I have to find the region of convergence and analyze the the pointwise and uniform convergence of the following sequence: $$\psi_n(z)=\frac{e^{-inz^2}}{(n+1)\sqrt{n}}$$
What I did:
What I did first was to find the real and imaginary part of $\psi_n(z)$: $$Re(\psi_n(z))=\frac{e^{2xy}\cos n(x^2-y^2)}{(n+1)\sqrt{n}}$$ $$Im(\psi_n(z))=-\frac{e^{2xy}\sin n(x^2-y^2)}{(n+1)\sqrt{n}}$$ It's easy to see that $Lim_{n\rightarrow \infty} Re(\psi_n(z))=0$ and $Lim_{n\rightarrow \infty} Im(\psi_n(z))=0$ for any pair $x$ , $y$ $\in R$. So the sequence $\psi_n(z)$ converges pointwise to $\psi(z)=0$ for any $z \in C$.
So, $$\left| \psi_n(z)-\psi(z) \right|=|\psi_n(z)|$$ $$\left| \frac{e^{-inz^2}}{(n+1)\sqrt{n}}\right|=\left| \frac{e^{2xy}}{(n+1)\sqrt{n}}\right|\rightarrow 0$$ When $n\rightarrow \infty$
Then $\psi_n(z)$ converges uniformly for $z\in C$. Is this process correct? Do I need more mathematical rigor?Thanks.
What you did is not correct. In order to see why, let $z$ be a square root of $-i$. Then $\psi_n(z)=\frac{e^n}{(n+1)\sqrt n}$ and therefore the limit $\lim_{n\to\infty}\psi_n(z)$ doesn't exist (in $\mathbb C$).
Note that $\bigl\lvert e^{-inz^2}\bigr\rvert=e^{\operatorname{Re}(-inz^2)}=e^{n\operatorname{Im}(z^2)}$. So, if $\operatorname{Im}(z^2)\leqslant0$, $\lim_{n\to\infty}\psi_n(z)=0$. Otherwise, the limit $\lim_{n\to\infty}\psi_n(z)$ doesn't exist (again, in $\mathbb C$).
It also follows from this that the convergence is uniform in $\{z\in\mathbb{C}\,|\,\operatorname{Im}(z^2)\leqslant0\}$.