Removable singularity equivalence proposition,complex analysis

Question

Say we have an analytic f in a deleted neighbourhood of z0. Then f has a removable singularity(defined as the b coefficients of its laurent series are all zero) if and only if f can be extended to z0 so that it is analytic at z0. The => direction is easy.How is the <= direction proven? I am studying Marsden Hoffman which says it is obvious and gives no proof.

Answer 1

If it has analytic extension, $$b_n = \oint_C\frac{f(z)}{(z-z_0)^{-n+1}}dz = 0$$ by Cauchy's integral formula(note for $n\ge 1$, the integrand is still analytic).