Question

If the lines $a\overline{z}+\overline{a}z+b=0$ and $c\overline{z}+\overline{c}z+d=0$ are mutually perpendicular, where $a$ and $c$ are non zero complex numbers, while $b$ and $d$ are real numbers, then

• A. $a\overline{a}+c\overline{c}=0$
• B. $a\overline{c}$ is purely imaginary
• C. $\arg \left(\frac{a}{c}\right)=\pm \frac{\pi }{2}$
• D. $\frac{a}{\overline{a}}=\frac{c}{\overline{c}}$

Answer 1

Answer: (B), (C)

$\arg \left(\frac{a}{c}\right)=\pm \frac{\pi }{2}$

Let $a=a_1+i{a}_2$ and $c=c_1+i{c}_2.$
Then, $\frac{a_1}{a_2}\times \frac{c_1}{c_2}=-1$
$\Rightarrow a_1{c}_1+a_2{c}_2=0$
$\Rightarrow \left(\frac{a+\overline{a}}{2}\right)\left(\frac{c+\overline{c}}{2}\right)+\left(\frac{a-\overline{a}}{2i}\right)\left(\frac{c-\overline{c}}{2i}\right)=0$
$\Rightarrow a\overline{c}+\overline{a}c=0\cdots \left(1\right)$
$\therefore a\overline{c}$ is purely imaginary.

From $\left(1\right)$, we have
$\frac{a}{c}=-\frac{\overline{a}}{\overline{c}}$
$\Rightarrow \frac{a}{c}+\frac{\overline{a}}{\overline{c}}=0$
$\Rightarrow \frac{a}{c}$ is purely imaginary.
$\therefore \arg \left(\frac{a}{c}\right)=\pm \frac{\pi }{2}$