How to prove a complex function is constant if it is given that complex function is entire function (Holomorphic function).

Question

Let $f$ be an entire function such that for all z with $\lvert z\rvert>M$, for $M\in \Bbb R$ there holds $$\lvert \text{Re}(f(z))\rvert \geq \lvert \text{Im}(f(z))\rvert$$ Prove that $f$ is constant.

Answer 1

Consider the function $$g(z) = e^{- f(z)^2}.$$ Then $$\lvert g(z) \rvert = e^{\Im(f(z))^2 - \Re(f(z))^2}.$$ By the given inequality, for $\lvert z \rvert > M$ we have $\lvert g(z) \rvert \leq 1$. Thus $g$ is bounded and entire, by Liouville constant and hence $f$ is also constant.

Answer 2

Real harmonic functions (on complex domains) cannot have any isolated zero (or any isolated value $a$ by taking $u-a$) since otherwise the zero would be a local minimum or maximum and non zero harmonic functions do not have such by the integral mean value property.

If they are defined on the full plane, non constant real harmonic functions must have arbitrarily large zeroes (in absolute value) since they cannot be bounded inferior or superior and being non zero from some $|z| > M$ on means function all positive or negative there and correlated with boundness on compact discs, we get one sided boundness and that is not possible.

But now the condition above implies that any zero of the real harmonic function $\Re f$ is a zero of the complex analytic $f$ from $M$ on and hence having such which are non isolated by the above implies $f$ constant zero, while not having such implies $\Re f$ constant non zero, hence $f$ constant nonzero too