Proving $f$ is a constant function

Let $f$ be a $2\pi$ periodic entire function satisfying $|f(z)|\leq 1+|{\rm Im}\; z|$.

I am trying to show $f$ is constant.

Initially I thought it is very easy that I can apply Louiville's Theorem. But I realized proving $f$ is bounded is not straight forward. This has to do something with that period of the function. I do not see what I can do with that.

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One approach: by the boundedness of its growth at infinity (less than linear), you know it's a polynomial, by periodicity you know it's constant – uncookedfalcon Dec 24 '12 at 1:54

Okay, first the bounded growth at infinity implies polynomial: suppose $|f(z)| \leqslant C|z|^n$ for $|z|$ sufficiently large, then I claim that $f^{n+1}(z) \equiv 0$. Indeed, to see this, for any $w \in \mathbb{C}$ we have $$f^{n+1}(w) = c\int_\gamma \frac{f(z)}{(z - w)^{n+2}} dz$$where we can take $\gamma$ to be a circle of radius $R$ about $w$. Now, the integral is bounded by $$2\pi R c \max_{|z-w| = R} \frac{|f(z)|}{|z^{n+2}|} = c \frac{\max_{|z - w| = R} |f|}{R^{n+1}}$$where I lumped together the constants. Now applying our bound on $f$ and taking $R$ to infinity gives the result.
(in our case we had $|f(x+iy)| \leqslant 1 + y \leqslant x^2 + y^2 = |z|^2$ for big enough $|z|$, so I guess we know $f$ is quadratic)
For the second bit, a periodic polynomial is constant: if it's not constant, it has a root $w$, so infinitely many roots $w + 2\pi \mathbb{Z}$.