# Show $z e^{\lambda-z}-1$ has only one real root in the unit disk.

$z e^{\lambda-z}-1$ 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.

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

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.