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?
On
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 $D(0,1)$.
I am not familiar with the notation $D(0;1)$. I am going to interpret it as $z_0$ is one of the root with multiplicity 2, which should exists if the solutions are not distinct.
Hence $z_0$ should satisfy both $f(z_0)=0$ and $f'(z_0)=0$. $$f(z_0)=e^{z_0}-3z_0^5=0 $$ $$f'(z_0)=e^{z_0}-15z_0^4=0$$
subtracting the equations gives us $$15z_0^4-3z_0^5=0$$
Since $f(0)=1 \neq 0$, $z_0 \neq 0$. Hence, dividing by $z_0$, we obtain $15-3z_0=0$ and we can conclude that $z_0=5$.
but let's check substitute $z_0=5$ into $e^{z_0}-3z_0^5$ and check if it is indeed a root.
$$e^5-3(5)^5 < (3)^5-3(5)^5=3(3^4-5^5)<0$$
Hence $z_0=5$ is not a root, which is a contradiction. Our assumption that a root with multiplicity 2 exists must be wrong. The roots have to be distinct.