When is an entire function a constant?

  • #1
Terrell
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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.
 

Answers and Replies

  • #2
FactChecker
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Complex numbers are not ordered, but Im(f(z)) is a real number, which are ordered.
 
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  • #3
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.

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.
 
  • #4
Terrell
317
26
Complex numbers are not ordered, but Im(f(z)) is a real number, which are ordered.
Thanks! I just realized my boo boo.
 

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