Use of Liouville's Theorem

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• Nov 26th 2007, 10:25 AM
smoothman
Use of Liouville's Theorem
Suppose f is an entire function such that $f(z) = f(z+2\pi)$
and $f(z)=f(z+2\pi i)$ for all z $\epsilon$ C. How can you use Liouville's theorem to show f is constant..

any help on that please to get me started off.. thnx a lot :)
• Nov 26th 2007, 10:27 AM
ThePerfectHacker
Quote:

Originally Posted by smoothman
Suppose f is an entire function such that $f(z) = f(z+2\pi)$
and $f(z)=f(z+2\pi i)$ for all z $\epsilon$ C. How can you use Liouville's theorem to show f is constant..

any help on that please to get me started off.. thnx a lot :)

Here is a start. By Liouville all you need to show is that $|f(z)|\leq A$ (meaning it is bounded).
• Nov 26th 2007, 10:34 AM
smoothman
would restricting f to a square also be applicable here as suggested by your bound :

so {z = x + iy : 0 <= x <= 2pi, 0 <= y <= 2pi}.

where would you go from there?
• Nov 26th 2007, 10:35 AM
ThePerfectHacker
Exactly! Because the function inside that square determines the values outside the square. But the function is entire so it is continous, but continous functions on compact sets are bounded. So this function is bounded everywhere. Does that make sense?
• Nov 26th 2007, 10:37 AM
smoothman
that does kind of make sense but ive just kind of started learning this lol. how would you show it as a proof mathematically though?
• Nov 26th 2007, 10:59 AM
ThePerfectHacker
Quote:

Originally Posted by smoothman
that does kind of make sense but ive just kind of started learning this lol. how would you show it as a proof mathematically though?

Let $R$ be the square $\{ x+iy: x\in [0,2\pi] \mbox{ and }y\in [0,2\pi]\}$. The function $f(z)$ is continous everywhere because entire functions (differenciable) are continous. Thus, $f(z)$ is bounded on $R$ because it is a closed and bounded set. But then $f(z)$ is bounded by that same number because the value of $f(z)$ is determined by within the rectangle. So for example, $f(100\pi ) = f(98\pi + 2\pi) = f(98\pi)=f(96\pi)=...=f(0)$. Thus, any value outside the rectangle is known from the value inside the rectangle. Since the function is bounded on this rectangle is is bounded outside it too because it has the same values as inside the rectangle.
• Nov 26th 2007, 11:14 AM
smoothman
thnx a lot for this. i understand this in the words you have set. but what i meant was how can u show this proof using the louville equations etc?