Section 7.4

Question 1: You can apply the standard precalculus techniques for inverting a function to do this problem. For example, you can solve the equation w = f_1(z) for z to obtain the inverse mapping z = f_1^{-1}(w). In this case, though, there is a substantial shortcut available, because of the identity f_1(z) = f_2(z+2). This means that

f_1^{-1}(f_2(z)) = f_1^{-1}(f_1(z-2)) = z-2.

Question 14: We first consider the simpler problem of whether the transformation (12) maps the unit circle |z|=1 to the unit circle |w|=1. In other words, we want to show that |w| = 1 whenever z is of the form z = exp(i beta). (Actually, because (12) is a Mobius transform, one only needs to check this claim for three values of z, such as z = 1, -1, and i, but for this question this trick won't save too much effort). If z is of the above form, then w = exp(i theta) (exp(i beta) - alpha) / (alpha* exp(i beta) - 1), where alpha* is the conjugate of alpha, and it is a routine matter of algebra to check that ww* = 1, so that |w| = 1, as desired.

To finish the question we have to show that the unit disk |z| < 1 maps to the unit disk |w| < 1. Since we knw the unit circle maps to itself, the inside of the unit circle must map to either the inside of the unit circle or the outside of the unit circle. So it suffices to check what happens to just one point of the unit disk, say the origin. This point maps to exp(i theta) alpha, which has magnitude less than one by assumption, so the unit disk must map to the unit disk.

Section 3.1

Question 5: (a) exp(2 + i pi/4) = e^2 (1/sqrt(2) + i/sqrt(2)) = e^2/sqrt(2) + i e^2/sqrt(2).

(c) sin(2i) = (exp(i 2i) - exp(-i 2i))/2i = (exp(-2) - exp(2)/2i = i (e^2 - e^{-2})/2.

(e) sinh(1 + pi i) = (exp(1 + pi i) - exp(-1 -pi i))/2 = (- exp(1) + exp(-1))/2 = - sinh(1).

Question 9(b): The normal rules of calculus apply, so the derivative is

- 2 sin(2z) - i sin(1/z) / z^2.

Question 11: (cos z / e^z) is entire, since cos z is entire and e^z is entire and never equal to zero. This means that Re(cos z / e^z) is harmonic in the entire plane, because the real part of an analytic function is automatically harmonic.

Question 12: (c) cosh(z1 + z2) = cos(iz1 + iz2) = cos(iz1) cos(iz2) - sin(iz1) sin(iz2) = cos(iz1) cos(iz2) + (-i sin(iz1)) (-i sin(iz2)) = cosh(z1)cosh(z2) + sinh(z2)sinh(z1)

Section 3.2

Question 1: (b) log 1-i = ln |1-i| + i arg(1-i) = ln sqrt(2) + i (-pi /4 + 2k pi), where k ranges over the integers.

(d) Log(sqrt(3)+i) = ln|sqrt(3)+i| + i Arg(sqrt(3)+i) = ln 2 + i pi/6.

Question 5: (a) Since e^z = 2i, z must be a logarithm of 2i, so z is one of the elements of log 2i = ln 2 + i (pi + 2k pi). So the general solution is z = ln 2 + i(pi + 2k pi).

(b) Since i pi/2 is a logarithm of z^2 - 1, z^2 - 1 must be exp(i pi/2) = i, so z^2 = 1 + i, so z = (1+i)^{1/2} = (sqrt(2) exp(i pi/4))^{1/2} = 2^{1/4} exp(i pi/8) or 2^{1/4} exp(i 9pi/8). One can verify that both of these numbers are solutions to the equation.

(c) Write w = exp(z), so that we have w^2 + w + 1 = 0. From the quadratic formula we have w = (-1 +- sqrt(-3))/2 = -1/2 +- i sqrt(3)/2. Taking logs as in (a) we see that z must be a logarithm of either -1/2 + i sqrt(3)/2 or -1/2 - i sqrt(3)/2, which after some computation becomes z = i (2 pi/3 + 2k pi) or i (-2 pi/3 + 2k pi) where k is an arbitrary integer.

Question 9: This function is only non-analytic when 4+i-z is zero or a negative real. In other words, f is analytic everywhere except at 4+i and on the rightward ray emenating from 4+i. Away from this discontinuity, the derivative of Log(z) is 1/z, just as in real-variable calculus, so we can use the chain rule and find that the derivative of f is -1/(4+i-z) away from 4+i and the rightward ray emenating from 4+i.

Question 13: A little bit of playing around shows that L_alpha (2z-1) is a branch of log(2z-1) everywhere in the complex plane except at 1/2 and on the ray at angle alpha from 1/2. Thus, we may take L_{-pi}(2z-1) = Log(2z-1) for (a), L_0(2z-1) for (b), and L_{pi/2}(2z-1) for (c). These aren't the only choices; for example, we could also pick L_{2 pi}(2z-1) for (b).

Section 3.3

Question 11: This question refers to the complex sine and cosine; it is not quite the same equation as sin x = cos x. If you use the real-valued trig functions to solve this problem, it turns out that you get the right answer, but for the wrong reasons. Solving sin x = cos x using the real-valued trig functions will only give you the real solutions to the equation; it doesn't say whether there are any extra complex solutions. For example, just because sin x = 2 has no solutions in real numbers doesn't mean that sin z = 2 has no solutions. So to solve it we should use the complex sine and cosine. Using the definition, we get

(e^{iz} - e^{-iz})/2i = (e^{iz} + e^{-iz})/2,
which after a bit of rearranging becomes
(1-i) e^{iz} = (1+i) e^{-iz}.
To solve for z we combine the two exponentials:
e^{2iz} = (1+i)/(1-i) = i
take logarithms:
2iz = log i = pi i/2 + 2 k pi i
and divide by 2i to get
z = pi/4 + k pi.
Yes, this is the same answer as in the real-variable case, but you can't take that for granted; you have to show it on a case by case basis.