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I am reviewing my complex analysis and I got stuck with an exercise about Rouché's theorem. It states: for $0 \leq C \leq \frac{1}{e}$, show that $Ce^z=z$ has exactly one root in the closed unit disc.

For $C < \frac{1}{e}$ I can use easily Rouché's theorem, checking that $z$ and $z-Ce^z$ have the same number of zeroes. Then, for the case $C=\frac{1}{e}$ I don't know what to do. I know the root I want is exactly in 1, so I was thinking about some contour approximating $\lbrace |z|=1 \rbrace$ which excludes 1, but anything I can think about does not give me a contour suitable for applying the theorem.

Am I going in the right direction? Any hint is welcome!

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  • $\begingroup$ Is the question, then, to solve $e^{z-1} = z$ for $z \in \mathbb{C}$? $\endgroup$ – Zubin Mukerjee Nov 21 '14 at 5:52
  • $\begingroup$ Yes, that's the equality giving me problems $\endgroup$ – Stefano Nov 21 '14 at 5:54
  • $\begingroup$ I've asked another question that will probably get more views if only because the title is perhaps less daunting. I'm not sure why yours hasn't attracted much attention. I tried to show that the only solution is $z=1$ using power series but I wasn't successful. $\endgroup$ – Zubin Mukerjee Nov 21 '14 at 6:12
  • $\begingroup$ I see. So far the only approach I tried is with Wouché's theorem, and I don't know whether it might be successful... $\endgroup$ – Stefano Nov 21 '14 at 6:17
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    $\begingroup$ I gave an elementary proof that $e^{z-1} = z$ has only the solution $z = 1$ in the unit disk at Zubin's linked problem. A contour could have worked, but the fact that $e^{z-1} - z$ has a double zero at $z = 1$ makes it messier. It should be possible to give a contour proof going something like: use the unit circle but dip inward near $z = 1$, then compute/approximate in two pieces to prove you get zero for the argument principle contour integral. $\endgroup$ – aes Nov 21 '14 at 7:50

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