Geometric description of points in the complex plane What does the following inequality look like if sketched in the complex plane:
$$\operatorname{Re}(az+b)>0$$
The $a$ and $b$ are both complex numbers (function parameters). I understand that $\operatorname{Re}(z)>0$ will yield all of the complex numbers to the right of the imaginary axis (quadrants 1 and 4). I also understand that $\operatorname{Re}(z+b)>0$ will shift the shaded region $\operatorname{Re}(b)$ to the left of the imaginary axis.
What does multiplying the $z$ by $a$ do to the inequality ($\operatorname{Re}(az)>0$)? Complex multiplication involves dilation and rotation, so would the shaded (included) region rotate?
 A: If $z=x+iy$, $$Re(az+b)=Re(a)x-Im(a)y+Re(b)$$ So, $$Re(az+b)>0\Rightarrow mx-ny+c>0$$ where  $m=Re(a),\  n=Im(a),\  c=Re(b)$ So it is the one half of $\mathbb{R}^2$ partitioned by the line  $\displaystyle mx-ny+c=0$ So, basically, it is $Re(z)>0$ rotated by $a$ and translated by $b$.
A: The mapping $z \mapsto az+b$ is just a special case of a Möbius transformation, $z \mapsto \frac{az+b}{cz+d}$ where $d = 1$ and $c = 0$. Mappings of this form are indeed dilations and rotations.
The best way to visualize this would be using the Riemann sphere. After the rotation, your region is one hemisphere. Invert the mapping to find the region that maps to it. The inverse mapping is easy, since the matrix representation of our Möbius transformation is invertible (assuming $a \neq 0$):
$$\begin{pmatrix} a & b \\ 0 & 1\end{pmatrix}.$$
If you really want, the Riemann sphere equations are easy to write down (depending on how you represent the sphere), so you can easily substitute your mapping into the equations!

Edit: For increased mathematical clarity.
One of the interesting features of the Riemann sphere is that Möbius transformations are automorphisms of the sphere.
When we look at the region $\textrm{Re}(w) > 0$, then we're looking at one hemisphere. In this case, it's the hemisphere that projects onto the positive real half-plane, including both the northern and southern portions. If you "colored" this part of the sphere, and then applied the inverse mapping, you'd get the region in $\mathbb{C}$ that satisfies the inequality. Then you can project this colored region back onto the plane to visualize it in the traditional Argand sense!
