 Solution

A simple zero of an analytic function has a local factorization $f(z)=(z-z_0)u(z)$, where $u$ is an analytic function and $u(z_0)=f^{\prime }(z_0)\ne 0$. Since $1/u$ is also an analytic function near $z_0$, its Taylor series givesThe nonzero coefficient of $(z-z_0{)}^{-1}$ proves that this is a simple pole, withMultiplying by the Taylor series $g(z)=g(z_0)+O(z-z_0)$ similarly givesThe assumption $g(z_0)\ne 0$ ensures that the pole is not canceled. In either formula the residue is precisely the coefficient of $(z-z_0{)}^{-1}$ in the Laurent series.