# Zeros of $e^{z}-z$, Stein-Shakarchi Complex Analysis Chapter 5, Exercise 13

This is an exercise form Stein-Shakarchi's Complex Analysis (page 155) Chapter 5, Exercise 13:

Prove that $$f(z) = e^{z}-z$$ has infinite many zeros in $$\mathbb{C}$$.

Attempt:

If not, by Hadamard's theorem we obtain $$e^{z}-z = e^{az+b}\prod_{1}^{n}(1-\frac{z}{z_{i}})$$ where $$\{z_{i}\}$$ are the zeros of $$f$$. How can we conclude ?

Are you allowed to use Picard's Theorem?

If yes here is a relative question:

Use Picard's Theorem to prove infinite zeros for $\exp(z)+Q(z)$

Note that $a=1$. Then rewrite your equation as

$$z=e^z P(z)$$

where $P(z)$ is a polynomial of degree $n$. For $|z|$ large enough, $P$ grows of order $|z|^n$. Therefore this equation implies upon taking absolut values that $e^{\mathrm{Re} z}$ decreases like $|z|^{1-n}$ as $|z|\rightarrow\infty$. This is clearly a contradiction.

Therefore you cannot apply Hadamard's theorem, i.e. $f$ has infinitely many zeros.

• Why can we assume a=1? – Sacha L'Heveder Nov 20 '19 at 20:51