Complex numbers, set of points satisfying an equality

**arbolis** · September 1st 2009, 02:52 PM

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?

**luobo** · September 1st 2009, 04:46 PM

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.

**mr fantastic** · September 1st 2009, 07:39 PM

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.

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