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$z e^{\lambda-z}-1$ has only one real root in the unit disk for all real $\lambda >1$

Usin calculus, I showed there is a root, but I can't see how I can use Rouche's theorem. I tried dividing and adding the different parts but it gets me nowhere. Would really appreciate any help.

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    $\begingroup$ $$ze^{\lambda - z} - 1 = 0 \iff ze^{\lambda} - e^z = 0$$ $\endgroup$ Jan 7, 2016 at 17:07
  • $\begingroup$ I took it into consideration, but $e^z$ doesn't seem to have any zeros. $\endgroup$
    – Meitar
    Jan 7, 2016 at 17:11
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    $\begingroup$ Right. But that's the term with smaller modulus on the unit circle. $\endgroup$ Jan 7, 2016 at 17:12
  • $\begingroup$ So you mean I looked at it from a wrong perspective? $\endgroup$
    – Meitar
    Jan 7, 2016 at 17:16

1 Answer 1

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Let $f(z)=ze^\lambda-e^z$, which, as the hint by Daniel Fischer notes, has the same zeros as your function. Let $g(z)=ze^\lambda$. Then $$|f(z)-g(z)|=e^z<|e^{\lambda-z}|+|e^\lambda|=|f(z)|+|g(z)|$$ holds on the boundary of the unit circle. Therefore $f(z)$, your function, and $g(z)$ all have the same number of complex zeroes inside the unit circle. $ze^\lambda$ very clearly has only one zero in this region.

Since you've proven there is at least one real root inside the unit circle, it follows that there is exactly one, since every real root is a complex root.

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