# Thread: 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 $\displaystyle \left | \frac{z-1}{z+1} \right |=c$, with $\displaystyle c\geq 0$ an arbitrary constant. (I guess in $\displaystyle \mathbb{R}$).

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

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

My attempt :
If $\displaystyle c=0$ then $\displaystyle z=1$, which makes a point on the complex plane.
If $\displaystyle c=1$ then $\displaystyle x=0$ and $\displaystyle y$ can be any number in $\displaystyle \mathbb{R}$ so we've a straight line in the complex plane ($\displaystyle x=0$) and I believe it will be the same for any other $\displaystyle c \neq 0 \in \mathbb{R}$.
Am I right?
$\displaystyle \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 $$\displaystyle \Rightarrow (z-1)(\bar{z}-1)=c^2(z+1)(\bar{z}+1)\Rightarrow (c^2-1)z\bar{z}+(c^2+1)(z+\bar{z})+(c^2-1)=0 - If \displaystyle c=0, is the point \displaystyle z=1. If \displaystyle c=1, z+\bar{z}=2x=0, is the imaginary axis. Otherwise, \displaystyle z\bar{z}+\frac{c^2+1}{c^2-1}(z+\bar{z})+1=0 \Rightarrow \left | z+\frac{c^2+1}{c^2-1} \right | = \frac{2c}{|c^2-1|}, is a circle. 3. Originally Posted by luobo \displaystyle \left | \frac{z-1}{z+1} \right |=c \Rightarrow \left | z-1 \right |=c\left | z+1 \right | \Rightarrow \left | z-1 \right |^2=c^2 \left | z+1 \right |^2$$\displaystyle \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 $\displaystyle c=0$, is the point $\displaystyle z=1$.
If $\displaystyle c=1, z+\bar{z}=2x=0$, is the imaginary axis.
Otherwise, $\displaystyle z\bar{z}+\frac{c^2+1}{c^2-1}(z+\bar{z})+1=0 \Rightarrow \left | z+\frac{c^2+1}{c^2-1} \right | = \frac{2c}{|c^2-1|}$, is a circle.
In fact, circles such as these have a special name: Circles of Apollonius.