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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!

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In the title you have $z^2e^z - z$; in the body you have $z^se^z-z$. Is the latter a typo? –  Arturo Magidin Mar 6 '11 at 3:09
    
It may be easier to study the equivalent problem of showing that $ze^z-1$ has $1$ root with modulus less than $2$. –  Jonas Meyer Mar 6 '11 at 3:29
    
Sorry, fixed the typo. And yes, I have been looking at that quite a bit Jonas, that's the essential difficulty here. –  Jon Beardsley Mar 6 '11 at 4:15

2 Answers 2

up vote 7 down vote accepted

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 $|x+iy|\geq|y|\geq\pi$.

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That's a very nice proof. Thankyou! –  Jon Beardsley Mar 6 '11 at 4:20

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 $f(z)$.

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