I'm trying to resolve the following exercise :
Show the existence of a $z_0 \in \mathbb{C}$ such that $z_0^2 = e^{- \vert z_0 \vert^2} (*)$.
More precisely, find the minimal radius of the disk containing the solutions of $(*)$.
My attempt is the following,
1) If $z_0$ is a solution, then $-z_0$ is too.
2) The solutions are real, for if we write $z \in \mathbb{C}$ in exponential form as $z = re^{i \theta}$, $r \in [0, \infty[$ and $\theta \in [0,2\pi[$, the equation $(*)$ becomes $r^2e^{2i \theta} = e^{-r^2} \in \mathbb{R}$ thus $e^{2i \theta} = 1$ (because the right hand term is positive).
3) One observes that $\displaystyle\lim_{\vert z \vert \rightarrow + \infty}\vert z \vert^2= +\infty$ and $\displaystyle\lim_{\vert z \vert \rightarrow + \infty}e^{- \vert z \vert^2}= 0$. Since $1< e^{-1}$, by the Intermediate Value Theorem, there exist a $r \in \mathbb{R}$ satisfying $r^2 = e^{-r^2}$, or equivalently there exists a $z_0 \in \mathbb{C}$ satisfying $(*)$.
In conlcusion, we know the existence of exactly two roots (since $r \mapsto r^2$ and $r \mapsto e^{-r^2}$ are increasing function when restricted to $r\in [0, \infty[$) and there is an equation giving the norm $r_0$ of the solution $z_0$.
Now, the questions are :
Is the argument valid ?
How can we solve $r^2 = e^{-r^2}$ ?
- Is there a more involved proof using tool from differential topology ? For instance, find the solutions of $(*)$ amounts to find the zeros of a limit of polynomials (in the real components) :
$$z^2 = e^{- \vert z \vert ^2} \quad \Longleftrightarrow z^2 -\left( 1 - \vert z \vert ^2 + \dfrac{\vert z \vert ^2}{2!} + \dots \right)=0$$
Any hint is welcome !
Your proof is fairly solid since having a real valued continuous function makes things easier. To solve for $r$,
$$r^2 = e^{-r^2} \implies r^2e^{r^2} = 1$$
At this point we apply the inverse of $xe^x$, called the Lambert W function or product log function, to both sides
$$W(1) = W\left(r^2e^{r^2}\right) = r^2$$
Thus $r = \sqrt{W(1)}$