2
$\begingroup$

I'm trying to solve this exercise from the book: Complex Analysis - Mathews & Howell

Let $f$ be an entire function that has the property $|f(z)| \geq 1$ $\forall z \in \mathbb{C}$. Show that $f$ is constant.

Sol.:

I have to apply Liouville's theorem. If I take $g=\frac{1}{f}$, then $|g(z)| \leq 1$ for all $z \in \mathbb{C}$. So $g$ is bounded, but how can I be sure that $g$ is still entire?

Maybe I should consider that if $|f(z)|\geq 1$, then $f$ has no zero in $\mathbb{C}$, and so $g$ is entire.

Does it works?

$\endgroup$
1
  • $\begingroup$ That's right. $f$ has no zero, so $1/f$ is analytic wherever $f$ is, that is, $g$ is entire. $\endgroup$
    – saulspatz
    Commented Feb 3, 2018 at 18:22

2 Answers 2

2
$\begingroup$

The inverse of a holomorphic function $f$ is holomorphic on the complement of the set where $f$ vanishes. Therefore your argument is perfect and $1/f$ is entire.

$\endgroup$
2
$\begingroup$

Your approach is fine. The function $g$ is analytic, since it is obtained as the composition of two analytic functions. Since the domain of $g$ is $\mathbb C$, $g$ is entire.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .