How to calculate $I= {\int_0^{2\pi} \frac{dt}{z-e^{it}}} \,(z\in \mathbb{C}, |z|>1)$ using Riemann sums?

Question

I came to study the Riemann sums and face to the following exercise. I have got no idea about how to solve it:

$I= {\displaystyle \int_0^{2\pi} \frac{dt}{z-e^{it}}} \quad (z\in \mathbb{C}, |z|>1)$

Do you have any clues to start?

Answer 1

To simplify a little, put $J=\int_0^1\frac{dt}{z-\exp(2i\pi t)}$. Then your integral is $2\pi J$.

The Riemann sum to consider is $$S_n=\frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{z-\exp(2i\pi k/n)}$$ Let $P(z)=z^n-1=\prod_{k=0}^{n-1}(z-\exp(2i\pi k/n))$. Then we have $$\frac{P^{\prime}(z)}{P(z)}=\sum_{k=0}^{n-1}\frac{1}{z-\exp(2i\pi k/n)}=nS_n$$

It is now easy to find the limit of $S_n$.