Rouche's theorem or degree theory Help. I haven't idea how to prove that 
$$z^2=e^{-|z|^2}$$ for some complex number $z\in\mathbb{C}$ ie there are two roots inside the unit circle $S^1$.
I can use Rouche's theorem or degree theory
 A: They can be equal only when $z^2$ is real, so $z$ is either real or a pure imaginary.  Since the right side is strictly positive and the square of a pure imaginary is nonpositive, they can be equal only if $z$ is real.  So finally we're looking for solutions of $x^2=e^{-x^2}$ for $x$ real.  As $x$ moves away from $0$ in either direction, the left side starts at $0$ and goes upward, reaching $1$ when $|x|=1$, and the right side starts and $1$ and moves down to $e^{-1}$ by the time $|x|$ reaches $1$.  So the intermediate value theorem does it.
A: Let $z = a + b i $,and $i = \sqrt{-1}$,then we get $z^{2} = |z|^{2} = z \overline{z} = x \geq 0 $.
Hence define a map $f $ from $S^{1} \rightarrow S^{1} $ by the following formula
\begin{equation}
f(x) = e^{-x}
\end{equation}
Since the derivative of $f(x)$ with $x \in S^{1}$,and we have 
\begin{equation}
|f{\,'}(x)| < 1 ,x \in S^{1}
\end{equation}
Using fixed point theory,we can get unique point $x_{0} \in S^{1}$ such that $f(x_{0}) = x_{0}$.
Meanwhile we have two point $z$ and $\overline{z}$ which satisfy $x_{0} = z \overline {z}$.
The foregoing thing is simple idea for you ,but for the details ,i think you should do it by yourself.
A: as OP mentioned Rouche and degree theory it may be that a winding number solution was desired.  Consider entire function given by $f(z) = z^2$ and continuous function on $\mathbb C$ given by $g(z)=z^2-e^{-|z|^2}$.  Tracing a curve around the unit circle with $\gamma:[0,1]\longrightarrow \mathbb C$ given by $\gamma(t)=\exp\big(2\pi i\cdot t\big)$ we have
$2=n\big(f\circ\gamma,0\big)=n\big(g\circ\gamma,0\big)$
by Dog on-a Leash Lemma since:  $e^{-1}= \big\vert f\circ\gamma(t)-g\circ\gamma(t)\big \vert\leq \big\vert f\circ\gamma(t)\big \vert=1$ for all $t$
$\implies 0\in g\big( D\big)$
(for unit disc $D$) by the topological form of the Argument Principle for a disc
(equivalently $0\not\in g\big( D\big)\implies n\big(g\circ\gamma,0\big)=0$)
Since $g$ is not analytic this doesn't tell us how many zeros it has in the unit disc, but notice $g(z)=0\implies g(-z)=0$ and since $g(0)\neq 0$ so we conclude (at least) 2 distinct zeros of $g$ exist in the unit disc.
