Question

Let C be the set of all complex numbers and C₀ be the set of all no-zero complex numbers. Let a relation R on C₀ be defined as

$z_1$ R $z_2\iff \frac{z_1-z_2}{z_1+z_2}$ is real for all $z_1,z_2\in$C₀.

Show that R is an equivalence relation.

Answer 1

(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

$$\begin{gathered} \Rightarrow -\left(\frac{z_1-z_2}{z_1+z_2}\right)=-x \\ \Rightarrow \left(\frac{z_2-z_1}{z_2+z_1}\right)=-x,\mathrm{is}\mathrm{also}\mathrm{a}\mathrm{real}\mathrm{number} \end{gathered}$$

So, $\left(z_2,z_1\right)\in R$

Hence, R is symmetric relation.

(iii) Test for transivity:

Let $\left(z_1,z_2\right)\in R\mathrm{and}\left({\mathrm{z}}_2,{\mathrm{z}}_3\right)\in \mathrm{R}$.

Then,

$$\begin{gathered} \frac{z_1-z_2}{z_1+z_2}=x,\mathrm{where}x\mathrm{is}\mathrm{a}\mathrm{real}\mathrm{number}. \\ \Rightarrow z_1-z_2=x{z}_1+x{z}_2 \\ \Rightarrow z_1-x{z}_1=z_2+x{z}_2 \\ \Rightarrow z_1\left(1-x\right)=z_2\left(1+x\right) \\ \Rightarrow \frac{z_1}{z_2}=\frac{\left(1+x\right)}{\left(1-x\right)}...\left(1\right) \end{gathered}$$

Also,

$$\begin{gathered} \frac{z_2-z_3}{z_2+z_3}=y,\mathrm{where}y\mathrm{is}\mathrm{a}\mathrm{real}\mathrm{number}. \\ \Rightarrow z_2-z_3=y{z}_2+Y{z}_3 \\ \Rightarrow z_2-y{z}_2=z_3+y{z}_3 \\ \Rightarrow z_2\left(1-y\right)=z_3\left(1+y\right) \\ \Rightarrow \frac{z_2}{z_3}=\frac{\left(1+y\right)}{\left(1-y\right)}...\left(2\right) \end{gathered}$$

Dividing (1) and (2), we get

$$\begin{gathered} \frac{z_1}{z_3}=\left(\frac{1+x}{1-x}\right)\times \left(\frac{1-y}{1+y}\right)=z,\mathrm{where}\mathit{}z\mathrm{is}\mathrm{a}\mathrm{real}\mathrm{number}. \\ \Rightarrow \frac{{\displaystyle z_1-z_3}}{{\displaystyle z_1+z_3}}=\frac{{\displaystyle z-1}}{{\displaystyle z+1}},\mathrm{which}\mathrm{is}\mathrm{real} \\ \Rightarrow \left(z_1,z_3\right)\in R \end{gathered}$$

Hence, R is transitive relation.

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

R is an equivalenve relation.