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?