Question 1: (a) The left hand side is |1+i|^2 = 1^2 + 1^2 = 2, while the right side is |1-i|^2 + |2i|^2 + 2 Re ( (1-i) (-2i)) = 1^2 + 1^2 + 2^2 + 2 Re( -2 -2i) = 6 - 4 = 2.

Alternatively, you can just do part (b) first and cite this as a special case.

(b) The left hand side is |z+w|^2 = (z+w)(z+w)* = zz* + zw* + wz* + ww* = |z|^2 + |w|^2 + (zw*) + (zw*)* = |z|^2 + |w|^2 + 2 Re (zw*).

Other solutions (e.g. using Cartesian co-ordinates) are certainly possible.

Question 2. (a) You can either check directly that u satisfies Laplace's equation, or (if you are a lucky guesser) you can check that u is the real part of cos z, and since cos z is obviously entire, u must be harmonic on the complex plane.

(b) In order for f = u + iv to be entire, u and v must satisfy the Cauchy Riemann equations. If you work them out, you obtain

dv/dy = sin x cosh y, dv/dx = cos x sinh y.
By inspection one can see that v(x,y) = sin x sinh y is a solution to the Cauchy-Riemann equations, and since u and v have continuous partial derivatives, u+iv is consequently entire. (It turns out that u+iv is equal to cos z). If you can't see v directly, you can integrate one of the equations and substitute the resulting equation into the second, to solve for v, and thus find f.

Question 3

(a) Solution A: we pick three points, say 1, -1, and i on the unit circle, and see where they map to under this map; in this case they map to 1/2, infinity, and 1/2 + 1/2 i. Since a circle must map to either a line or a circle, and only the lines go through infinity, the unit circle must map to the line through 1/2 and 1/2 - i/2, i.e. the line {Re z = 1/2}. To find out what happens to the inside of the circle we test it at one point; 0 goes to 0, so the closed unit disk goes to the closed half-plane to the left of {Re z = 1/2}, i.e. {Re z <= 1/2}.

Solution B: Since z/(z+1) = 1 - 1/(z+1), we can think of this map as a translation z_1 = z+1, followed by an inversion z_2 = 1/z_1, a reflection z_3 = -z_2, and a translation z_4 = 1 + z_3. Under these transformations the closed unit disk becomes (in turn) the disk of radius 1 centered at 1, the half-plane to the right of Re z = 1/2, the half-plane to the left of Re z = -1/2, and the half plane to the left of Re z = 1/2.

(b) We solve for the equation z/(z+1) = 2i. One way to do this is to write out z/(z+1) = 1 - 1/(z+1) as in solution B and work backward:

 1 - 1/(z+1) = 2i
 -1/(z+1) = 2i - 1
 1/(z+1) = 1 - 2i
 z+1 = 1/(1 - 2i) = 1/5 + 2/5 i
 z = 2/5 i - 4/5
Or you can use whatever algebraic manipulation you wish; they all give the same answer, of course.

(c) Incidentally, in mathematics a semi-circle is half of a circle - in other words, a 1-dimensional set. A half of a disk would be called a semi-disk, not a semi-circle.

Using 1(a), Solution A (or Solution B), we see that the right semi-circle maps to the interval between 1/2 - i/2 and 1/2 + i/2. So to move this to the interval we want we can shift left by 1/2 and then dilate by 2, giving us the transformation 2(z/(z+1) - 1/2) = (z-1)/(z+1). Alternatively, you can use the cross-ratio formula in the textbook.

Question 4 (a) (-1)^i = exp(i log (-1)). log (-1) = i(pi + 2 k pi), so the possible values of exp(i log (-1)) are exp(-pi - 2kpi), i.e. exp(-pi), exp(+pi), exp(-3pi), etc.

(b) exp(i L_0(z)) is a branch of z^i = exp(i log(z)) which is analytic everywhere except the positive real axis, so in particular it is analytic on the left half plane.

(c) we could try the same function f(z) = exp(i L_0(z)). However, if we compute f(-1), we get exp(i L_0(-1)) = exp(i(i pi)) = exp(-pi), which is not what we want. Since we want f(-1) to equal exp(5 pi), we should try a different branch of z^i, such as exp(i L_alpha(z)). We want to choose alpha such that i L_alpha(-1) = 5 pi, or in other words, i(i Arg_alpha(-1)) = 5 pi. This means we want Arg_alpha(-1) to equal -5 pi, so we can choose alpha = -6 pi. Now we have f(-1) = exp(5 pi) and f is analytic everywhere except when z has phase -6 pi, i.e. everywhere except the positive real line. So in particular it is analytic on the left half plane.

Question 5 (a) The Cauchy Riemann equations for f are 0 = x, and -2y = y. Since the partial derivatives of f are continuous, the function is differentiable exactly when the Cauchy Riemann equations are satisfied i.e. when x = y = 0.

(b) In order for a function to be analytic at a point, it has to be differentiable on some ball around that point. However, this function is only differentiable at a single point, and so cannot be differentiable on any ball, so the function is nowhere analytic.