Find the solutions of $z^2 = \exp(- \vert z^2 \vert)$. 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 !
 A: 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 $r = \sqrt{W(1)}$
A: Your arguments 1) and 2) seem fine, maybe for 2) you can say that not only is the right hand side positive, but r is real so that $^(−^2)$ lies on the real axis. However for 3) we actually have $e^(-1)$<1 (since e=2.7128....), since you're trying to show that one function is decreasing and another is increasing (from 0 to +infinity say), so they intersect at least once. 
By definition a complex number is of the form z=x+iy where both x and y are real and i is the imaginary unit (sqrt(-1)). So from this we have that $||^2$=zz*=(x+iy)(x-iy)=$x^2-y^2$ and also $z^2$=$(x+iy)^2$=$x^2-y^2+2ixy$. So substituting into the above equation $^2$=$^(−$||^2$)$ we obtain that $x^2-y^2+2ixy$=$e^($x^2-y^2$)$ and now we can compare both sides. Clearly e^($x^2-y^2)$ is a real number, so this means that we obtain $x^2-y^2$=e^($x^2-y^2)$ and 2xy=0 (by comparing the real and imaginary parts on both sides). Now the equation 2xy=0 clearly gives x=0, y=0 or both x=0 and y=0. So we have found the trivial solution (x,y)=(0,0). Now we can consider the cases when x=0 or y=0. When y=0, we are looking at $x^2$=$^(−x^2)$. 
At this point we can't compute the solutions of exactly. So we argue that solutions exists like you did. So we can consider x>0 only. Clearly $x^2$ goes to infinity as x goes to infinity and $^(−x^2)$ goes to zero as x goes to infinity. So they intersect at least once. The function $x^2$ is increasing (for x>o, take derivative and show it is positive on x>0) and the function is $^(−x^2)$ is decreasing on (0,infinity), since taking the derivative we have $-2xe^(-x^2)$ which is always negative in this interval. Hence one side is decreasing and going to 0 and the other is increasing and going to infinity, therefore they intersect exactly once. Similar for x<0 and when x=0. 
Using wolfram alpha, the minimal radius of the disk containing the solutions of (∗) is 0.753089164979674.....=$sqrt(W(1))$ (something to do with Lambert W function) https://www.wolframalpha.com/input/?i=x%5E2%3De%5E-%28x%5E2%29
https://en.wikipedia.org/wiki/Lambert_W_function
