Preparing for an exam, I have come up against this problem and I'm not capable to finish it. Could you give me a hand with this?
Given two complex numbers $\alpha, \beta$, and positive $\rho \neq 1$, justifies that set
$$\{z \in \mathbb{C}: \frac{|z-\alpha|}{|z-\beta|} = \rho\}$$
represents a circumference and calculate center and radius of that circumference.
On
This is the complexification of a classical definition of the circle. We can parameterise the solutions with a real $\theta$ viz.$$\frac{z-\alpha}{z-\beta}=\rho\exp i\theta\iff z=\frac{\alpha-\rho\beta\exp i\theta}{1-\rho\exp i\theta}.$$Define $\gamma:=\frac{\alpha+\rho\beta}{1+\rho},\,\delta:=\frac{\alpha-\rho\beta}{1-\rho}$ so$$z-\gamma=\frac{\alpha-\rho\beta\exp i\theta}{1-\rho\exp i\theta}-\frac{\alpha+\rho\beta}{1+\rho}=\frac{\rho(\alpha-\beta)(\exp i\theta-1)}{(1-\rho\exp i\theta)(1+\rho)}$$and$$z-\delta=\frac{\alpha-\rho\beta\exp i\theta}{1-\rho\exp i\theta}-\frac{\alpha-\rho\beta}{1-\rho}=\frac{\rho(\alpha-\beta)(\exp i\theta+1)}{(1-\rho\exp i\theta)(1-\rho)}=-i\left[\frac{1+\rho}{1-\rho}\cot\frac{\theta}{2}\right](z-\gamma).$$Since $\frac{z-\gamma}{z-\delta}$ is imaginary, $\angle \gamma z\delta=\frac{\pi}{2}$ for all $z$ in the locus, so it's a circle with $\gamma,\,\delta$ the ends of a diameter. Its centre is$$\frac{\gamma+\delta}{2}=\frac{\alpha-\rho^2\beta}{1-\rho^2},$$while the diameter is$$\left|\frac{\gamma-\delta}{2}\right|=\left|\frac{\rho(\beta-\alpha)}{1-\rho^2}\right|.$$
The set $C=\{w | |w| = \rho \}$ is a circle or radius $\rho$ centered at the origin.
The map $f(z) = {z-\alpha \over z-\beta}$ is a Möbius transformation (assuming $\alpha \ne \beta$) and has an inverse $g(z) = {\beta z -\alpha \over z-1}$.
Let $S= \{z | |f(z)| = \rho \}$, then $z \in S$ iff $f(z) \in C$ iff $z \in g(C)$.
A Möbius transformation maps (Riemann) circles to (Riemann) circles, so we see that $S$ is a Riemann circle.
If $z \in C$ we see that $z \neq 1$ (as $\rho \neq 1$) and so $g(z)$ is finite for all $z \in C$, hence $g(C)$ is a circle in the usual sense.
Alternative answer:
First, notice that a circle of radius $r$ centered at $c$ is defined by the equation $|z-c|=r$ which (when squared) yields $|z|^2 - 2 \operatorname{re} c \bar{z} +|c|^2 = r^2$.
As long as $\rho \neq 1$ and $\alpha \neq \beta$, we see that the set in question can be defined by $|z-\alpha| = \rho |z-\beta|$ and squaring gives $|z|^2 - 2 \operatorname{re} \alpha \bar{z} +|\alpha|^2 = \rho^2 (|z|^2 - 2 \operatorname{re} \beta \bar{z} +|\beta|^2)$ or $|z|^2 - 2 \operatorname{re} {(\alpha - \rho^2 \beta) \over 1 - \rho^2} \bar{z} +{ |\alpha|^2 - \rho^2 |\beta|^2 \over 1 - \rho^2 } = 0$.
If we compare with the above equation, this suggests that the centre is $c={\alpha - \rho^2 \beta \over 1 - \rho^2 }$ and so the equation can be written as $|z-c|^2 = |c|^2 - { |\alpha|^2 - \rho^2 |\beta|^2 \over 1 - \rho^2 } $.
If we expand $|c|^2$ we get $|c|^2 - { |\alpha|^2 - \rho^2 |\beta|^2 \over 1 - \rho^2 } = | {\rho(\beta-\alpha) \over 1-\rho^2} |^2$, so letting $r= | {\rho(\beta-\alpha) \over 1-\rho^2} |$ we get the desired result.