Integral along closed contour

Question

In the Laurent series, the coefficient

$$b_n = \frac{1}{2\pi i}\int_C\frac{f(z)}{(z - z_0)^{-n + 1}}dz,\qquad\left(\, n = 1,2,\ldots\,\right)$$

collapses to zero when $f(z)$ is analytic in the neighbourhood of a point $z_0$ enclosed by a closed contour $C$.

Why isn't that the same for

$$a_n = \frac{1}{2\pi i}\int_C\frac{f(z)}{(z - z_0)^{n + 1}}dz,\qquad\left(\, n = 0,1,2,\ldots\,\right)\,?$$

Answer 1

If $f$ is analytic, the function $$\frac{f(z)}{(z-z_0)^{-n+1}} = f(z)(z-z_0)^{n-1}$$ is analytic at $z_0$ whereas the function $$\frac{f(z)}{(z-z_0)^{n+1}}$$ may not be analytic at $z_0$ (it will be analytic at $z_0$ if and only if $f$ has a zero at $z_0$ of order at least $n+1$).

Answer 2

By Cauchy Integral Formula, $f^{(n)}(z_0)=\frac{n!}{2\pi i}\int_C\frac{f(z)}{(z-z_0)^{n+1}} dz$.