We need to integrate the following:
$$\oint_C \frac{1}{z-a} \ dz$$
Where $|a| < 1$, and $C$ the unit circle ($e^{it} \ | \ t \in [0,2\pi] $) . My idea was to find a geometric series, but I don't know how to find it for a function with two variables. Or does somebody have a better idea?
Best Answer
I'm sure later in your studies you will do this via the Cauchy Integral Formula, or via Cauchy's Residue Theorem or the like. For an elementary solution avoidng these, write $$\frac1{z-a}=\frac1{z}\frac1{1-a/z}=\sum_{n=0}^\infty\frac{a^n}{z^{n+1}}$$ (geometric series). This series is uniformly convergent over the unit circle, so we may interchange summation and integration to get $$\int_C\frac{dz}{z-a}=\sum_{n=0}^\infty a^n\int_C\frac{dz}{z^{n+1}}.$$ Then $$\int_C\frac{dz}{z^{n+1}}=i\int_0^{2\pi}e^{-int}\,dt$$ which is $2\pi i$ for $n=0$ and zero otherwise.