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

  • $\begingroup$ The right side is real and positive. Just use continuity. $\endgroup$ – anomaly Nov 17 '14 at 3:42
  • $\begingroup$ You definitely cannot use Rouché. :) $\endgroup$ – Ted Shifrin Nov 17 '14 at 4:27
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    $\begingroup$ @jimbo: Stop editing the tags for this question. It has nothing at all to do with differential geometry (the study of Riemannian metrics on manifolds and associated invariants) or differential topology (the study of diffeomorphism invariants between smooth manifolds and questions about the existence and uniqueness of smooth structures on topological spaces). $\endgroup$ – anomaly Nov 17 '14 at 5:28
  • $\begingroup$ degree theory is differential topology and if it relates. When one solution of an equation is easily found, degree theory can often be used to prove existence of a second, nontrivial, solution. $\endgroup$ – jimbo Nov 17 '14 at 13:49
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    $\begingroup$ This is not a question of degree theory; it's just very straightforward complex analysis. Compare this to other differential topology questions on the website, and note the lack of any differential topology in the answers given here. $\endgroup$ – anomaly Nov 17 '14 at 15:42

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.


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.

  • $\begingroup$ I am so sorry for no using Rouche's theory. $\endgroup$ – askuyue Nov 17 '14 at 4:45

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