# Complex Integration

1. Oct 16, 2007

### strangequark

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

Compute the following integrals using the principle value of $$z^{i}$$

a.
$$\int z^{i} dz$$ where $$\gamma_{1}(t)=e^{it}$$ and $$\frac{-\pi}{2}\leq t \leq \frac{\pi}{2}$$

b.
$$\int z^{i} dz$$ where $$\gamma_{1}(t)=e^{it}$$ and $$\frac{\pi}{2}\leq t \leq \frac{3\pi}{2}$$

2. Relevant equations

3. The attempt at a solution

There is a "hint" with the problem that says one of the integrals is easier than the other.
I don't see why, for part a, I can't use a branch cut along the negative real axis, so that $$z^{i}$$ will be analytic along the path.
And for part b, i don't see why I can't simply use a different branch cut, say one along the positive real axis, so that $$z^{i}$$ will then be analytic along that path.

Then for each, I can just take the antiderivative:

$$F(z)=\frac{z^{i+1}}{i+1}$$

and plug in the end points...

If I do, I get:

a. $$\int z^{i} dz = \frac{i^{i+1}}{i+1}+\frac{(-i)^{i+1}}{i+1} = (\frac{e^{\frac{\pi}{2}}}{2}+\frac{e^{\frac{\pi}{2}}}{2}i)-(-\frac{e^{\frac{\pi}{2}}}{2}-\frac{e^{\frac{\pi}{2}}}{2}i ) = cosh(\frac{\pi}{2})+cosh(\frac{\pi}{2})i$$

and I will get the same thing for b.

Is there something I am missing? Do I ned to parameterize the integral or is what I am doing correct?

I'm thinking that because $$f(z)=z^{i}$$ is entire, and that the region in which the curve lies will be simply connected.... then the anitderiv exists and since $$i$$ is just a constant, then the primitive of $$f(z)$$ will be $$F(z)=\frac{z^{i+1}}{i+1}$$...