Question

Let two non-zero distinct complex numbers $z_1$ and $z_2$ be represented respectively by points $A$ and $B$ in the complex plane $O$ as the origin. Then $OA$ is perpendicular to $OB$ if an only if

• A. $z_{1} + z_{2} = z_{2} + z_{1}$
• B. $z_{1} - z_{2} = z_{2} - z_{1}$
• C. $z_{1} z_{2} + z_{2}z_{1} = 0 $
• D. $z_{1} z_{2} - z_{2}z_{1} = 0 $

Answer 1

Answer: (C)

$OA$ and $OB $ are perpendicular iff slope of $OA \times$ slope of $OB = 1$ $\left(\frac{b_{1} -0}{a_{1} - 0}\right) \cdot \left( \frac{b_{2} - 0}{a_{2} - 0}\right) = -1 $
$\Rightarrow b_{1}b_{2} = - a_{1}a_{2} \Rightarrow b_{1}b_{2} + a_{1} a_{2} = 0 $


This is possible in case of $z_{1} z_{2} + z_{2} z_{1} = 0 $
i.e., $a_{1}a_{2} + a_{1} b_{2} i + b_{1} a_{2}i +b_{1}b_{2} i^{2} +a_2a_1+ a_{2} b_{1}i + b_{2} a_{1} i + b_{1}b_{2}i^{2}=0 $
$\Rightarrow 2a_{1}a_{2} - 2b_{1}b_{2} + \left(2a_{1} b_{2} + 2 b_{1} a_{2}\right)i = 0$
$ \Rightarrow 2a_{1}a_{2} - 2b_{1}b_{2} = 0$ and $\left(2a_{1} b_{2} +2b_{1} a_{2}\right) = 0$
$ \Rightarrow a_{1}a_{2} = b_{1}b_{2}$