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Let $f:\mathbb C\to\mathbb C$ is entire function. And $f(z+1)=f(z)$ and $f(z+i)=f(z)$ then what can you say about $f$ ?

I guessed that it must be a constant because evaluating the function on a translation of $z$ by 1 unit and $i$ result the same. so it must be a constant. But I need a concrete argument. Thanks in a bunch.

David
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1 Answers1

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Note that $f$ is bounded, because it is determined by value on square $\{(x+iy \; : x \in [-0,1] \; y \in [0,1]\}$, because:

$$f(a+bi)=f(\{a\}+i\{b\})$$

where $\{y\}=y-\lfloor y \rfloor$

It can be proven by induction.

Because $f$ is bounded entire function it must be constant.

agha
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  • am sorry i could not understand your argrment.@agha – David Oct 09 '14 at 19:46
  • Set ${(x+iy ; : x \in [-0,1] ; y \in [0,1]}$ is a compact set, so $F$ is bounded on this set. The function doesn't take other values than on this set, right? Finally there is theorem that if entire function is bounded then it's constant. – agha Oct 09 '14 at 19:57
  • am sorry again.. i could not understand that it cannot take values other than the unit square.@agha – David Oct 09 '14 at 20:00
  • Ok, for example consider value for $2,5+3,5i$ then by $f(z)=f(z+1)$ you have $f(2,5+3,5i)=f(1,5+3,5i)=f(0,5+3,5i)$. By $f(z)=f(z+i)$ you have $f(0,5+3,5i)=f(0,5+2,5i)=f(0,5+1,5i)=f(0,5+0,5i)$. It's an idea. – agha Oct 09 '14 at 20:03
  • oh, great... i understood clearly.. now only i am realising that i can apply the relation given recursely...thanks a bunch.@agha – David Oct 09 '14 at 20:07