# Prove that set with complex numbers is circumference

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.

• Presumably $\alpha \neq \beta$? Commented Jan 3, 2020 at 18:08
• Can be $\alpha = \beta$ Commented Jan 3, 2020 at 18:11
• If $\alpha = \beta$ then the set is empty. Commented Jan 3, 2020 at 18:11
• There is a geometric way of looking at this problem that will give the center and radius fairly easily... Pick two points $\alpha, \beta$ on the plane and draw a line between them. Commented Jan 3, 2020 at 18:12
• And analitically? Commented Jan 3, 2020 at 18:19

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.

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.

• Try to explain the solution to a person that is beginning with Complex Variable hahaha Commented Jan 3, 2020 at 18:05
• What sort of topics have you covered so far? Commented Jan 3, 2020 at 18:06
• Until Residue Theorem, but it's a problem from first topic of course. Commented Jan 3, 2020 at 18:09

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$$\left|\frac{\gamma-\delta}{2}\right|=\left|\frac{\rho(\beta-\alpha)}{1-\rho^2}\right|.$$