TITLE: $e^z=3z^5$ - Rouche's theorem
QUESTION [1 upvotes]: Question : Show that the equation $e^z=3z^5$ possesses five distinct
  real roots.

In using the Rouche's theorem with the function $f(z)=-e^z+3z^5$ and $g(z)=-3z^5$, I succeeded to prove the first part. How could I prove that the five solutions are distincts?
My teacher explain : Consider that $z_0 \in D(0;1)$ be of multiplicity $\geq 2$. So we must have $$f(z_0)=e^{z_0}-3z_0^5=0$$ and $$f'(z_0)=e^{z_0}-15z_0^4=0$$ that imply $z_0=5$. Impossible.
Is anyone could explain what the teacher explain?

REPLY [0 votes]: Assume by contradiction that one of the roots $z_0$ is not simple.
This means that 
$$f(z)=(z-z_0)^2g(z)$$
for some $g$ which is analytic at $z_0$.
Then 
$$f(z_0)=0 \\
f'(z_0)=2(z_0-z_0)g(z_0)+(z_0-z_0)^2g'(z_0)=0$$
This gives $15z_0^4-3z_0^5=0$.
Therefore $z_0=0$, which is not possible, or $z_0=5$ which doesn't lie in $D(0,1)$.
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