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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| 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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    In other words, forget about complex analysis and just use tools of intro calculus to study the equation.2012-04-23

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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. 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.