Question

If $z_1$ and $z_2$ are two complex numbers such that $\left|\frac{z_1-z_2}{z_1+z_2}\right|=1$, $\frac{i{z}_1}{z_2}=k$, where k is a real number. Find the angle between the lines from the origin to the points $z_1+z_2$ and $z_1-z_2$ in terms of k.

• A. $\theta =ta{n}^{-1}\left(\frac{2k}{k^2+1}\right)$
• B. $\theta =ta{n}^{-1}\left(\frac{2k}{k^2-1}\right)$
• C. $\theta =ta{n}^{-1}\left(\frac{2k}{k-1}\right)$
• D. $\theta =ta{n}^{-1}\left(\frac{2k}{k+1}\right)$

Answer 1

Answer: (B)

$\theta =ta{n}^{-1}\left(\frac{2k}{k^2-1}\right)$

(i) Given $\left|\frac{z_1-z_2}{z_1+z_2}\right|=1$
$\Rightarrow \left|\frac{\frac{z_1}{z_2}-1}{\frac{z_1}{z_2}+1}\right|=1$
$\Rightarrow |\frac{z_1}{z_2}-1|=|\frac{z_1}{z_2}+1|$
squaring both sides
${\left|\frac{z_1}{z_2}\right|}^2+1-2Re\left(\frac{z_1}{z_2}\right)={\left|\frac{z_1}{z_2}\right|}^2+1+2Re\left(\frac{z_1}{z_2}\right)$
$\Rightarrow 4Re\left(\frac{z_1}{z_2}\right)0\Rightarrow \frac{z_1}{z_2}$ is purely imaginary number $\frac{z_1}{z_2}$ can be written as $i\frac{z_1}{z_2}=k$ where k is real number.
(ii) Let $\theta$ is the angle between $z_1-z_2$ and $z_1-z_2$, then
$\theta =Arg\left(\frac{z_1+z_2}{z_1-z_2}\right)$
$=Arg\left(\frac{\frac{z_1}{z_2}+1}{\frac{z_1}{z_2}-1}\right)$
$=Arg\left(\frac{-ik+1}{-ik-1}\right)$
$=Arg\left(\frac{-1+ik}{1+ik}\right)$
$=Arg\left(\frac{k^2-1+2ik}{k^2+1}\right)$
$\theta ={\tan }^{-1}\left(\frac{2k}{k^2-1}\right)$
Ans: B