Limit of imaginary part divided by real part of an entire function on the complex plane

Question

Let $f\colon \mathbb{C}\to \mathbb{C}$ be holomorphic. I assume that the $lim_{z\to \infty} \frac{\Im(f(z))}{\Re(f(z))}$ should be bounded. But is this the case, and if yes, how can I prove it? Or could I make a statement about the $\limsup$ ?

Answer 1

This is not true. Take the function $f(z)=z$ and let $z\to\infty$ along the curve $z(t)=1/t+it$.

$$\lim_{t\to\infty}\frac t{1/t}=\infty.$$

Answer 2

In fact, this always fails (unless $f$ is constant): If $f$ is entire and nonconstant, then there exists a sequnce $z_n \to \infty$ such that

$$(1)\,\,\,\,\frac{\text {Im} \,f(z_n)}{\text {Re}\, f(z_n)} \to \infty.$$

Proof: $f(\{|z|>R\})$ is dense in $\mathbb {C}$ for each $R>0.$ Thus, for each $n\in \mathbb {N}$ there is $z_n, |z_n|>n$ such that $f(z_n)\in D(1+ni,1).$ It follows that $\text {Im}\,f(z_n) > n - 1,0< \text {Re}\,f(z_n) < 2.$ This gives (1).