$f(z)$ is the branch $z^{-1 + i} = e^{(-1 +i)\ln{z}}$ such that $|z| > 0$ and $0 < arg(z) < 2\pi$.
I'm to integrate $f(z)$ over the contour $e^{i\theta}$ (just the unit circle).
... I have no idea what I'm supposed to be doing.
My thoughts:
$$f(\phi(\theta)) = e^{{(i \theta)(-1 + i)}}~\text{where}~\phi(\theta) = i\theta$$
So the integral is $$\int_0^{2\pi} f(\phi(\theta))\phi'(\theta)~~d\theta = i\int_0^{2\pi}e^{(i\theta)(-1+i)}~d\theta$$
But this gives me the incorrect answer.
I feel as if it's not so much that I've reached a point where I have to think carefully about the mathematics—just that I don't actually know what I'm supposed to be doing or looking for here. I can fiddle around with a lot of things—finding reasons why $z = -1$ or $z = 1$, but I don't actually know what I'm supposed to do with it all.
Could someone shed some light on this problem?
In your parametrization, $\phi(\theta)$ should be $e^{i\theta}$, not $i\theta$. So
$$\int_{|z| = 1} f(z)\, dz = \int_0^{2\pi} e^{(-1 + i)\log(e^{i\theta})} ie^{i\theta}\, d\theta.$$
For $\theta \in (0,2\pi)$, $\operatorname{arg}(e^{i\theta}) = \theta$, in which case $\log(e^{i\theta}) = \ln|e^{i\theta}| + i\operatorname{arg}(e^{i\theta}) = i\theta$. The integral now reduces to
$$\int_0^{2\pi} e^{(-1 + i)i\theta}ie^{i\theta}\, d\theta = i\int_0^{2\pi} e^{(-1 + i)i\theta + i\theta}\, d\theta = i\int_0^{2\pi} e^{-\theta}\, d\theta = i(1 - e^{-2\pi}).$$