Q. Find the uniform convergence region of a following complex function sequence
$$f_n(z)= \frac{z}{z^2+n^2}$$
I want to know how to find the uniform convergence region of a complex function. It is simple when z is real, but it is very difficult when z is a complex number because of the imaginary part of z. Finally, I want to know a strategy for finding a uniform convergence region of a general complex function.
Let $f_n(z)=\frac{z}{z^2+n^2}$. Evidently, the domain for $f_n(z)$ is $\mathbb{C}\setminus \{\pm in\}$.
Now, note that the sequence $f_n(z)$ converges pointwise with $\lim_{n\to \infty}f_n(z)=0$. Moreover, if $|z|\le R$, then we see that for $n>R$
$$\begin{align} |f(z)|&=\left|\frac{z}{z^2+n^2}\right|\\\\ &\le \frac{|z|}{\left|n^2-|z|^2\right|}\\\\ &\le \frac{R}{n^2-R^2}\to 0 \end{align}$$
So, we see that $f_n(z)$ converges uniformly on any closed disk $|z|\le R$.
We can extend the region for which $f_n(z)$ converges uniformly. Note that if $\left|\text{Im}(z)\right|\le R$, then for $n>4R$
$$\begin{align} |f(z)|&=\left|\frac{z}{z^2+n^2}\right|\\\\ &=\frac{|z|}{\sqrt{(|z|^2+n^2)^2-4n^2\left(\text{Im}(z)\right)^2}}\\\\ &\le \frac{|z|}{|z|^2+\frac12n^2}\\\\ &\le \frac{1}{\sqrt{2}\,n}\to 0 \end{align}$$
And we see that $f_n(z)$ converges uniformly in any region for which $\left|\text{Im}(z)\right|\le R$.