TITLE: Solve $e^{z-1}=z$ with $|z| \leq 1$
QUESTION [5 upvotes]: I'm looking for solutions to $$e^{z-1}=z$$ when $z \in \mathbb{C}$ with $|z| \leq 1$. 

The obvious solution is $z=1$, but I don't know how to show that there aren't any others.
This question is related to this one.

REPLY [2 votes]: Just look at the imaginary parts of each. Let $z = a + bi$.
$\mathrm{Im}(e^{z-1}) = e^{a-1} \mathrm{sin}(b)$
$\mathrm{Im}(z) = b$
For $a < 1$ and $b \neq 0$ we have $|e^{a-1} \mathrm{sin}(b)| < |\mathrm{sin}(b)| < |b|$. Thus for $\mathrm{Im}(z) \neq 0$ we have $\mathrm{Im}(e^{z-1}-z) \neq 0$.
Thus it reduces to the real case, which another poster has discussed.