Holomorphic 1/z

And it seems like I forgot at 2 in the exponent, but the result of the integral did not change. I'm a lucky guy!

What if I did this instead?
[3exp(it)-exp(-it)] /[3exp(it)+exp(-it)]
= [3exp(it)-exp(-it)]exp(-it) /[(3exp(it)+exp(-it))*exp(-it)]
= [3-exp(-2it)] /[3+exp(-2it)]
= - [-3+exp(-2it)] /[3+exp(-2it)]
= - [3+exp(-2it) -6] /[3+exp(-2it)]
= -1 -6/[3+exp(-2it)]

Then I'd end up with -2πi when I integrate. Does that mean the integral of -6i/[3+exp(-2it)] from 0 to 2π is 4πi?

That's a shame. Professors love interesting and curious students, and you do seem like one.

Unfortunately, my professors like students who can do the practice problems. I have finals on Monday and Tuesday. I'm pretty screwed.

The problem you run into when you integrate method 3 is not only with your parametrization, which you need to be more careful about, but also the fact that you never specify a branch cut. Method 1 sort of implies that you've done a branch cut, by integrating over [0,2π). But you don't even know it.

How do I select a branch cut? I can't even find anything online to visualize complex functions.

Ugh, I have Method 3 redone as an example for you at home. When I get my internet back up there I'll post it. Blame comcast. Wikipedia isn't that much help for this concept, if you ask me.

._.