# When is an entire function a constant?

Terrell

## Homework Statement

Let ##f(z)## be an entire function of ##z \in \Bbb{C}##. If ##\operatorname{Im}(f(z)) \gt 0##, then ##f(z)## is a constant.

n/a

## The Attempt at a Solution

I don't get how the imaginary part of ##f(z)## would be greater than any number. Aren't complex numbers not ordered? The proof is one line and uses Louiville's Theorem, but I think I don't understand this question in the first place.

## Answers and Replies

Gold Member
Complex numbers are not ordered, but Im(f(z)) is a real number, which are ordered.

• Terrell and PeroK
Homework Helper
Gold Member
2021 Award

## Homework Statement

Let ##f(z)## be an entire function of ##z \in \Bbb{C}##. If ##\operatorname{Im}(f(z)) \gt 0##, then ##f(z)## is a constant.

n/a

## The Attempt at a Solution

I don't get how the imaginary part of ##f(z)## would be greater than any number. Aren't complex numbers not ordered? The proof is one line and uses Louiville's Theorem, but I think I don't understand this question in the first place.

Any line in the complex plane can be ordered: it's essentially the same as the Real line. That's also true of the imaginary line.

In any case, the imaginary part of a complex number can be seen as a function from ##\mathbb{C}## to ##\mathbb{R}##.

And, in fact, ##Im(z) > 0## simply means that ##z## lies in the upper half of the complex plane.

• Terrell
Terrell
Complex numbers are not ordered, but Im(f(z)) is a real number, which are ordered.
Thanks! I just realized my boo boo.