# When is an entire function a constant?

## 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.

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

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