The maximum modulus of $e^{z^2}$ on the set $S=\{z\in \mathbb{C}: 0\leq Re(z)\leq1, 0\leq Im(z)\leq1\}$ is
My attempt: We know $|e^{z^2}|\leq e^{|z|^2}$ so maximum of $|z|=\sqrt{2}$ since $z$ can be $1+i$, so $|e^{z^2}|\leq e^{|z|^2}=e^2$, so $4$ is right? Is my solution correct? If it's not then how to solve this? Thanks.
You've shown that $e^2$ is an upper bound for $|e^{z^2}|$, but you haven't shown that $|e^{z^2}| = e^2$ for some $z \in S$. If you can find such a $z$, then $e^2$ would be the maximum of $|e^{z^2}|$ on $S$. However, if no such $z$ exists, then the maximum is smaller.
Hint: Let $z = x+iy$ where $x,y$ are real. Then, you have:
$e^{z^2} = e^{(x+iy)^2} = e^{x^2+2ixy+i^2y^2} = e^{x^2-y^2+i \cdot 2xy} = e^{x^2-y^2}e^{i \cdot 2xy}$.
Hence, $|e^{z^2}| = |e^{x^2-y^2}| \cdot |e^{i \cdot 2xy}| = e^{x^2-y^2} \cdot 1 = e^{x^2-y^2}$.
Now, what is the maximum of $e^{x^2-y^2}$ over the region $0 \le x \le 1$ and $0 \le y \le 1$?
