 1, dz = −dt. So Z Z 1 1 1 I3 = z dz = − ((1 − t) + i) dt = −(1 + i) + = − − i. 2 2 [z3 , z4 ] 0 To evaluate I4 , parametrize [z4, z1] by z = γ4(t) = (1 − t)i, 0 ≤ t ≤ 1, dz = −idt. So Z Z 1 1 1 I4 = z dz = −i i(1 − t) dt = 1 − = . 2 2 [z4 , z1 ] 0 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 of the next section, you will realize that all our work in this solution is unnecessary to conclude that the answer is 0. 25. In our computation, we will suppose that a is an integer 6= 0. Then the curve is traced in its entirety if we take 0 ≤ t ≤ 4π. We apply the definition of the path integral, with a −i a2 t γ(t) = aeit + be−i 2 t , 0 ≤ t ≤ 4π, dz = (aieit − i ab )dt: 2e Z Z 4π a a ab z dz = (aeit + be−i 2 t )(aieit − i e−i 2 t ) dt 2 γ 0 Z 4π 2 a ab −iat a2b i(1− a )t 2 ) dt = 0, = (ia2e2it − i + iabei(1− 2 )t − i e e 2 2 0

54

Chapter 3

Complex Integration

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 Z

(x2 + y 2 ) dz =

Z

[z1 , z2 ]

1

((−3 − 2i)t + 2 + i)((−3 + 2i)t + 2 − i)(−3 − 2i)dt 0

= (−3 − 2i)

Z

1

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

0

13 (−3 − 2i)(2 − i) (2 + i)(−3 + 2i) 8 = (−3 − 2i)( + + + 5) = −4 − i. 3 2 2 3 t

t

33. We have γ(t) = (et − t) + 4ie 2 , γ 0(t) = (et − 1) + 2ie 2 , q t t t 0 |γ (t)| = (e − 1) + 2ie 2 = (et − 1)2 + (2e 2 )2 p p = e2t + 1 + 2et = (et + 1)2 = et + 1. So L=

Z

1 0

|γ (t)| dt =

Z

0

1 0

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

√ 37. We apply the M L-inequality (29). We have L = l([z1, z2 , z3 , z1 ]) = 2 + 2 5, as can be easily verified by plotting the points z1 , z2 , and z3 . Now |z −5 | = |z|−5 , and for z on the 1 path [z1 , z2 , z3 , z1 ], we have |z| ≤ 2 2 . So |z|−5 ≤ 25 for z on the path [z1 , z2 , z3 , z1]. Thus, by the M L-inequality, Z √ 5 −5 z dz ≤ 2− 2 (2 + 2 5). [z1 , z2 , z3 , z1 ] 41. We have Z

f (z) dz = Γ

Z

f (z) dz Z Z = f (z) dz + f (z) dz γ −γ Z Z = f (z) dz − f (z) dz = 0. (γ, −γ)

γ

γ

Section 3.3

Independence of Path

55

Solutions to Exercises 3.3 1. An antiderivative of f(z) = z 2 +z −1 is simply F (z) = 13 z 3 + 12 z 2 −z +C where C is an arbitrary complex constant. The function F is entire and so we can take Ω = C. 5. To find an antiderivative of partial fractions, write

1 , (z−1)(z+1)

we proceed as we would have done in calculus. Using

1 = (z − 1)(z + 1) 1 = (z − 1)(z + 1) 1 =

A B + z−1 z +1 A(z + 1) + B(z − 1) (z − 1)(z + 1) A(z + 1) + B(z − 1).

Taking z = −1, it follows that B = − 12 . Taking z = 1, it follows that A = 12 . Hence 1 1 1 = − . (z − 1)(z + 1) 2(z − 1) 2(z + 1) An antiderivative of this function is F (z) =

1 Log (z − 1) − Log (z + 1) + C, 2

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 in C\(−∞, −1]. So the function F (z) = 12 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−1 z+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 that the linear fractional transformation w = z−1 takes the interval [−1, 1] onto the half-line (−∞, 0]. z+1 All other values of z outside the interval [−1, 1] are mapped into C \ (−∞, 0] and so the composition Log 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) and f(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 composition f(g(z)) is equal to the function z−1 , which is analytic except at the points z = ±1. According to z+1 Theorem 4, Sec. 2.3, the function g(z) is analytic in C \ [−1, 1]. 9. An antiderivative of z sinh z 2 is tiderivative is valid for all z.

