# Methodology for evaluating Contour Integrals

1. Apr 22, 2012

### NewtonianAlch

1. The problem statement, all variables and given/known data
I'm a bit uncertain as to how to do these types of integrals.

Let γ be any contour from 1 - i to 1 + i. Evaluate the following:

∫ 4z^3 dz

3. The attempt at a solution

I did this in three different methods, two of them gave the correct answer, although this could just be a fluke.

I first parametrised the line going from (1 - i) to (1 + i):

z = (1 - i) + t(2i), dz = 2i

Using Maple I substitute this in for z, expanded it out, multiplied by 4 and multiplied by dz. Integrated the result and substituted for t = 1, and t = 0, the result is 0, which is correct.

I tried a different parametrisation method which a book uses, the line was parametrised as simply z = 1 + it, since it's only changing through the y-axis. This method gives an incorrect answer, my question here is why are these two parametrisations giving different answers? Clearly they are different, but does it matter how it is being parametrised?

My last method which the lecture notes uses at times is a lot simpler.

The integral goes from (1 - i) to (1 + i). Integrate 4z^3.dz to get z^4, substitute in the bounds of the integral to get (1 + i)^4 - (1 - i)^4 => 0

So the first and last method gave the same answer, are these both viable methods? What went wrong with the variant of the parametrisation? My guess is that the first method although correct is a bit computationally intensive, in fact I didn't bother to expand it myself because it was time-consuming and there was a chance of errors kicking in, so I just used MAPLE to check whether it was going to be correct or not.

Last edited: Apr 22, 2012
2. Apr 22, 2012

### LCKurtz

I get the same answer for the second method. Did you use the correct limits t=-1 and t=1?

3. Apr 22, 2012

### NewtonianAlch

Ah...I completely forgot it was varying from - 1 to 1 and not 0 to 1, that's where I went wrong.

Thanks for pointing that out.

4. Apr 22, 2012

### hamsterman

It should be noted that these three aren't really different methods. They're just different substitutes. The point of substitutions is to simplify, so only the third (no substitution) makes much sense.

5. Apr 22, 2012

### NewtonianAlch

Fair point. I didn't quite realise it at first since I'm new to this. I was just fiddling with different examples to see what would happen, but yes, the third makes the most sense here.

Thanks!