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.

Homework Equations

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.

 

[FactChecker]

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

 

[PeroK]

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.

Homework Equations

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.