Question 1: (a) f(z) = 2z/(z+1); f(0) = 2 * 0/(0+1) = 0; f(1) = 2 * 1/(1+1) = 1; f(-1) = 2(-1)/(-1+1) = infinity

(b) {z: Re(z) < 1} (The open half plane to the left of 1)

(c) (-1+2i)/5

(d) Simple pole at -1; removable singularity at infinity (Or: if f(infinity) is defined to be 2, then there is no singularity at infinity).

Question 2: (a) If 0 < |z| < 1:
1/z^3 - 1/z^2 + 1/z - 1 + z - z^2 + ...

If |z| > 1:

1/z^4 - 1/z^5 + 1/z^6 - ....

(b) Triple pole at 0; simple pole at -1; Res(0) = 1; Res(-1) = -1

(c) 0

Question 3: (a) Differentiable at (0,0); nowhere analytic

(b) No such function exists. (u is not harmonic).

(c) Errata for this question: It should be v, not u. Yes, a function does exist: h(z) = (-x^2-2xy+y^2) + i(x^2 - 2xy - y^2) = (i-1)z^2 will work.

Question 4: (a) 1/2 ln(2) + 3pi i/4 + 2kpi i (k is any integer)

(b) Everywhere except when Im(z)=1 and Re(z) >= 0 (i.e. the ray to the right of i)

(c) Many examples work, e.g. L_{-2 pi} (i - z). To see why this works, we compute

L_{-2 pi} (i - 1) = ln |i-1| + i arg_{(-2pi,0)] (i-1)
= ln sqrt(2) + i (-5 pi/4)

Question 5: If |f(z)| > M for all z, then |1/f(z)| < 1/M for all z Since f is never zero, and f is entire: 1/f(z) is entire

1/f(z) is entire and bounded, hence by Louville's theorem it is constant

Hence f(z) is constant.

Question 6: pi cos(1) / (3 e^3)
Question 7: pi sqrt(3)/3. Note: the calculations will be easier if you make the change of variables z = exp(2 i theta) instead of z = exp(i theta).
Question 8: (a) Simple zero at -pi/2, removable singularity at pi/2 (residue = 0)

(b) Essential singularity at 0 (residue = 1/2)

(c) Removable singularity at 0 (residue = 0); once singularity is removed, we have a simple zero

(d) Triple pole at 0 (residue = 2).

Question 9: (This was a toughie!)

(a) pi ln(2)/2 + i pi^2/4

(b) How big is L_{-pi/2}(z) / (z^2 + 4)?

L_{-pi/2}(z) = ln(|z|) + i Arg_{-pi/2} z has real part ln(epsilon), imaginary part between -pi/2 and 3pi/2, so

| L_{-pi/2}(z) | <= |ln(epsilon)| + 3 pi/2.
Also,
4 - epsilon^2 <= |z^2 + 4| <= 4 + epsilon^2.
Hence
L_{-pi/2}(z) / (z^2 + 4) <= (|ln(epsilon)| + 3 pi/2) / (4 - epsilon^2)
and
|integral| <= pi epsilon (|ln(epsilon)| + 3 pi/2) / (4 - epsilon^2)
As epsilon -> 0, the right hand side looks like epsilon |ln(epsilon)|, which goes to zero (L'hopital's rule).

(c) How big is L_{-pi/2}(z) / (z^2 + 4)?

L_{-pi/2}(z) = ln(|z|) + i Arg_{-pi/2} z has real part ln(R), imaginary part between -pi/2 and 3pi/2, so

| L_{-pi/2}(z) | <= ln(R) + 3 pi/2.
Also,
R^2 - 4 <= |z^2 + 4| <= R^2 + 4.
Hence
L_{-pi/2}(z) / (z^2 + 4) <= (ln(R) + 3 pi/2) / (R^2 - 4)
and
|integral| <= pi R (ln(R) + 3 pi/2) / (R^2 - 4)
As R -> infinity, the right hand side looks like ln(R)/R,, which goes to zero (L'hopital's rule).

(d) If z is on gamma_3, then z = x+i0 for some positive x

So L_{-pi/2}(z) = ln(x) + i0

int_{gamma_3} f(z) dz = int_epsilon^R  ln(x)/(x^2 + 4) dx

If z is on gamma_1, then z = x+i0 for some negative x

So L_{-pi/2}(z) = ln|x| + i pi

int_{gamma_1} f(z) dz = int_{-R}^{-epsilon}  (ln|x| + i pi)/(x^2 + 4)
dx
Symmetry (integrand is even)
int_{gamma_1} f(z) dz = int_{epsilon}^{R}  (ln|x| + i pi)/(x^2 + 4)
dx
Putting the two equations together we are done.

(e) The first integral is pi ln(2)/4 (it's half of the real part to the answer for (a)), and the second integral is pi/4 (it's 1/pi of the imaginary part to the answer for (a), by part (d)).