Determine the number of distinct solutions of the equation
\[e^{z^2} = 5z^5\]
in the unit disk \(\{z \in \mathbb{C} : |z| \leq 1\}\).
Solution
There are exactly \(\boxed{5}\) distinct solutions. Let \(F(z)=e^{z^2}-5z^5\).
1. There are five zeros inside the unit circle counted with multiplicity, and none on the circle.
For \(|z|=1\), \[ |e^{z^2}|=e^{\operatorname{Re}(z^2)}\leq e<5=|5z^5|. \] The exponential series and \(n!\geq2^{n-1}\) for \(n\geq2\), with strict inequality for \(n\geq3\), give \(e=2+\sum_{n\geq2}1/n!<3<5\). Rouché’s theorem therefore gives the same interior zero count for \(F\) and \(-5z^5\), namely five with multiplicity [@SS03]. The strict inequality also shows that \(F\) is nonzero on \(|z|=1\).
2. Every zero in the disk is simple.
A zero \(z\) of \(F\) is nonzero since \(F(0)=1\). Substituting \(e^{z^2}=5z^5\) into the derivative yields \[ F'(z)=2z e^{z^2}-25z^4 =5z^4(2z^2-5). \] For \(|z|\leq1\) one has \(|2z^2|\leq2<5\), so this derivative is nonzero. Thus no counted zero has multiplicity greater than one, by its Taylor expansion [@SS03]. The five zeros in step 1 are therefore distinct and give all the solutions in the requested closed disk.