Section 4.4

Question 10: (a) The numerator and denominator are analytic, and the denominator is zero exactly when z is +5i or -5i, so the domain of analyticity is C - {5i, -5i}. Thus the function is analytic on and inside the contour |z| = 2, and Cauchy's theorem applies. (Note that the domain of analyticity is not simply connected, but we could make a smaller domain, e.g. {|z| < 3} which is simply connected and still contains the contour.

(c) Similar reasoning to (a), except that the two bad points are 3+i and 3-i.

(e) sec(z/2) = 1 / cos(z/2); both the numerator and denominator are analytic, and the denominator is zero only when z = +-pi, +-3pi, +-5pi, ... (as can be seen by e.g. writing cos(z/2) = (exp(iz/2) + exp(-iz/2))/2 and solving). So the domain of analyticity is all the points which are not odd multiples of pi, so the function is analytic on and inside the contour.

Question 13: Although this question precedes the Cauchy integral formula, the CIF is by far the best way to compute these integrals. (I don't know why the question is placed where it is). For (a), apply CIF with f(z) = 1/(z+i) and z_0 = i to get 2pi i 1/(i+i) = pi. For (b), split Gamma_2 into two contours, one circling i, and one circling -i. The former contour gives an integral of pi, and the latter gives an integral of -pi, for a total integral of 0. For (c), the right-hand loop has a zero integral by Cauchy's theorem, so the only thing which contributes is the left-hand loop, which has an integral of -pi by the CIF.

Section 4.5

Question 2: We may assume of course that Gamma is positively oriented. Let z_0 be any point inside Gamma. Applying the CIF to both f and g, we see that

f(z_0) = 1/(2pi i) int_Gamma f(z)/(z-z_0) dz

g(z_0) = 1/(2pi i) int_Gamma g(z)/(z-z_0) dzBut the right hand sides are equal since f(z)=g(z) on Gamma. Thus the left hand sides are equal as well, and we are done.

Question 3: (b) Apply the CIF with f(z) = z exp(z)/2 and z_0 = 3/2 to obtain 2 pi i f(3/2) = 3 pi exp(3/2) i/2.

(c) Apply the CIF with f(z) = cos z / (z^2 + 9) [which is analytic on and inside the contour] and z_0 = 0 to obtain 2 pi i cos 0 / (0^2 + 9) = 2 pi i/9.

(e) Apply the GCIF with f(z) = exp(-z), z_0 = 0, and m=2 to obtain 2 pi i f'(0)/1! = -2 pi i.

Question 4: (a) Apply the GCIF with f(z) = (z+i)/(z+2), z_0 = 0, and m=2 to obtain 2 pi i f'(0)/1! = 2 pi i (1/2 - i/4) = pi/2 + pi i.

(b) Apply the CIF with f(z) = (z+i)/z^2 and z_0 = -2 to obtain 2 pi i (-2 + i)/(-2)^2 = -pi/2 - pi i.

(c) Apply Cauchy's theorem to obtain 0.

You should be able to see why we have to use a different formula for each situation; the preconditions for each of the Cauchy theorems are very important.

Question 5: (a) 1 is inside the ellipse C, so the CIF applies and we obtain G(1) = 2 pi i (1^2 - 1 + 2) = 4 pi i.

(b) By similar reasoning to (a) we have that G(z) = 2 pi i (z^2 - z + 2) for all z in C. (When z is outside C, G(z) = 0. Why?) So by differentiating this at i, we get G'(i) = 2 pi i (2i - 1) = - 4pi - 2 pi i. Alternatively, we can differentiate the formula for G(z) under the integral sign and then use the GCIF.

(c) Differentating our formula for G(z) in (b) twice we get that G''(z) = 4 pi i when z is inside C, so G''(-i) = 4 pi i. (What would G''(5) be?)

Question 7: The inner loop gives a contribution of 0 by Cauchy's theorem, so we can just loop at the outer loop, which is a simple closed positively oriented contour. We apply GCIF with f(z) = cos z / (z-3), which is analytic inside the contour, z_0 = 0, and m=2 to obtain 2 pi i f'(0) = 2 pi i (0/(-3) - 1/(-3)^2) = - 2 pi i /9.

Question 8: Apply the GCIF to the contour

z(theta) = z_0 + r exp(i theta), 0 <= theta <= 2piwith the given function f(z) and the given point z_0, and use the parameterization z = z(theta), dz = r i exp(i theta) d theta, to write the resulting contour integral as a real integral.

Section 4.6

Question 5: If Re f(z) <= M for all z, then |e^f(z)| <= e^M for all z. Thus e^f is both entire and bounded, hence constant by Louiville's theorem. Differentiating with respect to z, we get that f'(z) e^f(z) = 0 for all z. e^f(z) is never zero, so we can cancel it and conclude that f'(z) = 0, i.e. f is constant.