In the following problem I am going to use terms (denoted with quotation marks) I couldn't properly translate - if you happen to know the particular terms please feel free to edit them. Hopefully you will, using my solution, understand what am I referring to. Moreover this is a preparation for a larger problem about the error function.
Sketch the "lines of constant absolute value and lines of constant phase" of $\mathbb C\ni z\mapsto \exp(-z^2)$.
Let $z=a+\mathrm ib$ be our complex number then
$$\exp(-(a+\mathrm ib)^2) = \exp(-a^2-2\mathrm iab+b^2) = \exp(-a^2+b^2)\exp(-2\mathrm iab)=R\exp(\mathrm i\varphi).$$
In the first case we are interested in all complex numbers $z=a+\mathrm ib$ which have the same constant absolute value $|\exp(-z^2)|=\exp(-a^2+b^2)=R$. It is fairly obvious that two possible lines are described by $a=b$ and $a=-b$. There are definitely more of those lines but I had no good idea how to find them hence I did ask WolframAlpha which yields
Is there any approach to find some of those lines other that just trial-and-error by letting $a$ be some specific value and trying out to find suitable $b$ s.t. $\exp(-a^2+b^2)$ is constant?
Similiarly one can argue about the constant phase $\operatorname{Arg}(\exp(-z^2))=-2ab = \varphi$ which is very similiar to above with $a=0,b$ arbitrary and the other way round as two trivial solutions. The others apparently being the following ones
Once again: how may I properly find those lines?
For the first part, note that: $$\exp(-a^2+b^2)=R \Rightarrow b = \pm \sqrt{a^2 + \ln(R)}.$$ For each value of $R>0$, you get two functions of $b$ in terms of $a$. In particular, $R=1$ gives you the straight lines you mentioned above.
Recarding the second case, you can proceed analogously: $$-2ab = \varphi \Rightarrow b=-\frac{\varphi}{2a}.$$ With $\varphi>0$ you obtain the curves on the first and third quadrants; $\varphi=0$ gives you the straight lines and $\varphi<0$, the curves on the second and fourth quadrants.