1 2

cosh z 2 + C, as you can verify by differentiation. The an-

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 hint from calculus and the fact that an antiderivative of ln x is −x+x ln x+C (as you can verify directly or integrate ln x 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 of f2 (z) = log π2 z is F2 (z) = −z + z log π2 z + 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 + log π2 z + 1z is −2z + z log0 z + z log π2 z + 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 π2 . 17. The integrand is entire and has an antiderivative in a region containing the path. So, by

56

Chapter 3

Complex Integration

Theorem 1, Z

2

z dz

γ( π ) 1 3 4 1 π π = z (ei 4 + 3e2i 4 )3 − (ei·0 + 3e2i·0)3 3 γ(0) 3 2π π π π 3π 1 i 3π e 4 + 9ei 4 ei 2 + 27ei 4 e2i 2 + 27ei 2 − (4)3 3 ! √ √ √ √ 1 − 2 2 2 2 3 +i − 9 − 27( +i ) − 27i − (4) 3 2 2 2 2 √ √ 1 (−14 2 − 73 + i(−13 2 − 27) 3

=

γ

= = =

21. Evaluate using Theorem 1: Z sin z dz

=

γ( π2 ) − cos z γ(0)

γ

= =

i·0

π

cos(2e ) − cos(2ei 2 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, Z z Log z dz = 0. [z1 , z2 , z3 , z1 ]

29. Part (a) is clear. For (b), the path γ1 lies completely in a region on which the function Log z is analytic. Since Log z is an antiderivative of z1 , it follows from Theorem 1 that Z 1 dz = Log (z1 ) − Log (z2 ). γ1 z Similarly, the path γ2 lies completely in a region on which the function log0 z is analytic. Since log0 z is an antiderivative of z1 , it follows from Theorem 1 that and Z 1 dz = log0 (z2 ) − log0 (z1 ). γ2 z Now

Z CR (z0 )

1 dz = z

Z γ1

1 dz + z

Z γ2

1 dz = Log z1 − Log z2 + log0 z2 − log0 z1. z

Recall that z1 is on the positive y-axis, so Log z1 = log0 z1. Also z2 is on the negative y-axis, so arg 0 z2 = 3 π2 and Arg z2 = − π2 . Hence log0 z2 − Log z2 = 2πi. Putting this information together, we get Z 1 dz = 2πi. CR (z0 ) z 33. If f is analytic in a region Ω, then obviously f is an antiderivative of f 0 in Ω. Let z0 and z1 be any two points in Ω. Since Ω is connected, we can join z0 to z1 by a polygonal path γ, with initial point z0 and terminal point z1 . By Theorem 1, Z f 0 (z) dz = f(z1 ) − f(z0 ). γ 0

But f (z) = 0, so f(z1 ) − f(z0 ) = 0 implying that f(z0 ) = f(z1 ). Since z1 is arbitrary, it follows that 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 are mutually 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 of minor 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 construct the desired mapping of the unit square. We take γ0 as described above and γ1 (t) = 3i. Then, for 0 ≤ t ≤ 1 and 0 ≤ s ≤ 1: H(t, s)

= = =

13. The function f(z) =

ez z+2

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

ez dz = 0. z +2

z

e 17. The function f(z) = z+i has a problem at z = −i. Hence it is analytic inside an on the simple it path γ(t) = i + e , 0 ≤ t ≤ 2π (circle, centered at i with radius 1). By Theorem 5, Z ez dz = 0. γ z+i

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) = z 2 Log z is analytic inside an on the simple path [z1 , z2 , z3 , z1 ], and so by Theorem 5, Z z 2 Log z dz = 0.

[z1 , z2 , z3 , z1 ]

25. The path C2 (0) contains both roots of the polynomial z 2 − 1. We will evaluate the integral by using the method of Example 5. We have z A B = + z2 − 1 z−1 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 = 12 . Hence z 1 1 = + , z2 − 1 2(z − 1) 2(z + 1)

58

Chapter 3

and so

Complex Integration

Z C2 (0)

z 1 dz = z2 − 1 2

Z C2 (0)

1 1 dz + z−1 2

Z C2 (0)

1 1 1 dz = 2πi + 2πi = 2πi, z+1 2 2

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 Z Z Z 1 1 1 dz = dz + dz = I1 + I2 . 2 (z 2 + 1) 2 (z 2 + 1) 2 (z 2 + 1) (z + 1) (z + 1) (z + 1) γ γ1 γ2 The partial fraction decomposition of the integrand is 1 (z +

1)2(z 2

+ 1)

=

1 1 1 1 − + − . 2 2(z + 1) 2(z + 1) 4(z + i) 4(z − i)

