Bounded Function

  • Thread starter alvielwj
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  • #1
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Main Question or Discussion Point

How to show that if f is an entire function,such that f(z) = f(z + 2π ) and f(z) = f(z + 2π i)
for all z belong to C.
π is pi.
 

Answers and Replies

  • #2
HallsofIvy
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How to show that if f is an entire function,such that f(z) = f(z + 2π ) and f(z) = f(z + 2π i)
for all z belong to C.
π is pi.
How to show "if ...."

but where is your conclusion? What do you want to prove?
 
  • #3
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need to prove f(z) is constant.
first show f is bounded,then by the Liouville's theorem, f is constant
 
  • #4
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let me post the whole question
Suppose that f is an entire function such that f(z) = f(z + 2π ) and f(z) = f(z + 2π i)
for all z belong to C. Use Liouville's theorem to show that f is constant.
Hint: Consider the restriction of f to the square {z = x + iy : 0 <x < 2π ; 0 < y <2π }
 
  • #5
HallsofIvy
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Looks like a good hint! Although wasn't it [itex]0\le x\le 2\pi[/itex], [itex]0\le y\le 2\pi[/itex]? The "=" part is important because that way the set is both closed and bounded and so any continuous function is bounded on it. Since f is "periodic" with periods [itex]2\pi[/itex] and [itex]2\pi i[/itex], the bounds on that square are the bounds for all z.
 
  • #6
20
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Thank you for your answer..
I finally know how to use the hint..
At the begining i really dont know how to start..
 

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