# Homework Help: Complex integration

1. Sep 12, 2009

### oddiseas

1. The problem statement, all variables and given/known data

1)a Let D = C\{-i,i} and let γ be a closed contour in D. Find all the possible
values of :

(∫(1/(1+z²))dz around γ)

b)
If σ is a contour from 0 to 1, determine all possible
values of:
(∫(1/(1+z²))dz ( around σ)

2. Relevant equations

3. The attempt at a solution
for part a the denominator factorises to (z+i)(z-i). Now these points are NOT in the domain. If they are not in the domain does that mean i cant evaluate the integral or does it mean that i can evaluate the integral because the points i and -i are discontinuities, but since f(z) is analytic in the domain i can evaluate it and it should give me zero, by cauchys theorm. Anyway when i evaluate it i get the following:

((-1)/(2i)){∫(1/(z+i))-∫(1/(z-i)) and this gives me zero but i am not sure if i am right.

b)
If σ is a contour from 0 to 1, determine all possible
values of:
(∫(1/(1+z²))dz ( around σ)

I am not sure how to try this, or how to represent a contour from zero to 1. Should the contour be a quarter circle moving clockwise, or should it be a line on the real axis? or doesnt it matter what contour i use as long as the endpoints agree. When i calcvulated this using the partial fraction representation above i get zero again. So i think i am stuffing up somewhere. Can someone show me the procedure they use forevaluating these integrals

2. Sep 12, 2009

### Dick

Suppose your contour winds around one of the poles +/-i in a counterclockwise direction? Suppose it winds around twice. Suppose it winds around one clockwise? Suppose you don't wind around any poles? Etc etc. There are lots of possible values for both questions. Think about the Cauchy integral formula and the residue theorem.

3. Sep 12, 2009

### oddiseas

we havnt done the residue theorm. we are doing it at the end of next week;
I am thinking that if part a is equal to 0, then it can take on any 2pi multiple of this. Am i on the right trach?

4. Sep 12, 2009

### Dick

Roughly. Can you say why you would think that exactly, using the Cauchy integral formula, maybe?