Solve $e^{z-1}=z$ with $|z| \leq 1$

Question

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.

Answer 1

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.