1
$\begingroup$

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 ?

$\endgroup$
1
$\begingroup$

Are you allowed to use Picard's Theorem?

If yes here is a relative question:

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

$\endgroup$
3
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.