Modulus of $\exp(-(A+iB)^2)$

Question

I know modulus of $e^z$ is just $e^{\Re(z)}$, and that $|A+iB| \geq A^2$. What I want to know is if the following holds true $$ |e^{-(A+iB)^2}| \leq e^{-A^2} $$

Answer 1

In the case when $A, B \in \mathbb{R},$ the answer is: "No".
The given condition holds if $B=0$. Because, $|e^{-(A+iB)^2}| = |e^{-(A^2-B^2+2iAB)}| = |e^{(B^2-A^2-2iAB)}| = |e^{(B^2-A^2)}*e^{-2iAB}|= |e^{(B^2-A^2)}|*|e^{-2iAB}|$
Using $|e^{i*x}|=1$, for real $x,$ therefore $|e^{-2iAB}|=1,$ So $|e^{-(A+iB)^2}| = e^{(B^2-A^2)}. $ $|e^{-i*2AB}| \leq e^{-A^2} $ if and only if $e^{(B^2-A^2)} \leq e^{-A^2}.$
Therefore, $B^2-A^2 \leq -A^2,$ Hence $B^2 \leq 0.$
Since $B \in \mathbb{R},$ then $B=0.$ So, the given condition $|e^{-(A+iB)^2}| \leq e^{-A^2} $ is true if $B=0.$

Counter-example has been given by Kavi Rama Murthy in the comment section just below the question.