Does an entire function $f$ with a pole at infinity implies $|Re(f)| \to \infty$ or $|Im(f)| \to \infty$?

Question

Let $f$ be an entire function with a pole at infinity.

Does this imply $\lim_{|z|\to \infty}|Ref(z)|=\infty$ or $\lim_{|z|\to \infty}|Im(z)|=\infty$?

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Here is my idea:

We have that $f(z)=Ref(z)+iIm(z)$. Hence, $\infty=\lim_{|z|\to \infty}|f(z)|=\lim_{|z|\to \infty}|Ref(z)+iIm(z)| \leq \lim_{|z|\to \infty}|Ref(z)|+|Im(z)|$.

Assume by contradiction that $\lim_{|z|\to \infty}|Ref(z)| \neq \infty$ and $\lim_{|z|\to \infty}|Im(z)| \neq \infty$. Then these limits are bounded by some numbers $M \in \Bbb R$. So we get $\infty \leq M \in \Bbb R$, which is a contradiction.

I fill like I'm missing something. I would be grateful for any comment or advice to either confirm the validity of my proof or to point out why it is wrong.

Answer 1

No.

Remember that $f(z)=z$ has a pole at infinity, but $|\operatorname{Im} f(z)|$ does not tend to infinity along the real axis, and $|\operatorname{Re} f(z)|$ does not tend to infinity along the imaginary axis.

Answer 2

If $f$ has a pole at $ \infty$, then $f$ is a (non-constant) polynomial !

For example $f(z)=z^2$. We have

$u(x,y):=Re f(z)= x^2-y^2$

Now consider $u(x,x)$.

Can you now get it, that we do not have $\lim_{|z|\to \infty}|Ref(z)|=\infty$ ?