Real solution to a complex equation

Question

I have some trouble solving this equation. Let $a\in\mathbb R$, $a>1$. I want to show that there is a unique solution of

$ze^{a-z} = 1$,

with $|z|<1$ and that this solution is real and positive.

It is easy to prove the existence part of this problem. Indeed, taking log on both side, we get $\log x + a -x = 0$. However, $\log 1 + a-1>0$ and $\lim_{x\to 0} \log x + a- x < 0$. Thus, by the mean value theorem, there is a positive real solution. How can one show that this is the unique solution in the unit disk? Any help would be appreciated. Thanks!

You can get a solution in terms of Lambert W function. See my answer here. — Mhenni Benghorbal, Aug 06 '13 at 19:15
You are welcome. Your problem is a special case of the problem. — Mhenni Benghorbal, Aug 06 '13 at 19:25

score 3 · Accepted Answer · 2013-08-06T19:48:25.330

3

Define $f(z) = ze^{a - z}$, $g(z) = -1$. Then for $z$ on the boundary of the disk,

$$|f(z)| = |z| |e^{a - z}| = |e^{a - z}| = exp({\Re(a - z)}) > 1 = |g(z)|$$

where the last inequality follows from the fact that $a > 1$. Hence by Rouche's Theorem, $f + g$ and $f$ have the same number of zeros in the disk, and $f$ clearly has a unique zero at $0$.

edited Aug 06 '13 at 19:48

answered Aug 06 '13 at 19:22

I am not sure I see the first inequality, since $|z|<1$. – Gary Aug 06 '13 at 19:25
@Gary The setup for Rouche requires the estimate to hold on the boundary of the disk, and $\partial \mathbb{D} = {z : |z| = 1}$. – Aug 06 '13 at 19:33
I see! Thanks, this is a really good proof! – Gary Aug 06 '13 at 19:45

Real solution to a complex equation

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