Problem: Let $g(z)$ be an analytic function in a neighborhood of $z_0\in \mathbb{C} $. Suppose that $g\left( z_0 \right) =0$. Prove that $\frac{g(z)}{z-z_0}$ extends to an analytic function at $z=z_0$.
Solution: Given that the function $g$ is analytic in a neighborhood of $z_0$. Consider the following function: \[ f(z) = \begin{dcases} \frac{g(z)}{z-z_0}, &\text{ if } z\neq z_0 ;\\ g^\prime (z_0), &\text{ if } z=z_0. \end{dcases} \] Since, $g$ is analytic function in a neighborhood of $z_0$, the function $f$ will also be analytic in the same neighborhood. We need to check whether this is analytic at $z_0$. We recall here Riemann's removable singularity theorem.
In order to prove that the function $f(z)$ is analytic $z_0$, we need to check if it is continuous at $z=z_0$. Note that \begin{align*} \lim_{z \to z_0} f(z) & = \lim_{z \to z_0} \frac{g(z)}{z-z_0} \\ & = \lim_{z \to z_0} \frac{g(z)-g(z_0)}{z-z_0} \\ & = g^\prime (z_0)\\ & = f(z_0). \end{align*} Therefore, $f$ is continuos at $z_0$ and hence $f$ is an analytic extension of $\frac{g(z)}{z-z_0}$ at $z=z_0$.