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**Dinkydoe** I want to show that $\displaystyle ze^{a-z}-1=0$, with $\displaystyle a> 1$, has exactly one root in the open disc $\displaystyle D(0,1)$. I'm trying to do this with Rouche's theorem, that is:

finding a suitable function $\displaystyle g(z)$ with exactly one zero in $\displaystyle D(0,1)$, that satisfies $\displaystyle |f(\zeta)-g(\zeta)|<|f(\zeta)|+|g(\zeta)|$ for all $\displaystyle \zeta\in D(0,1)$

I believe it's not all too hard to find such function, like $\displaystyle g(z)=ze^{a-z}$ however showing the desired inequality is for me problematic. Any help/suggestions?