Absolute value of complex $\exp(z^2)$

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This question is about how the absolute value function works with the complex exponential.

We have to determine, what $$|\exp (z^2)|$$ is. Since we know that $$z=x+i y$$ and $$|\exp(z)| = \exp(Re(z))$$, after some calculation, it arises that $$|\exp(z^2)| = \exp(x^2 - y^2)$$.

Does the absolute value of the left side of the equation influence the right side? How I take it, since $$|\exp(z)| = \exp(Re(z))$$ i.e. it doesn't. But as I said, we're unsure.

P.S: Yes, I know that the right side still needs to be calculated further, but right now, I'd primarily like to know, how the abs. value works in this situation.

• I don't understand the question. Does the absolute value on the LHS influence the RHS? What do you mean? I mean, yes, if it weren't for the absolute value, both the sides would be complex in general, and not real. – Paulo Mourão May 24 at 22:49
• Are you asking how to deduce the equation $\left| \exp(z) \right| = \exp(Re(z))$? – Lee Mosher May 24 at 22:54
• @PauloMourão Yes, kind of like that. See, my fellow student's calculation led to exp($x^2$ +$y^2$), which they justified with the absolute value turning the minus to a plus. But I think the RHS is unaffected by the abs.value. I'm sorry, if I'm wording this confusingly. – Gandeon May 24 at 23:06
• The absolute value does, of course, influence the right hand side, but it does so precisely in the way that you showed in your question. – Paulo Mourão May 24 at 23:22

Not many calculations, actually. You know that, for every $$w\in\mathbb{C}$$, $$\lvert\exp(w)\rvert=\exp(\operatorname{Re}(w))$$ If $$z=x+yi$$ and $$w=z^2=(x^2-y^2)+2xyi$$, you immediately get $$\lvert\exp(z^2)\rvert=\exp(x^2-y^2)$$ You might write this as $$\lvert\exp(z^2)\rvert=\exp(\operatorname{Re}(z)^2-\operatorname{Im}(z)^2)$$ in order to express the result only in terms of $$z$$.