Show Boundedness of Entire Function f: f(z) = f(z + 2π ) & f(z + 2π i)

  • Thread starter Thread starter alvielwj
  • Start date Start date
  • Tags Tags
    Bounded Function
AI Thread Summary
The discussion focuses on proving that an entire function f, which satisfies f(z) = f(z + 2π) and f(z) = f(z + 2πi) for all z in C, is constant. To demonstrate this, it is suggested to first show that f is bounded by restricting it to a closed and bounded square in the complex plane. The periodic nature of f with respect to the given periods implies that the bounds established in this square apply to all z. By applying Liouville's theorem, which states that a bounded entire function must be constant, the conclusion follows. The hint about using the square is crucial for establishing the boundedness of f.
alvielwj
Messages
20
Reaction score
0
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.
 
Mathematics news on Phys.org
alvielwj said:
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?
 
need to prove f(z) is constant.
first show f is bounded,then by the Liouville's theorem, f is constant
 
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 }
 
Looks like a good hint! Although wasn't it 0\le x\le 2\pi, 0\le y\le 2\pi? 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 2\pi and 2\pi i, the bounds on that square are the bounds for all z.
 
Thank you for your answer..
I finally know how to use the hint..
At the beginning i really don't know how to start..
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top