Question

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 ${{z}_{1}}R{{z}_{2}}\Leftrightarrow \dfrac{{{z}_{1}}-{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}$ is real for all ${{z}_{1}},{{z}_{2}}\in {{C}_{0}}$ . Show that $R$ is an equivalence relation.

Answer 1

Hint: For solving this question first we will find the value of $\dfrac{{{z}_{1}}-{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}$ for any general value of ${{z}_{1}},{{z}_{2}}\in {{C}_{0}}$ then we will prove that the given relation is reflexive, symmetric and transitive then the relation will be automatically an equivalence relation.

Complete step-by-step answer:
Given:
Two complex numbers ${{z}_{1}}$ and ${{z}_{2}}$ have relation such that ${{z}_{1}}R{{z}_{2}}\Leftrightarrow \dfrac{{{z}_{1}}-{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}$ is real for all ${{z}_{1}},{{z}_{2}}\in {{C}_{0}}$ .
Now, let ${{z}_{1}}={{x}_{1}}+{{y}_{1}}\text{i}$ and ${{z}_{2}}={{x}_{2}}+{{y}_{2}}\text{i}$ are two non-zero complex numbers. Then,
$\begin{align}
  & \dfrac{{{z}_{1}}-{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}=\dfrac{\left( {{x}_{1}}+{{y}_{1}}\text{i} \right)-\left( {{x}_{2}}+{{y}_{2}}\text{i} \right)}{\left( {{x}_{1}}+{{y}_{1}}\text{i} \right)-\left( {{x}_{2}}+{{y}_{2}}\text{i} \right)}=\dfrac{\left( {{x}_{1}}-{{x}_{2}} \right)+\left( {{y}_{1}}-{{y}_{2}} \right)\text{i}}{\left( {{x}_{1}}-{{x}_{2}} \right)+\left( {{y}_{1}}-{{y}_{2}} \right)\text{i}} \\
 & \Rightarrow \dfrac{{{z}_{1}}-{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}=1 \\
\end{align}$
Thus, from the above result, we can say that for any two non-zero complex numbers ${{z}_{1}},{{z}_{2}}\in {{C}_{0}}$ the value of $\dfrac{{{z}_{1}}-{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}$ is constant and is equal to 1. Thus, now we will prove that the given relation is reflexive, symmetric, and transitive.

1. If ${{z}_{1}}={{z}_{2}}=a$ where $a$ is any non-zero complex number then also ${{z}_{1}}R{{z}_{2}}\Leftrightarrow \dfrac{{{z}_{1}}-{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}$ is real as the value of $\dfrac{{{z}_{1}}-{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}$ is constant and is equal to 1 that’s why $\left( a,a \right)\in R$. Thus, the given relation is reflexive.

2. If ${{z}_{1}}=a$ and ${{z}_{2}}=b$ are two non-zero complex numbers then ${{z}_{1}}R{{z}_{2}}\Leftrightarrow \dfrac{{{z}_{1}}-{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}$ is real as the value of $\dfrac{{{z}_{1}}-{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}$ is constant and is equal to 1 that’s why $\left( a,b \right)\in R$. Now, if ${{z}_{1}}=b$ and ${{z}_{2}}=a$ then also then ${{z}_{1}}R{{z}_{2}}\Leftrightarrow \dfrac{{{z}_{1}}-{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}$ is real as the value of $\dfrac{{{z}_{1}}-{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}$ is constant and is equal to 1 that’s why $\left( b,a \right)\in R$. Thus, now $\left( a,b \right)\in R$ and also $\left( b,a \right)\in R$ . So, the given relation will be symmetric also.

3. If ${{z}_{1}}=a$ and ${{z}_{2}}=b$ are two non-zero complex numbers then ${{z}_{1}}R{{z}_{2}}\Leftrightarrow \dfrac{{{z}_{1}}-{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}$ is real as the value of $\dfrac{{{z}_{1}}-{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}$ is constant and is equal to 1 that’s why $\left( a,b \right)\in R$. And if ${{z}_{1}}=b$ and ${{z}_{2}}=c$ then also ${{z}_{1}}R{{z}_{2}}\Leftrightarrow \dfrac{{{z}_{1}}-{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}$ is real as the value of $\dfrac{{{z}_{1}}-{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}$ is constant and is equal to 1 that’s why $\left( b,c \right)\in R$. Now, if ${{z}_{1}}=a$ and ${{z}_{2}}=c$ are two non-zero complex numbers then ${{z}_{1}}R{{z}_{2}}\Leftrightarrow \dfrac{{{z}_{1}}-{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}$ is real as the value of $\dfrac{{{z}_{1}}-{{z}_{2}}}{{{z}_{1}}-{{z}_{2}}}$ is constant and is equal to 1 that’s why $\left( a,c \right)\in R$. Thus, now $\left( a,b \right)\in R$ and $\left( b,c \right)\in R$ then $\left( a,c \right)\in R$. So, the given relation will be transitive also.

Now, as we have proved that the given relation is reflexive, symmetric and transitive. Thus, the relation will be an equivalence relation.
Hence, proved.

Note: Here, the student must try to understand the relation given in the problem then prove that the relation is reflexive, symmetric, and transitive separately and don’t mix up their conditions with each other to avoid the confusion.