Question

A relation R on the set of complex numbers is defined by $z_1$ R$z_2$ if and only if $(z_1-z_2)/(z_1+z_2)$ is real. Show that R is an equivalence relation.

Answer 1

Here $z_1$ R $z_2$ for all complex number $z_1$.
For we have $\frac{z_1-z_1}{z_1+z_2}=0$ is real, then $\frac{z_1-z_1}{z_1+z_2}$ is real, then $\frac{z_2-z_1}{z_2+z_1}$ is also real.
Hence $z_{1}R{z}_2\to z_{2}R{z}_1$.
We now show that R is transitive.
Let $z_1=x_1+i{y}_1,z_2=x_2+i{y}_2$ and $z_3=x_3+i{y}_3$ be three complex numbers such that $z_{1}R{z}_2$ and $z_{2}R{z}_3$.
Now $z_{1}R{z}_2\Rightarrow \frac{z_1-z_2}{z_1+z_2}$ is real
$\Rightarrow \frac{(x_1-x_2)+i(y_1-y_2)}{(x_1+x_2)+i(y_1+y_2)}$ is real
$\Rightarrow \frac{\left[\right(x_1-x_2)+i(y_1-y_2\left)\right]\left[\right(x_1+x_2)-i(y_1-y_2\left)\right]}{(x_1+x_2{)}^2+(y+y_2{)}^2]}$ is real
$\Rightarrow (y_1-y_2)(x_1+x_2)-(x_1-x_2)-(y_1+y_2)=0$
$\Rightarrow 2x_2{y}_1-2y_2{x}_1=0\Rightarrow \left(x_1/y_1\right)=\left(x_2/y_2\right)$ ........(1)
Similarlly $z_{2}R{z}_3=\frac{x_2}{y_2}=\frac{x_3}{y_3}$ ...(2)
From (1) and (2), we have $\frac{x_1}{y_1}=\frac{x_3}{y_3}$
Hence $z_{1}R{z}_3$.
Thus $z_{1}R{z}_2$ and $z_{2}R{z}_3\Rightarrow z_{1}R{z}_3$.
Hence R is transitive. It follows that R is an equivalence relation.
Note: If we consider R is a relation on the set of all complex number but 0(~R)0 for we have $\frac{0-0}{0+0}=\frac{0}{0}$ which is indeterminate. Hence in order that R may be an equivalence relation the set on which R is defined must be non-zero complex numbers.