An entire function, such that $f^{-1}(\Bbb R)=\Bbb R$ must be linear

Question

Let $f$ be an entire function, such that $f^{-1}(\Bbb R)=\Bbb R$. Prove that $f=az+b$ for $a, b \in \Bbb C$, $a \neq 0$.

Here is what I was able show:

1. $f$ maps the upper/lower plane to to the upper/lower plane (which implies using the argument principle, but I'm not sure how).
2. from 1, we get $\forall_{z \in \Bbb R} f^{'}(z)=0$ by looking at $Im(f(z))^{'}$ and the Cauchy-Riemann equations.

I'm not sure how to continue. I Any clues?

Answer 1

Hint/Answer -

1. W.L.O.G assume that $f$ maps the upper half-plane to itself. Take a circle of radius $r$ about $0$, say $C_r$, on which $f$ has no zeroes. The winding number of $f(C_r)$ about $0$ is at most $1$, since it only crosses the real line at most two times. Conclude that $f$ has at most one zero by using the Argument Principle.

2. If $\infty$ is a removable singularity, use Liouville's Theorem. If not, consider the case when it is an essential singularity. Now, Picard's Theorem implies that $f$, on a punctured nbd. of $\infty$, must take every value, except possibly one, infinitely many times. This is impossible! Thus, $f$ has a pole at $\infty$.

3. Since a constant function cannot possibly be the answer, the only possibility that remains is a linear function.