Complex equation has two roots inside $|z|=1$ Prove that the equation $z^3[\exp(1-z)]=1$ has exactly $2$ roots inside $|z|=1$.
I have tried applying Rouche Theorem , without any result...

Put $g(x)=x^3\exp(1-x)$. We show easily that $g$ is strictly increasing on $[1,3]$, hence $g(x)>1$ for $x>1$ close to $1$.
Now let $\rho>1$ close to $1$. Put $h(z)=z^3\exp(1-z)$, $f(z)=h(z)-1$. On $|z|=\rho$, we have $|f(z)-h(z)|=1$, and with $z=x+iy$, $|h(z)|=\rho^3\exp(1-x)\geq \rho^3\exp(1-\rho)>1$. Hence Rouché's theorem show that $f(z)=0$ has exactly $3$ solutions in $|z|\leq \rho$. But one of them is $z=1$ (it is easy to see that $z=1$ is a simple root), it remains two solutions. As $\rho$ can be close to $1$, we see that these two solutions verify $|z|\leq 1$. Suppose that for one we have $|z|=1$. Then $|z^3\exp(1-z)|=1=\exp(1-x)$ show that $x=1$, and $z=1$. Hence the two solutions are in $|z|<1$. We are done.   
