How to find the radius of convergence for $\sum_{n=0}^{\infty}{a^{n^2}}\cdot z^n$ with $a\in \mathbb{C}$?
I have attempted to use the "root test": $$ \frac{1}{\sqrt[n]{a^{n^2}}}=\frac{1}{a^n}$$ So it does converge for $a>0$, is that valid though since $a\in \mathbb{C}$?
By the Cauchy-Hadamard theorem,
$$ \frac{1}{R}=\limsup_{n\to\infty}\sqrt[n]{|c_n|},\quad c_n=a^{n^2},\tag{1} $$ where $R$ is the radius of convergence and we use the convention in this context that $\frac{1}{\infty}=0$ and $\frac{1}{0}=\infty$.
As you observed, $$ \sqrt[n]{|c_n|}=|a|^n. $$ Thus $$ \limsup_{n\to\infty}\sqrt[n]{|c_n|}=\lim_{n\to \infty}|a|^n=\begin{cases} 0,&|a|<1\\ 1,&|a|=1\\ \infty,&|a|>1. \end{cases} $$ Now you can apply (1) to find the radius of convergence, which depends on the absolute value of the fixed complex number $a$.