Proving that $\DeclareMathOperator{\Re}{Re}\DeclareMathOperator{\Im}{Im}|e^{z^2}| = e^{(\Re z)^2-(\Im z)^2}$

Question

I want to show that: $|e^{z^2}| = e^{(\Re z)^2-(\Im z)^2}$.

$|e^{z^2}| = |e^{(x+iy)^2}| = |e^{x^2+2xiy+(iy)^2}| = |e^{x^2+2xiy+i^2y^2}| = |e^{x^2-y^2+i2xy}| = |e^{x^2-y^2}| = e^{(\Re z)^2-(\Im z)^2}$.

Using $i^2 = - 1$ and |$e^{ix}| = 1$.

Is this correct?

Answer 1

Yes, what you did is correct. $$z^2 = (x + iy)^2 = x^2 - y^2 + 2xy i$$ Then $$e^{z^2} = e^{ x^2 - y^2 + 2xy i} = e^{x^2 - y^2}.e^{2xyi}$$ So $$\vert e^{z^2} \vert = \vert e^{x^2 - y^2}\vert .\vert e^{2xyi}\vert $$ But $$\vert e^{2xyi}\vert =1$$ So $$\vert e^{z^2} \vert = \vert e^{x^2 - y^2}\vert. $$ But $$ e^{x^2 - y^2} \in \mathcal{R}^+$$ So $$\vert e^{z^2} \vert = e^{x^2 - y^2} $$ where $x = Re(z)$ and $y = Im(z)$.

Answer 2

Yes it is correct indeed from here we have

$$|e^{x^2+2xiy+i^2y^2}| = |e^{x^2-y^2}||e^{i2xy}|=|e^{x^2-y^2}|$$

since $\forall \theta$ we have $|e^{i\theta}|=1$.