# If $G_n(z)=(z-z_0)g_n(z)$ converges uniformly to $f(z)-f(z_0)$, what can I say about the convergence of $g_n(z)$?

Let $$G_n(z) = (z-z_0)g_n(z)$$ be analytic on a region $$\Omega$$ for $$n=1, 2,\dots$$, with $$z_0 \in \Omega$$ and suppose $$G_n(z)$$ converges uniformly to $$f(z)-f(z_0)$$ on $$\Omega$$ for some analytic function $$f$$. Does $$g_n(z)$$ converge uniformly to $$\frac{f(z)-f(z_0)}{z-z_0}$$ on $$\Omega$$? I know by the uniform convergence of $$G_n(z)$$ that if $$z\in \Omega$$ with $$|z-z_0|> \delta$$, then there exists $$N>0$$ such that $$|g_n(z)-\frac{f(z)-f(z_0)}{z-z_0}|< \epsilon$$ for all $$n \geq N$$, but I am unsure how to show the uniform convergence inside $$|z-z_0|<\delta$$. I wanted to show $$\{g_n(z)\}$$ was uniformly bounded on $$|z-z_0|<\delta$$ but have gotten stuck. There may be something simple that I am not considering; any help is appreciated.

$$\frac{f(z)-f(z_0)}{z-z_0}$$ has a removable singularity at $$z=z_0$$, so that the maximum modulus principle can be applied to $$h_n(z) = g_n(z)-\frac{f(z)-f(z_0)}{z-z_0} \, .$$
It follows that if $$h_n \to 0$$ uniformly on $$|z-z_0| \ge \delta$$ the same is true in the disk $$|z-z_0| < \delta$$.