I have an integral I wish to evaluate and I wish to do so using Cauchy's theorems for complex analysis. I haven't been using these theorems for very long so I i might be using them incorrectly so please bare with me, in advance.
evaluate $\int_{C(0,1)} \frac{1}{(z-a)(z-b)}dz$
a) if $|a|,|b|>1$
this implies that $f(z)= \frac{1}{(z-a)(z-b)}$ is holomorphic on all of C
$\therefore$ by Cauchy's integral theorem $\int_{C(0,1)}\frac{1}{(z-a)(z-b)}=0$
b)$|a|<1, |b|>1$
here we note that $f(z)=\frac{1}{z-b}$ is holomorphic on all of C
$\therefore f(a)= \frac{1}{2\pi i}\int_C \frac{f(z)dz}{z-a} \Rightarrow \int_C \frac{1}{(z-a)(z-b)}dz=2\pi if(a)$
c)$|a|,|b|<1$ here I'm not sure wht to do as I cant see a way to decompose this into a holomorphic function. Any hints for this part would be much appreciated. Thank you all in advance :)