Let C be the set of all complex numbers and C0 be the set of all non-zero complex numbers. Let a relation R on C0 be defined asz1Rz2⇔z1−z2z1−z2 is real for all z1,z2∈C0. Show that R is an equivalence relation.

Question

Let C be the set of all complex numbers and C0​ be the set of all non-zero complex numbers. Let a relation R on C0​ be defined asz1​Rz2​⇔z1​−z2​z1​−z2​​ is real for all z1​,z2​∈C0​. Show that R is an equivalence relation.

Filo · Accepted Answer

(i)Test for reflexivity:Since ,z1​+z1​z1​−z1​​=0, which is a real number.So,(z1​,z1​)∈RHence,R is a reflexive relation.(ii)Test for symmetric:Let (z1​,z2​)∈R⇒z1​+z2​z1​−z2​​=x, where x is real⇒−z1​+z2​z1​−z2​​=−x⇒z2​+z1​z2​−z1​​=−x is also a real number.So,(z2​,z1​)∈RHence,R is symmetric relation.(iii)Test for transitivity:Let (z1​,z2​)∈R and (z2​,z3​)∈RThen z1​+z2​z1​−z2​​=x⇒z1​−z2​=xz1​+xz2​⇒z1​(1−x)=z2​(1+x)⇒z2​z1​​=1−x1+x​         .......(1)Also,z2​+z3​z2​−z3​​=y⇒z2​−z3​=yz2​+yz3​⇒z2​(1−y)=z3​(1+y)⇒z3​z2​​=1−y1+y​         .......(2) Dividing (1) and (2), we getz3​z1​​=1−x1+x​×1+y1−y​=z where z is a real number.z1​+z3​z1​−z3​​=z+1z−1​, which is real.⇒(z1​,z3​)∈R             .......(3)From (1),(2) and (3) we have as R is an equivalence relation.

Question Text	Let $C$ be the set of all complex numbers and $C_{0}$ be the set of all non-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.
Answer Type	Text solution:1
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Let $C$ be the set of all complex numbers and $C_{0}$ be the set of all non-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.

Text solutionVerified

Practice questions from similar books

Let C be the set of all complex numbers and C0​ be the set of all non-zero complex numbers. Let a relation R on C0​ be defined asz1​Rz2​⇔z1​−z2​z1​−z2​​ is real for all z1​,z2​∈C0​. Show that R is an equivalence relation.

Text solutionVerified

Practice questions from similar books

Let $C$ be the set of all complex numbers and $C_{0}$ be the set of all non-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.