Let $\gamma: [-\pi,\pi] \rightarrow \mathbb{C}$ be given via $\gamma(t) = 3e^{it}$. I am supposed to find $\int_\gamma \frac{1}{z^2-4} dz $
So I am unsure if I am supposed to use a theorem from Complex Analysis or brut force it. (this is my main question really)
I tried to find an anti-derivative for $f(z) = \frac{1}{z^2-4}$ Then just plug in my limits but it gets messy,
Then I tried
$\int_\gamma \frac{1}{z^2-4} dz = \int_{-\pi}^{\pi} \frac{1}{(3e^{it}+2)(3e^{it}-2)}dt$ And I get stuck, unfortunately I took the calculus series about a decade ago so my integration is rusty, I am enrolled in a complex analysis course at CSULB. Thank you in advance, just need like a hint really.
The poles inside the contour are $2$ and $-2$. The residues at these points are $\frac 1 4$ and $-\frac 1 4$. Hence (by the Residue Theorem) the integral is $2\pi i (\frac 1 4-\frac 1 4)=0$.