It is easy to understand that $z=a+bt$ is a straight line. Here, $a$ and $b$ are complex numbers and $b\neq0$; the parameter $t$ runs through all real values.
But how to understand $Im \left[\frac{z-a}{b}\right] <0 $ is a half plane? Notice that $b$ is a complex number here.
This a question from Ahlfors' Complex Analysis. And I show you the origin of the question below. I have trouble understanding the statement with red line.
I'm using $\Im$ to denote the imaginary part (MathJax "\Im"). Consider the usual representation with the real axis horizontal and the imaginary axis vertical.
$\Im (z) < 0$ is the half plane down of the real axis.
$\Im (z-a) < 0$ moves the above half plane, it is now the half plane down of a horizontal axis going through $a$.
Division by $b$ is the same as multiplication with $1/b$. Geometrically, if $1/b=re^{i\phi}$, that is rotation around the origin with angle $\phi$, then stretching with factor $r$. Both of these operations transform a half plane into a half plane, so indeed
$$\Im (\frac{z-a}b) < 0$$
is a half plane, but which? It should be clear that the boundary of that half plane is where $\Im (\frac{z-a}b) = 0$ holds, but that is exatly the set of points with $z=a+bt$, because exactly then we have $\frac{z-a}b = t \in \mathbb R$ and $t \in \mathbb R \iff \Im(t)=0$.
Which one is the left/right half plane is a bit harder to determine. Let's say I trust Ahlfors on that one, maybe somebody else can make a good argument why that is true.