We have

Z Z Z Z dz dz dz dz 1 1 1 1 − + − 2 γ1 z + 1 2 γ1 (z + 1)2 4 γ1 z + i 4 γ1 z − i 1 1 1 1 π = · 0 + · 0 − · 0 − · 2πi = − i, 2 2 4 4 2 where the first 3 integrals are 0 because the integrands are analytic inside an on γ1 , and the fourth integral follows from Example 4. Similarly, Z Z Z Z dz dz dz dz 1 1 1 1 I2 = − + − 2 γ2 z + 1 2 γ2 (z + 1)2 4 γ2 z + i 4 γ2 z − i 1 1 1 1 = · 2πi + · 0 − · 0 − · 0 = πi, 2 2 4 4 where the first, third, and fourth integrals follow from Example 4, and the second integral follows from Example 4, Sec. 3.2. Thus, the desired integral is equal to π I1 + I2 = i. 2 I1

=

33. (a) Let z1 , z2 , . . . , zn be distinct complex numbers (n ≥ 2). Reducing to the common denominator in the partial fraction decomposition 1 A2 An A1 + +···+ , = (z − z1 )(z − z2 ) · · · (z − zn ) z − z1 z − z2 z − zn we obtain 1 p(z)

=

A1 (z − z2 ) · · · (z − zn ) A2(z − z1 )(z − z3) · · · (z − zn ) + + p(z) 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 coefficient of z n−1 in the numerator on the right side is A1 + A2 + · · · + An. Comparing with the left side, we see 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 Z Z Z Z 1 dz dz dz + A2 + · · · + An dz = A1 C (z − z1 )(z − z2 ) · · · (z − zn ) C z − z1 C z − z2 C 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 3.5 1. An example of a closed and bounded subset K of the real line and a bounded subset S of the real line such that K and S are disjoint but the distance from K to S is 0: Let K = [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: p (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 p |(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) p = (Aa − Bb)2 + (Ab + Ba)2 + (Ac − Bd)2 + (Ad + Bc)2 p = (A2 + B 2 )(a2 + b2 + c2 + d2 ) p p = A 2 + B 2 a2 + b 2 + c 2 + d 2 = |α| |(z, w)|. (c) We will prove the triangle inequality: |(z1, w1) + (z2 , w2)| ≤ |(z1, w1)| + |(z2, w2)| by following 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 for complex numbers z1 = a1 + ib1, z2 = a2 + ib2, w1 = c1 + id1, and w2 = c2 + id2, we have p |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 p ≤ |z1 |2 + |w1|2 + |z2|2 + |w2|2 + 2 (|z1|2 + |w1|2)2 · (|z2 |2 + |w2|2 )2 1 1 2 = |z1 |2 + |w1|2 2 + |z2|2 + |w2|2 2 . The desired inequality follows upon taking square roots on both sides.

60

Chapter 3

Complex Integration

9. Pick y0 in the interval (c, d). We have to sow that 1 h→0 h lim

Z

b

u(x, y0 + h) − u(x, y0 ) dx =

a

Z

b a

∂ u(x, y0 ) dx. ∂y

Equivalently, we have to show that the difference Z b Z b 1 ∂ u(x, y u(x, y + h) − u(x, y ) dx − ) dx 0 0 0 h a a ∂y ∂ can be made arbitrarily small for all small values of h. Since ∂y u is continuous in R, it is uniformly continuous. So, given > 0, we can find a δ > 0 such that if 0 < h < δ, then ∂ u(x, s) − ∂ u(x, y0 ) < for all y0 < s < y0 + h. ∂y ∂y

