This is probably something quite simple, but I can't get my head around this. Consider the following holomophic funtion : $$f(z)=1-|z|^2$$ This function is zero all along the unit circle C.
So I would say that for every complex number $a$ within the open unit disc, $$\int_{C}\frac{f(z)}{z-a}dz=0$$
But according to Cauchy's Integral formula, that would mean that $f$ is $0$ everywhere inside the open unit disc, which is obviously false.
Am I missing something obvious here ?
Your function is not holomorphic. Try to think what it series might be.
More formally, zeros of a holomorphic function are isolated, which is not the case with your function.