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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.

## 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.