Complex Zeros of $z^2e^z-z$

Can anyone give me a hint on showing (in a relatively elegant way, as I know the answer from WolframAlpha), that the complex valued function $z^2e^z-z$ has at most 2 roots with norm less than 2? Obviously it has one root at $z=0$. The other root is the number $W_0(1)$, where $W_0(1)$ is the principal branch of the product log function at 1 (the inverse of the function $ze^z$, and is approximately equal to .56. I have seen that the next possible zero is outside of the ball of radius 2, but I cannot show it algebraically. I am primarily trying to use Rouche's Theorem, and to find a suitable function $g(z)$ such that $z^2e^z-z-g(z)|<|g(z)|$, however I have not been able to find a good one yet.

Any help would be much appreciated, especially ideas about what sort of function to consider in cases like this, rather than just "here's the function..."

Thanks!

This is different from the approach you were thinking about, and perhaps not "relatively elegant." It uses no complex analysis.

There is a unique nonzero real solution, because if $x$ is negative then so is $xe^x$, while for positive $x$, $xe^x$ increases without bound. Since $1e^1\gt1$, the real solution has modulus less than $1$. It remains to be shown that there are no nonreal solutions of modulus less than 2.

If $ze^z=1$, then the imaginary part of $ze^z$ must be zero, which means that if $z=x+iy$, then $x\sin(y)+y\cos(y)=0$. Thus $x=-y\cot(y)$ if $\sin(y)\neq0$. The real part of $ze^z$ must be $1$, which after substituting $-y\cot(y)$ for $x$ means $-\frac{y}{\sin(y)}e^{-y\cot(y)}=1$. But for $0<|y|<\pi$ this is impossible because $-\frac{y}{\sin(y)}e^{-y\cot(y)}$ is negative. Thus if $x+iy$ is a nonreal solution to the equation, then $|x+iy|\geq|y|\geq\pi$.

As I understand the question, you are also interested in how one finds functions for Rouché. That is why I give this answer, even though I cannot prove the last inequality (which is however true).

For Rouché you want a function which is close enough to $f(z) = z^2 e^{z}-z$ for $z=2e^{i\phi}$, but simple enough such that you know how many zeros there are in circle with $|z| \leq 2$. I cannot give you a general recipe. But here is how I proceed to find it:

The first obvious choice $g(z) =z^2 e^{z}$ (for which we know the number of zeros) fails for $z$ around $-2$, simply because the $g(z)$ becomes exponential small whereas in $f(z)$ the $-z$ term dominates. Therefore, you want some other term in $g(z)$ which takes around $-2$ whereas you still need to have an $e^{z}$ in order that the function approximates $f(z)$ around $2$ correctly ($f$ is quite large for $z=2$ and then quickly falls off before at some value the $z$ term takes over).

My second guess therefore is $g(z)= z^2 ( e^{z} + c)$ with some $c>0$. Playing around a bit, I see that the value $c=1/2$ seems to work.

Now what one has to proof: we have to show that $|f(z) - g(z)| < |g(z)|$ for $z=2 e^{i\phi}$. The left hand side is $\left|z(\frac{1}{2}+z)\right| = 4 \left| \cos (\phi/2)\right|$. The right hand side can be simplified to $$|g(z)|^2= 16 \left|e^{z} +\frac{1}{2}\right|^2 = 16 \left[\frac{1}{4} + e^{4 \cos \phi} + e^{2 \cos \phi} \cos(2 \sin \phi)\right].$$

Taking the difference $$ \frac{|g(z)|^2 -|f(z) - g(z)|^2}{16} = \frac{1}{4} + e^{4 \cos \phi} + e^{2 \cos \phi} \cos(2 \sin \phi) - \cos^2 (\phi/2) \geq 0.07 .$$ The last estimate was obtained by plotting the function. The minimum is attained around $\phi=1.9$. Too make the argument complete one should prove that the difference is larger than zero (which I couldn't do so far). Then we know that $f(z)$ and $g(z)$ have the same number of zeros with $|z| \leq 2$. As $g(z)$ has only two zeros (two times $z=0$), the same hold for $f(z)$.