Find the residue(s) of this function at each pole that lies in the contour?

Question

Going through past papers and found this residue question I can't do. The question asks you to find the residue at each pole that lies in the contour shown.

I've got as my answer for the poles but can't figure out how to move on from here :S

Answer 1

The poles are at the solutions to $z^5+32=0$, or equivalently, the solutions to $z^5=-32$.

Since $|z|^5 = |z^5| = |-32| = 32$, we have $|z| = 2$. So every pole will be of the form $z = 2\exp {i \theta}$. Then, $z^5 = 2^5\exp \left(5 i\theta\right) = -32,$ or $$\exp\left(5i\theta\right) = -1.$$

Since $\exp \left(i\pi +2ik\pi\right)= -1$ for integer values of $k$ then $5i\theta = i\pi + 2ik\pi$. Solving this, we get $\theta = \frac{\pi+2k\pi}{5},$ which leads to the unique values of:

$$\theta = \left\{\frac{\pi}{5},\frac{3\pi}{5},\pi,\frac{7\pi}{5},\frac{9\pi}{5}\right\}.$$

If you plot these, it becomes obvious that only the pole at $\theta=\frac{\pi}{5}$ lies within your contour, as long as $R > 2$.

Since this is a simple pole, we can easily evaluate the residue using $$\textrm{Res}\left(\frac{g(z)}{h(z)}; z=\alpha\right) = \frac{g(\alpha)}{h'(\alpha)},$$

so we get

$$\textrm{Res}\left(\frac{1}{z^5+32}; 2\exp\left(i \frac{\pi}{5}\right)\right)=\frac{1}{5\left(2\exp\left[i\frac{\pi}{5}\right]\right)^4}=\frac{1}{80\exp\left[i\frac{4\pi}{5}\right]}.$$

Simplify this, and you are done.