# complex zeros of an equation inside the unit ball

I found this exercise and I don't know where i do wrong: Let $a > e$ be a real number. Prove that the equation $a z^4 e^{−z} = 1$ has a single solution in $D(0, 1)$, which is real and positive.

Well, the equation is equivalent to finding the numer of zeros of the function $az^4-e^z$. In $|z|=1$ we should have $|e^z|<e<|az^4|$ so by Rouche's theorem the number of zeros it's the same as $az^4$ which is 4! (One is real positive since calculating the function in 0 and in 1 it gives a negative and a positive value.) So what is wrong with my proof?

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your use of Rouche's theorem seems right but perhaps that the question is to prove that there is only one solution real and positive in $D(0,1)$ (so that others could exist in $D(0,1)$!). – Raymond Manzoni Apr 22 '12 at 21:32
If it was so, how could i say there's only that one? – balestrav Apr 22 '12 at 21:34
Perhaps simply studying the real functions $ax^4$ and $e^x$ on $(0,1)$... Concerning the complex problem it requires the Lambert-W function as solved for $a=3$ for example by Alpha. – Raymond Manzoni Apr 22 '12 at 21:48
In other words, forget about complex analysis and just use tools of intro calculus to study the equation. – Gerry Myerson Apr 23 '12 at 3:22

## 1 Answer

You have shown that there is at lease one real solution in $(0, 1)$. All that is left is to show there is at most one. Since $f(x)=ax^4e^{-x}-1$ is differentiable for all x, we can apply Rolle's theorem.

Suppose there exist two such roots $a,b\in(0,1)$ with $a<b$. Then there exists a number $c\in(a, b)$ such that $f'(c)=0$. But $$f'(x)=\frac{ax^3(4-x)}{e^x}>0$$ for all $x\in(0,1)$. Therefore there cannot exist two such roots.

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