Number of zeros inside unit disk

Question

What is the number of zeros of the equation $ze^{3-z}-5^{2}=0$ inside the unit disk?

I believe that that the answer is zero, by some kind of Rouche Theorem. But, I am not %100 sure about that. A collague of mine keeps asking this question to the students. It became a nightmare question. It seems also that all roots of that equation can be found by De Lambert function. Wolfram Alpha found them. But, it didn't give the values explicitly as decimal expansions.

I posted my solution. Finally.

Answer 1

Recalling that $$ e^z=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\dots $$ you can write $$ 25e^z-e^3z=\underbrace{25\left(1+\frac{z^2}{2!}+\frac{z^3}{3!}+\dots\right)}_{f(z)}+\underbrace{(25-e^3)z}_{g(z)}\;. $$ If we prove that $|g(z)|<|f(z)|$ on the unit circle $|z|=1$, then (by Rouchè theorem) we have that $f+g$ (our original expression) and $f$ have the same number of zeroes in $|z|<1$. But $f(z)=25(e^z-z)$ vanishes if and only if $z=-W(-1)$ where $W$ is the Lambert W function. Since $|W(-1)|>1$, we conclude that $25e^z-e^3z$ never vanishes inside the unit disk.

Let's then show that $|g(z)|<|f(z)|$ on $|z|=1$. Trivially $$ |g(z)|=|(25-e^3)|z||\le5 $$ when $|z|=1$. Now observe that $$ \left|\frac{z^2}{2!}+\frac{z^3}{3!}+\dots\right|\le e-2<1 $$ therefore \begin{align*} \left|1+\frac{z^2}{2!}+\frac{z^3}{3!}+\dots\right| &\ge\left|1-\left|\frac{z^2}{2!}+\frac{z^3}{3!}+\dots\right|\right|\\ &=1-\left|\frac{z^2}{2!}+\frac{z^3}{3!}+\dots\right|\\ &\ge1-(e-2)\\ &=3-e\;. \end{align*} Since $$ |f(z)|\ge25(3-e)>5\ge|g(z)| $$ (just by the crude hi school approximation $e\approx2.71$) we conclude.

Answer 2

I think a simpler solution than Joe's is in the following way: The real part $\Re(25e^z-e^3z)=25e^xcosy-e^3x$ can not be zero for $x\leq0$ and $-1\leq y\leq 1$. So, it is enough to consider the contour $$\Gamma=\{z : x=0, -1\leq y\leq 1\}\bigcup\{z : x\geq0, |z|=1\}.$$ On $\Gamma$, it is not difficult to show that the analytic functions $f(z)=25e^z$ and $g(z)=-e^3z$ satisfy the inequality: $|g(z)|<|f(z)|$. Hence, by Rouche's Theorem, $f(z)+g(z)=25e^z-e^3z$ and $25e^z$ have the same number of zeros inside $\Gamma$, that is, no zeros.

This is still not a direct application of Rouche's Theorem on the unit circle. But, I think it is ok.