How can I show that an entire function where the sum of real and imaginary part are less or equal to the absolute of the function is constant?

Question

Suppose there is an entire function f with

$\Re(f(nz))+\Im(f(nz))\leq|f(z)|,\qquad \qquad \forall z\in\partial\mathbb{D}, \forall n\in\mathbb{N} $,

How could I show that $f$ is constant?

Answer 1

Letting $M=max|f(z)|, z\in\partial\mathbb{D}$ we get $\Re f(w) + \Im f(w) \le M, |w| = n, n \ge 1$. Since $\Re f + \Im f$ is real harmonic, applying the maximum principle on the domains $n \le |w| \le n+1$ leads to $\Re f(w) + Im f(w) \le M, |w| \ge 1$ which means that the image $f(\mathbb C)$ is disjoint of the set $|w| >2M, \Re w >0, \Im w >0$ and that is impossible for non-constant $f$ as $\infty$ cannot be an essential singularity, hence $f$ is polynomial, but the fundamental theorem of algebra implies $f$ constant then

Answer 2

Can't we just use $|f(nz)| \leq |f(z)|$ for each $n, z,$
and the Maximum Modulus Principle to say $f$ is bounded by $\max_{z \in \partial D} |f(z)| = M < \infty?$

If $\forall w \in \mathbb{C}$ we find a larger radius $n > |w|,$ MMP bounds $|f(w)| \leq |f(nz)| \leq M.$ Such $f$ should be a constant by Liouville.