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So the problem states that if $f(z)$ is entire, and satisfies the relation $f(z+i) = f(z)$ and $f(z+1) = f(z)$, show that $f(z)$ is constant. So I was thinking that since any point in $\mathbb{C}$ can be written as $\alpha * 1 + \beta * i $ we can say that $f(z + z_0) = f(z) $ in which case it is constant, but I'm having trouble breaking down the steps, and using the fact that f is entire, which makes me feel like I'm missing something. What should I review to figure this out?

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Might try showing $f(\mathbb{C})$ is the same as $f(I\times I)$, where $I$ is the unit interval. Then of course since $I\times I$ is compact, the image is bounded, and Liouville saves the day. –  user641 May 4 '11 at 0:32
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max modulus principle –  yoyo May 4 '11 at 0:45
    
Thanks @yoyo, that was really helpful to review as well. It applies to another question that I was also stuck on. –  I Love Cake May 4 '11 at 0:49
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2 Answers

I think you can just use Liouville's theorem. The thing is that $f$ is bounded on compact subsets of the complex plane, so in particular it is bounded in the unit square $S = \{ a + bi \in \mathbb{C} \mid a, b \in [0, 1] \}$.

So then you can use the two relations $f(z+1) = f(z)$ and $f(z+i) = f(z)$ to prove that the values of $f$ are determined by its values in the unit square because you can write $z = x + iy$ and then

$$f(z) = f(x + iy) = f(a + bi + (n + mi)) = f(a + bi)$$

where $0 \leq a, b \leq 1$ and $n, m \in \mathbb{Z}$. In this case $n = \lfloor x \rfloor$ and $m = \lfloor y \rfloor$. Then of course $f$ is bounded in the whole complex plane so Liouville's theorem applies, and $f$ has to be constant.

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How did you get from (x+iy) = (a+bi +(n+mi)) to f(a+bi)? –  I Love Cake May 12 '11 at 22:38
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@Cake: You said that $1$ and $i$ are periods of your $f(z)$; it stands to reason that adding/subtracting any integer multiple of your periods from the argument of $f$ doesn't change its value; much like saying that $\sin(x+2\pi m)=\sin\,x$ –  J. M. May 13 '11 at 0:41
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Here is a Hint: Liouville Theorem.

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