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Please Solve RD Sharma Class 12 Chapter Relation Exercise 1.2 Question 17 Maths Textbook Solution.

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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.
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
\frac{z_{1}}{z_{2}}=\frac{1+x}{1-x }                                                                                                                                                  ...(i)
Also, \frac{ z_{2}-z_{3}}{z_{2}+z_{3}}=y
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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