Number of solutions of $e^z=ez^3$ inside the unit circle.

Question

I want to show that the number of solutions of the equation $e^z=ez^3$ inside the unit circle
$S^1=\{|z|<1\}$ is $2$ (counted with multiplicities). I know how to prove in general that the equation $e^z=az^n$, for some $a>e$ has exactly $n$ zeroes inside $S^1$ using Rouchè' Theorem.

I have thought of two ways of proving this but I get stuck in both:

• Since the function $f(z)=e^z-ez^3$ does not have any poles, I can use the Argument Principle which, in this case, states that
  $$\frac{1}{2\pi i}\oint_{|z|=1}\frac{f'(\zeta)}{f(\zeta)}d\zeta=N_{f}(0)\ ,$$
  where $N_{f}(0)$ denotes the number of zeroes of $f$ in $S^{1}$.
  My problem here is that I don't know how to compute this integral.

• The derivative of $f$ is $f'(z)=e^z-3ez^2$, which I know that it has $2$ zeroes in $S^1$. So if I prove that $f$ has the same number of zeroes with its derivative $f'$ then I am done. But I don't know how to prove this also.

Answer 1

As per the hint in comments by Conrad, choose $g(z)=e^z$ and $h(z)=ez^3$ so that you have the inequality
$|g(z)|<|h(z)|$
$\implies ez^3-e^z>0$
Inside the circle $|z|=1+\delta$ where $\delta>0$, above inequality reduces to $ (1+\delta)^3-e^{\delta}>0$ which can be proved using little calculus on the function $p(\delta)=(1+\delta)^3-e^\delta$.

Observe that $p(0)=0$ and $p'(0)=2>0$ tells that $p(\delta)$ is increasing in the vicinity of zero. Using Rouche's theorem, as $h(z)$ has $3$ zeros counting multiplicities inside $|z|=1+\delta$, hence $g(z)+h(z)=f(z)$ too has $3$ zeros inside the same circle, one of them is at $z=1$ so you have exactly $2$ zeros inside the unit circle.