Now using the fact that u(x, y0 + h) − u(x, y0 ) =

Z

y0 +h y0

∂ u(x, s) ds, ∂y

we find that for 0 < h < δ Z b Z b 1 ∂ u(x, y0 + h) − u(x, y0 ) dx − u(x, y0 ) dx h a a ∂y Z b Z y0 +h Z b 1 ∂ ∂ = u(x, s) ds dx − u(x, y0 ) dx h a y0 ∂y a ∂y 0 in the w-plane such that f(z) is not in BR (w0 ) for all z. Let g(z) = f (z)−w . 0 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 distance to w0 is always greater than R. That is, |f(z) − w0 | ≥ R. But this implies that |g(z)| ≤ R1 , which in turn implies that g is constant, by Liouville’s theorem. Since g(z) 6= 0, this constant is obviously 1 , hence f(z) = C1 + w0 is constant. not 0. So C = f (z)−w 0 21. Suppose that f(z) = f(x + i y) is periodic in x and y, and let T1 > 0 and T2 > 0 be such that f((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 values repeat 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 in C. 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 constant M and all z in R. But since f takes on all its values in R, we concude that |f(z)| ≤ M for all z in C, 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 f is 2π-periodic in the x variable. By considering f(iz) = sin(iz), we obtain a function that is constant in 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 realvalued for all |z| = 1. Consider g(z) = eif (z) . Then g is analytic and nonvanishing in |z| ≤ 1, because the complex exponential function is never 0. Moreover g is continuous on |z| = 1. By Corollary 3, g attains 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 inside the unit disk. Hence, for all z inside the unit disk, f(z) = c + 2kπ, where c is a constant and k is an integer. Now it is easy to see that since f is continuous, its values cannot hop around a discrete set. (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 have that |f(z)| ≤ 1 for all |z| ≤ 1 Given z in the open unit disk, we can form a circle centerd at z with radius 1 − |z|. This circle is contained in the unit disk. Since f is analytic on C1−|z|(z), we have by Cauchy’s generalized formula, Z f(ζ) n! f (n) (z) = dζ. 2πi C1−|z|(z) (ζ − z)n+1 For ζ on C1−|z|(z), we have |ζ − z| = 1 − |z|. So, since |f(ζ| ≤ 1, using the familiar inequality for path integrals, we obtain n! Z f(ζ) n! 2π(1 − |z|) n! (n) |f (z)| = dζ = . ≤ 2πi C1−|z| (z) (ζ − z)n+1 2π (1 − |z|)n+1 (1 − |z|)n

Section 3.8

Applications to Harmonic Functions

67

Solutions to Exercises 3.8 1.

(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 v on 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 0 (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 and only if f is constant. Note that one implication is immediate: If f is constant, then |f| is constant and hence it is harmonic. It is the other implication that is more interesting. We will offer two proofs! 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 in Ω 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 harmonic on BR (z0 ), applying the mean value property, we find that for any 0 < r < R, Z 2π 1 0 = |f(z0 )| = |f(z0 + reit )| dt. 2π 0 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 on the circle of radius r and center at z0 . Since this is true for all 0 < r < R, it follows that f = 0 on BR (z0 ). Lemma 1 Suppose f is analytic in a region Ω and |f| is harmonic in Ω. Either |f| is identically 0 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 and Ω1 ∩ Ω2 = ∅. If we can show that both Ω1 and Ω2 are open, then by the connectedness of Ω, either Ω1 = Ω or Ω2 = Ω. Since |f| is continuous, it follows that Ω2 is open, because Ω2 is the inverse image of the open set (0, ∞). The fact that Ω1 is open follows from Lemma 1, because every z0 in Ω1 is contained in a disk on which |f| vanishes, i.e., the disk is contained in Ω1 . This proves Lemma 2. 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 h i (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 .

68

Chapter 3

Complex Integration

Since φ is harmonic, we obtain 0

= =

=

φxx + φyy

h i (u2 + v2 )−3/2 (uvx − vux )2 + (u2 + v2 ) uuxx + vvxx h i +(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 =0

= =

=0

h z }| { z }| { i (u + v ) (uvx − vux )2 + (uvy − vuy )2 + (u2 + v2 ) u (uxx + uyy ) +v (vxx + vyy ) h i (u2 + v2 )−3/2 (uvx − vux )2 + (uvy − vuy )2 ; 2

2 −3/2

⇒ (uvx − vux )2 = 0 and (uvy − vuy )2 = 0 ⇒ u2(vx )2 + v2 (ux )2 − 2uvux vx = 0 and u2(vy )2 + v2 (uy )2 − 2uvvy uy = 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 − 2uvux vx = 0 and u2(ux )2 + v2 (uy )2 + 2uvux vx = 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 and uy = 0, which in turn implies that u is identically constant in Ω. Consequently, v is identically constant 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 = eu ux and φxx = eu (ux )2 + eu uxx . Similarly, φy = eu uy and φyy = eu (uy )2 + eu uyy . 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)| is the 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 ) and check that ∆φ = 0. This will imply that φ is harmonic and hence that ln |f(z)| is harmonic. To simplify notation, let ψ(x, y) = u2 + v2 so φ = ln ψ. Let’s compute: φx

=

φxx

=

φyy

=

φxx + φyy

=

ψx ; ψ ψxx ψ − ψx2 ; ψ2 ψyy ψ − ψy2 ; ψ2 (ψxx + ψyy )ψ − (ψx2 + ψy2 ) . ψ2

Section 3.8


69

But (ψxx + ψyy )ψ

=

2u2x + 2uuxx + 2vx2 + 2vvxx + 2u2y + 2uuyy + 2vy2 + 2vvyy ψ

=

z }| { z }| { 2u2x + 2u2y + 2u(uxx + uyy ) + 2v(vxx + vyy ) ψ

=0

=0

=

2(u2x + u2y + vx2 + vy2 )(u2 + v2 )

=

4(u2x + u2y )(u2 + v2 ),

because by the Cauchy-Riemann equation ux = vy and uy = −vx . Also ψx2 + ψy2

=

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

2

2

=

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

=

4(u2x + u2y )(u2 + v2 ).

