Complex polynomial that sends real line to real line, positive imaginary part to positive imaginary part, negative imaginary part to... Let p be a polynomial that sends points on the real line to points on the real line, sends points with positive imaginary part to points with positive imaginary part, and sends points with negative imaginary part to points with negative imaginary part. Prove that p must be of the form $p(z)=az+b$ where a is a positive number and b is a real number.
Since p maps the real line to the real line, all coefficients are real. From what else we're given, we can also deduce that all zeros are real. This is where I get stuck. I thought about the fundamental theorem of algebra but couldn't get anywhere with it. I feel like I'm missing something obvious. Any hints?
 A: $p$ nonconstant clearly and if $x_1,..x_n$ are its roots the property of $p$ means $x_k \in \mathbb R$ since $p(x_k)=0$ and then $P(z)=a(z-x_1)..(z-x_k)$ and $a \in \mathbb R, a \ne 0$ by taking $z=x$ not a root and using $0 \ne P(x) \in \mathbb R$
But then there is a large $C$ st $|p'(x)| >0, |x| >C$ hence $p$ is strictly monotonic on $\mathbb R -[-C,C]$ and $|p| \to \infty$ at $\pm \infty$; if the degree of $p$ is even it follows that $p$ cannot be surjective on the reals (has either a global min or a global max depending whether $a>0$ or $a<0$ as the limits at $\pm \infty$ of $p$ are same sign infinity) so since by the fundamental theorem of algebra $p(z)=b$ has roots for all real $b$ there is such $b$ for which the roots are complex non real which is a contradiction
If $p$ has odd degree, then outside $p([-C,C]$) compact set $p$ can take any value only once so again if the degree is $3$ or higher there are complex non real roots to  $p(z)=b$ for some real $b$ with $|b|$ large enough and that is not possible.
Hence $p=az+b,a.b \in \mathbb R$ and then the orientation property (upper plane to upper plane) shows $a>0$
