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Find all entire functions such that $\Re f + \Im f$ is bounded.

So there exists $M \ge 0$ such that $\Re f + \Im f \lt M$. We now try to find $g$ such that the given bounded expression is similar to the modulus of $g$.

Let $$g(z) = e^{f(z)} \cdot e^{-if(z)}$$ $$\vert g(z) \vert = e^{\Re f} \cdot e^{\Im f} = e^{\Re f + \Im f}$$

But since we have $\Re f + \Im f \lt M$, $$\implies e^{\Re f + \Im f} \lt e^M$$

Now it is evident that $g$ is an entire function (being a composition of entire functions) and it is bounded so by Liouville's theorem $g(z) = const \implies f(z)-if(z) = f(z) (1-i) = const \implies f=const$

Is this correct?

lojle
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    You are essentially using that the real part of $F = (1-i)f$ is bounded. It is “well-known” that this implies that $F$ is constant, see for example https://math.stackexchange.com/q/229312/42969 or https://math.stackexchange.com/q/660131/42969. – Martin R Feb 06 '21 at 22:00

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