Show $\int_{0}^{2\pi}\frac{1}{2-e^{it}}\, dt = \pi$ using the Cauchy Integral Formula.

Question

Using the Cauchy Integral Formula show $$\int_{0}^{2\pi}\frac{1}{2-e^{it}}\, dt = \pi$$

I am fairly sure I can do this with the residue theorem (as I did complex analysis two years ago but I've forgotten everything), but I want to do it using just the Cauchy Integral Formula because that is apparently possible. I'm sure it's simple but I have forgotten it all. I was trying to do keyhole contour arguments but I always end up needing to using the residue theorem ... how do I do it using just Cauchy integral formula and elementary arguments?

Answer 1

Hint: Observe we have \begin{align} \int^{2\pi}_0 \frac{1}{2-e^{it}}\ dt = \frac{1}{i}\int_{|z|=1}\ \frac{dz}{z(2-z)} = \frac{1}{2i}\int_{|z|=1} \frac{1}{z}+\frac{1}{2-z}\ dz \end{align} then use Cauchy integral formula + Cauchy's Theorem.

Answer 2

If no restriction of using CIF then one can use Cauchy's Residue theorem also f(z)=1/i[z(2-z)], hence desired contour integral along mod z=1 is 2πi[Res (f(z)) at 0]=2πi[1/2i]=π