(a) The integrand e^(-z)/(z - (pii/2)) is of the form f(z)/(z-z_0) where f(z) = e^(-z) is analytic everywhere on and inside C, and z_0 = pi i/2 is a point inside C. Therefore, by Cauchy's Integral Formula,
-z
/ e dz - pi i/2
| ------------- = 2 pi i f(pi i/2) = 2 pi i e = (2 pi i)(-i)
/C z - (pi i/2)
= 2 pi.
(b) The integrand is of the form f(z)/(z-z_0) where f(z) = (cos z)/(z^2 + 8) is analytic everywhere on and inside C (the singularities are at +/- 2sqrt(2)), and z_0 = 0 is a point inside C. Therefore, the integral equals 2 pi i f(0) = 2 pi i(cos 0)/8 = pii/4.
The other parts are similar; I won't write them all out. For example, the integrand in (d) is of the form f(z)/(z-z_0)^2 where f(z)=tan(z/2) is analytic on and inside C, and z_0 = x_0 is inside C; therefore, the integral is 2 pii f'(x_0)/1! = i pisec^2(x_0/2).
Page 125 #2
(a) The integrand equals 1/(z-2i)(z+2i) which is of the form f(z)/(z-z_0) where f(z) = 1/(z+2i) is analytic everywhere on and inside C (the singularity at -2i is outside C), and z_0 = 2i is a point inside C. Therefore, the integral equals 2 pi i f(2i) = (2 pi i) 1/(4i) = pi/2.
(b) The integrand equals 1/(z-2i)^2(z+2i)^2 which is of the form f(z)/(z-z_0)^2 where f(z) = 1/(z+2i)^2 is analytic everywhere on and inside C and z_0 = 2i is a point inside C. Therefore, the integral equals 2 pi i f'(2i)/1! = (2 pi i) (-2)/(2i + 2i)^3 = pi/16.
Page 125 #3
2
/ 2z - z - 2
g(2) = | ---------- dz.
/C z - 2
The integrand is of the form f(z)/(z-w) where f(z) = 2z^2 - z - 2
is analytic everywhere on and inside C, and w=2 is a point inside C.
Therefore, the integral equals 2 pi i f(2) = (2 pi i)((2)(2^2) - 2 - 2)
= 8 pi i.
When |w| > 3 the integrand (2z^2 - z - 2)/(z-w) is analytic for all z on and inside the contour C, and hence the integral is zero, so g(w) = 0 when |w| > 3.
Page 125 #4
When w is outside C, the integrand is analytic for all z on and inside C. (The only singularity is at the point w, which is outside C). Therefore, by the Cauchy-Goursat Theorem, the integral is zero. In other words, g(w) = 0 when w is outside C.
When w is inside C, the integrand is of the form f(z)/(z-w)^3 where f(z) = z^3 + 2z is analytic on and inside C, and w is a point inside C. Thefore, by the extension of Cauchy's Integral Formula, the integral is (2pi i) f"(w) / 2! = (pi i)(6w) = 6 pii w.
Page 132 #4 (6th edition: page 136 #7)
Let g(z) = exp(-f(z)). Then g(z) is the composition of analytic functions and is therefore analytic on R. The maximum modulus principle guarantees that |g(z)| has a maximum value on R which occurs on the boundary not in the interior. Now |g(z)| = |exp(-u(x,y) - i v(x,y))| = exp(-u(x,y)). If |g(z)| attains its maximum at z=z_0=x_0+iy_0, that means |g(z)| <=|g(z_0)| for all z in R, so e^(-u(x,y)) <=e^(-u(x_0,y_0)) for all (x,y) in R, so -u(x,y) <=u(x_0, y_0), so u(x,y) >=u(x_0,y_0) for all (x,y) in R.
This proves that the place(s) where |g(z)| attains its maximum are the same as the places where u(x,y) attains its minimum. Therefore, u(x,y) has a minimum value on R which occurs on the boundary not in the interior.
Page 132 #6 (6th edition: page 136 #2)
Let g(z) = exp(f(z)). Then
|g(z)| = |exp(f(z))| = exp(Re [f(z)]) = exp(u(x,y)) <=exp(u_0)for all points z. This proves g(z) is bounded. It is also an entire function since it is the composition of two entire functions (exp and f). Liouville's Theorem says that any bounded, entire function is constant, so it follows that g(z) is constant. Since g(z) = exp(f(z)), it follows that f(z) must be constant also (for 0 = g'(z) = f'(z) exp(f(z)) and exp(f(z)) is never zero, so f'(z) must be zero). Finally, since f(z) is constant and u(x,y) = Re[f(z)], it follows that u(x,y) is constant.
Page 132 #8 (6th edition: page 136 #1)
We will show that f"(z_0) = 0 at every point z_0. To do this, consider a circle C_R given by |z-z_0|=R. On C_R, we have |f(z)| <= A|z| = A |(z-z_0)+z_0| <=A |z-z_0| + A |z_0| = A (R + |z_0|). Then the generalization of Cauchy's Integral Formula, together with the fact that |z-z_0| = R and |f(z)| <= A(R+|z_0|) on C_R, gives
|2 pi i / f(z) | A (R + |z0|)
|f''(w)| = |------ | -------- dz | <= pi(------------) (length of C )
| 2! /C (z-z0)^3 | R^3 R
R
= A pi (R + |z0|) 2 pi R / R^3
= 2 A pi^2 ( 1/R + |z0|/R^2)
which goes to zero as R goes to infinity. Therefore, for any positive number
e > 0, you can choose an R for which the above bound is less than
e, proving that that |f"(w)| < e.
This means |f"(w)| is smaller than every positive number, and hence is
zero.
Since f" is zero everywhere, f' must be a constant a, so f'(z) = a. To show that this implies f(z) = az, let g(z) = az and consider the difference h(z) = f(z)-g(z). We have h'(z) = f'(z)-g'(z) = a-a = 0 so h(z) is a constant. However, h(0) = f(0)-g(0) = 0-0 = 0 (f(0)=0 because we know |f(0)| <=A 0 = 0). Therefore, h(z) must be the constant zero, showing that f(z)-g(z) = 0 for all z, and hence f(z) = g(z) = az.