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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...

Michael Hardy
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herashefat
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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.

Kelenner
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  • Thanks for your response: But Rouche's would imply that f(z) and h(z) have that same number of zero's inside |z|< rho , but why does but why does h(z)=0 have 3 solutions inside |z|<rho ?? – herashefat Nov 27 '15 at 21:31
  • Since $\rho\to1$, $h(z)$ have 3 zeros in $|z|=1$. Minus $z=1$, it has 2 zeros in $|z|<1.$ – Eugene Zhang Nov 28 '15 at 00:21
  • @herashefat: In Rouché's theorem, the zeros are counted with their multiplicity. Here the zero $z=0$ is counted three times in $h(z)$. The zeros of $f(z)$ are all simple (compute the derivative).Try the simpler example $z^3,z^3-1$ inside the disc $|z|<2$. – Kelenner Nov 28 '15 at 06:24
  • @Kelenner , thanks for your response , although I do understand your explanation , the whole idea of the proof seems mysterious to me : why is rho chosen to be greater than 1 (does less than 1 also work?? for example)...A bit more explanation on the thought process would be much appreciated.... – herashefat Nov 30 '15 at 17:25
  • @herashefat: In fact this is not so mysterious: If we think that the answer is true, and we want to use $h(z)=z^3\exp(1-z)$, we cannot apply Rouché's thm with a $\rho<1$: The success of that give that $f$ has $3$ zeros in $|z|\leq \rho<1$ counting multiplicities, a contradiction with the answer. So if we want to use Rouché's th, we can only choose $\rho>1$. The key point is that there is an obvious zero, namely $1$, and that on $|z|=1$ there is no other zero. – Kelenner Nov 30 '15 at 17:54