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).
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
(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.
(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)
1/f(z) is entire and bounded, hence by Louville's theorem it is constant
Hence f(z) is constant.
(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).
(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) dxSymmetry (integrand is even)
int_{gamma_1} f(z) dz = int_{epsilon}^{R} (ln|x| + i pi)/(x^2 + 4) dxPutting 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)).