text: #### Please Solve RD Sharma Class 12 Chapter Relation Exercise 1.2 Question 17 Maths Textbook Solution.
  
  Answer:  Hence prove, R is an equivalence relation.
  Hint: To prove an equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.
  Explanation:
  Given:  $\text{C\: be\: the\: set\: o\! f\: all\: complex \: numbers.}$
  $C_{0}\: \text{be\: the\: set\: o\! f\: all\: non-zero\: complex\: numbers.}$
  $\text{Relation R on }C_{0} \text{ be defined as}$
  $z_{1}\: R\: z_{2}\Leftrightarrow\frac{ z_{1}-z_{2}}{z_{1}+z_{2}} \text{ is real for all } z_{1}, z_{2}\: \epsilon \: C_{0}$
  (i) Test for reflexivity
  $Since, \frac{ z_{1}-z_{1}}{z_{1}+z_{1}}=0, which\: is\: a \: real \: number$
  $So, (z_{1}, z_{1})\: \epsilon \: R$
  Hence, R is a reflexive relation.
  (ii) Test for symmetric
  $\frac{z_{1}-z_{2}}{z_{1}+z_{2}}=x , where\: x\: is\: real$
  $-\frac{ z_{1}-z_{2}}{z_{1}+z_{2}}= -x$
  $\frac{z_{2}-z_{1}}{z_{2}+z_{1}}= -x \: is\: also\: a\: real\: number$
  $So, (z_{2},z_{1}) \: \epsilon \: R$
  Hence, R is a symmetric relation.
  (iii) Test for transitivity
  $Let (z_{1}, z_{2}) \: \epsilon \: R \: and\: (z_{2}, z_{3} ) \: \epsilon \: R$
  $Then\: \frac{z_{1}-z_{2}}{z_{1}+z_{2}}= x$
  $z_{1}-z_{2}=xz_{1}+xz_{2}$
  $\frac{z_{1}}{z_{2}}=\frac{1+x}{1-x }$                                                                                                                                                  $...(i)$
  $Also, \frac{ z_{2}-z_{3}}{z_{2}+z_{3}}=y$
  $z_{2}-z_{3}=yz_{2}+yz_{3}$
  $z_{2}\left ( 1-y \right )=z_{3}(1+y)$
  $\frac{z_{2}}{z_{3}}=\frac{1+y}{1-y }$                                                                                                                                                    $...(ii)$
  Dividing (i) and (ii), we get
  $\frac{z_{1}}{z_{3}}=\frac{1+x}{1-x}\times \frac{1+y}{1-y}$
  $=z\: where\: z\: is\: a\: real \: number.$
  $\frac{z_{1}-z_{3}}{z_{1}+z_{3}}=\frac{z-1}{z+1}, which \: is\: real$
  $(z_{1}, z_{3})\: \epsilon \: R$                                                                                                                                                             $...(iii)$
  From (i), (ii) & (iii)
  $\text{Hence, R\: is\: transitivity\: relation.}$
  $\text{Hence\: proved, R \: is\: an\: equivalence \: relation.}$
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