# Math Help - Complex numbers, set of points satisfying an equality

1. ## Complex numbers, set of points satisfying an equality

Find the set of points satisfying the equality $\left | \frac{z-1}{z+1} \right |=c$, with $c\geq 0$ an arbitrary constant. (I guess in $\mathbb{R}$).

My attempt :
If $c=0$ then $z=1$, which makes a point on the complex plane.
If $c=1$ then $x=0$ and $y$ can be any number in $\mathbb{R}$ so we've a straight line in the complex plane ( $x=0$) and I believe it will be the same for any other $c \neq 0 \in \mathbb{R}$.
Am I right?

2. Originally Posted by arbolis
Find the set of points satisfying the equality $\left | \frac{z-1}{z+1} \right |=c$, with $c\geq 0$ an arbitrary constant. (I guess in $\mathbb{R}$).

My attempt :
If $c=0$ then $z=1$, which makes a point on the complex plane.
If $c=1$ then $x=0$ and $y$ can be any number in $\mathbb{R}$ so we've a straight line in the complex plane ( $x=0$) and I believe it will be the same for any other $c \neq 0 \in \mathbb{R}$.
Am I right?
$\left | \frac{z-1}{z+1} \right |=c \Rightarrow \left | z-1 \right |=c\left | z+1 \right | \Rightarrow \left | z-1 \right |^2=c^2 \left | z+1 \right |^2$ $\Rightarrow (z-1)(\bar{z}-1)=c^2(z+1)(\bar{z}+1)\Rightarrow (c^2-1)z\bar{z}+(c^2+1)(z+\bar{z})+(c^2-1)=0$

- If $c=0$, is the point $z=1$.
If $c=1, z+\bar{z}=2x=0$, is the imaginary axis.
Otherwise, $z\bar{z}+\frac{c^2+1}{c^2-1}(z+\bar{z})+1=0 \Rightarrow \left | z+\frac{c^2+1}{c^2-1} \right | = \frac{2c}{|c^2-1|}$, is a circle.

3. Originally Posted by luobo
$\left | \frac{z-1}{z+1} \right |=c \Rightarrow \left | z-1 \right |=c\left | z+1 \right | \Rightarrow \left | z-1 \right |^2=c^2 \left | z+1 \right |^2$ $\Rightarrow (z-1)(\bar{z}-1)=c^2(z+1)(\bar{z}+1)\Rightarrow (c^2-1)z\bar{z}+(c^2+1)(z+\bar{z})+(c^2-1)=0$

If $c=0$, is the point $z=1$.
If $c=1, z+\bar{z}=2x=0$, is the imaginary axis.
Otherwise, $z\bar{z}+\frac{c^2+1}{c^2-1}(z+\bar{z})+1=0 \Rightarrow \left | z+\frac{c^2+1}{c^2-1} \right | = \frac{2c}{|c^2-1|}$, is a circle.
In fact, circles such as these have a special name: Circles of Apollonius.