2

=0 by Cauchy-Riemann eq.

z }| { + 8uuxvvx + 8uuy vvy

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 θ and 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 the boundary temperature vary between −2 and 3. So if T < −2 or T > 3, the isotherm corresponding to T is empty. 21.

The solution of the Dirichlet problem on the unit disk with boundary data 100 if 0 ≤ θ ≤ π, f(θ) = 0 if π < θ < 2π

is given by the Poisson formula (12): u(r, θ)

= = =

Z 1 − r2 2π f(φ) dφ 2π 1 − 2r cos(θ − φ) + r2 0 Z 1 − r2 π 100 dφ 2 2π 0 1 − 2r cos(φ − θ) + r Z 1 − r2 π−θ dt 100 dt, 2π 1 − 2r cos t + r2 −θ

70

Chapter 3

Complex Integration

where in the last integral we have changed variables: t = φ − θ, dt = dφ. By Exercise 20, we have Z dt 1+r 2 t −1 dt = tan tan( ) + C. 1 − 2r cos t + r2 1 − r2 1−r 2 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 a continuous antiderivative is to use the following formula: ( 2 t −1 1+r if − π < t < π, 1−r 2 tan 1−r tan( 2 ) F (t) = F (t − 2π) + π otherwise. (A good way to see that this formula yields a continuous function, just plot it.) This formula says that F (t ± 2π) = F (t) + π. So between two consecutive intervals of length 2π, the values of F increase 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 π−θ 100 100 100 u(r, θ) = = F (t) F (π − θ) − F (−θ) = F (π − θ) + F (θ) . −θ π π π To write down the formula explicitly, we distinguish two cases: If 0 < θ < π, then 0 < π − θ < π, and 100 h 1+r 1+r π θ θ i −1 −1 u(r, θ) = tan tan( − ) + tan tan( ) π 1−r 2 2 1−r 2 i 100 h 1 + r 1 + r θ θ = tan−1 cot( ) + tan−1 tan( ) , π 1−r 2 1−r 2 where we have used the identities tan( φ2 − α) = cot α 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 i 100 h 1+r 1+r θ θ u(r, θ) = tan−1 cot( ) + tan−1 tan( ) + π . π 1−r 2 1−r 2 It is useful at this point to check that u verifies the boundary conditions. If 0 < θ < π, then 0 < θ2 < π2 , hence tan θ2 and cot θ2 are both positive, and so lim

r→1

1+r θ tan( ) = +∞ 1−r 2

and

lim

r→1

1+r θ cot( ) = +∞. 1−r 2

Consequently, if 0 < θ < π, lim u(r, θ) =

r→1

100 π π 100 tan−1(+∞) + tan−1 (+∞) = + = 100, π π 2 2

which is the correct boundary value for u. If π < θ < 2π, then by a similar reasoning, we find that limr→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 Z 2π 1 f(φ)Pr (θ − φ) dφ. 2π 0

Section 3.8


71

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 stated formulas are special cases of formulas given in Sec. 7.5. In particular, see (13) and Corollary 2 of Sec. 7.5.

72

Chapter 3

Complex Integration

Solutions to Exercises 3.9 1. Consider calculus of a real variable and take the function f(x) = x2 sin(1/x) if x 6= 0 and f(0) = 0. (a) For x 6= 0, 1 1 1 1 1 f 0 (x) = 2x sin − x2 2 cos = 2x sin − cos . x x x x x (b) If x = 0, then f(0 + h) − f(0) 1 f 0 (0) = lim = lim h sin = 0, h→0 h→0 h h because h → 0 and sin h1 is bounded between −1 and 1, so the limit is 0 by the squeeze theorem. (c) Even though f 0 (x) exists for all x, it is not continuous at 0. First note that limx→0 cos x1 does 2 not exist. To see this, take x = xn = (2n+1)π , then cos

1 (2n + 1)π = cos = (−1)n, xn 2

and clearly the limit as n → ∞ does not exist. Now if limx→0 f 0 (x) does exist, then because f 0 (x) = 2x sin x1 − cos x1 and limx→0 2x sin x1 = 0 by the squeeze theorem, this would imply that limx→0 cos x1 exists, which is a contradiction. So limx→0 f 0 (x) does not exist and hence f 0 is not continuous at x = 0.