A relation R on the set of complex numbers defined by Z1RZ2⇔Z1−Z2Z1+Z2 is real then which of the following is not true

Z1RZ1⇒Z1−Z1Z1+Z1=0∴R is Reflexive because 0 is real. Z1RZ2⇒Z1−Z2Z1+Z2 is real ⇒−Z1−Z2Z1+Z2 is real ∴ R is symmetric Z1=a1+ib1; Z2=a2+ib2Z1RZ2⇒Z1−Z2Z1+Z2 is real ⇒a1+a2b1−b2−a1−a2b1+b2=0⇒2a2b1−2b2a1=0⇒a1b1=a2b2∴Z2RZ3⇒a2b2=a3b3therefore a1b1=a3b3∴ Z1RZ3⇒R is transitive.

A relation R on the set of complex numbers defined by Z1RZ2⇔Z1−Z2Z1+Z2 is real then which of the following is not true

Z1RZ1⇒Z1−Z1Z1+Z1=0∴R is Reflexive because 0 is real. Z1RZ2⇒Z1−Z2Z1+Z2 is real ⇒−Z1−Z2Z1+Z2 is real ∴ R is symmetric Z1=a1+ib1; Z2=a2+ib2Z1RZ2⇒Z1−Z2Z1+Z2 is real ⇒a1+a2b1−b2−a1−a2b1+b2=0⇒2a2b1−2b2a1=0⇒a1b1=a2b2∴Z2RZ3⇒a2b2=a3b3therefore a1b1=a3b3∴ Z1RZ3⇒R is transitive.

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A relation R on the set of complex numbers defined by $Z_{1} R Z_{2} \Leftrightarrow \frac{Z_{1} - Z_{2}}{Z_{1} + Z_{2}} is real then which of the following is not true$

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R is reflexive

R is symmetric

R is transitive

R is not equivalence

answer is D.

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Detailed Solution

$\begin{array}{l} Z_{1} R Z_{1} \Rightarrow \frac{Z_{1} - Z_{1}}{Z_{1} + Z_{1}} = 0 \\ ∴ R is Reflexive because 0 is real. \end{array}$

$\begin{array}{l} Z_{1} R Z_{2} \Rightarrow \frac{Z_{1} - Z_{2}}{Z_{1} + Z_{2}} is real \Rightarrow - (\frac{Z_{1} - Z_{2}}{Z_{1} + Z_{2}}) is real \\ ∴ R is symmetric \end{array}$

$\begin{array}{l} Z_{1} = a_{1} + i b_{1}; Z_{2} = a_{2} + i b_{2} \\ Z_{1} R Z_{2} \Rightarrow \frac{Z_{1} - Z_{2}}{Z_{1} + Z_{2}} is real \\ \Rightarrow (a_{1} + a_{2}) (b_{1} - b_{2}) - (a_{1} - a_{2}) (b_{1} + b_{2}) = 0 \\ \Rightarrow 2 a_{2} b_{1} - 2 b_{2} a_{1} = 0 \\ \Rightarrow \frac{a_{1}}{b_{1}} = \frac{a_{2}}{b_{2}} \\ ∴ Z_{2} R Z_{3} \Rightarrow \frac{a_{2}}{b_{2}} = \frac{a_{3}}{b_{3}} \end{array}$