# Let C be the set of all complex numbers and C0 be the set of all no-zero complex numbers.

Question:

Let $C$ be the set of all complex numbers and $C_{0}$ be the set of all no-zero complex numbers. Let a relation $R$ on $C_{0}$ be defined as

$z_{1} R z_{2} \Leftrightarrow \frac{z_{1}-z_{2}}{z_{1}+z_{2}}$ is real for all $z_{1}, z_{2} \in C_{0}$.

Show that R is an equivalence relation.

Solution:

(i) Test for reflexivity:

Since, $\frac{z_{1}-z_{1}}{z_{1}+z_{1}}=0$, which is a real number.

So, $\left(z_{1}, z_{1}\right) \in R$

Hence, R is relexive relation.

(ii) Test for symmetric:

Let $\left(z_{1}, z_{2}\right) \in R$.

Then, $\frac{z_{1}-z_{2}}{z_{1}+z_{2}}=x$, where $x$ is real

$\Rightarrow-\left(\frac{z_{1}-z_{2}}{z_{1}+z_{2}}\right)=-x$

$\Rightarrow\left(\frac{z_{2}-z_{1}}{z_{2}+z_{1}}\right)=-x$, is also a real number

So, $\left(z_{2}, z_{1}\right) \in R$

Hence, R is symmetric relation.

(iii) Test for transivity:

Let $\left(z_{1}, z_{2}\right) \in R$ and $\left(\mathrm{z}_{2}, \mathrm{z}_{3}\right) \in \mathrm{R}$.

Then,

$\frac{z_{1}-z_{2}}{z_{1}+z_{2}}=x$, where $x$ is a real number.

$\Rightarrow z_{1}-z_{2}=x z_{1}+x z_{2}$

$\Rightarrow z_{1}-x z_{1}=z_{2}+x z_{2}$

$\Rightarrow z_{1}(1-x)=z_{2}(1+x)$

$\Rightarrow \frac{z_{1}}{z_{2}}=\frac{(1+x)}{(1-x)} \quad \ldots(1)$

Also,

$\frac{z_{2}-z_{3}}{z_{2}+z_{3}}=y$, where $y$ is a real number.

$\Rightarrow z_{2}-z_{3}=y z_{2}+Y z_{3}$

$\Rightarrow z_{2}-y z_{2}=z_{3}+y z_{3}$

$\Rightarrow z_{2}(1-y)=z_{3}(1+y)$

$\Rightarrow \frac{z_{2}}{z_{3}}=\frac{(1+y)}{(1-y)} \quad \ldots(2)$

Dividing (1) and (2), we get

$\frac{z_{1}}{z_{3}}=\left(\frac{1+x}{1-x}\right) \times\left(\frac{1-y}{1+y}\right)=z$, where $z$ is a real number.

$\Rightarrow \frac{z_{1}-z_{3}}{z_{1}+z_{3}}=\frac{z-1}{z+1}$, which is real $\Rightarrow\left(z_{1}, z_{3}\right) \in R$

Hence, R is transitive relation.

From (i), (ii), and (iii),

R is an equivalenve relation.