Question

If $z_1,z_2$ are two complex numbers such that $\left|\frac{z_1-z_2}{z_1+z_2}\right|=1$ and $t{z}_1=k{z}_2$ where $k\in R$, the angle between $(z_1-z_2)$ and $(z_1+z_2)$ is

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

Answer 1

The correct option is C $-2{\tan }^{-1}k$

Interpreting according Coni's theorem.

Let the angle between the lines joining $z_1,z_2$ and $z_1,-z_2$ be $\alpha$

$\therefore \left|\frac{z_1-z_2}{z_1+z_2}\right|=\cos \alpha +i\sin \alpha$

Using componendo and dividendo, we have

$\frac{2z_1}{-2z_1}=\frac{1+\cos \alpha +i\sin \alpha }{\cos \alpha -1+i\sin \alpha }$

$\Rightarrow \frac{z_1}{z_2}=\frac{2{\cos }^2\left(\frac{\alpha }{2}\right)+i2\sin \left(\frac{\alpha }{2}\right)\cos \left(\frac{\alpha }{2}\right)}{-2{\sin }^2\left(\frac{\alpha }{2}\right)+i\sin \left(\frac{\alpha }{2}\right)\cos \left(\frac{\alpha }{2}\right)}$

$\Rightarrow \frac{z_1}{z_2}=-\cot \left(\frac{\alpha }{2}\right)\Rightarrow i{z}_1=-\cot \left(\frac{\alpha }{2}\right)z_2$

But $i{z}_1-k{z}_2$ {where, $k=-\cot \left(\frac{\alpha }{2}\right)$(say)}

Now, $k=-\cot \left(\frac{\alpha }{2}\right)\Rightarrow \cot \left(\frac{\alpha }{2}\right)=-k\Rightarrow \tan \alpha =\frac{2k}{k^2-1}$

$\Rightarrow \tan \alpha =-\frac{2k}{1-k^2}\Rightarrow \alpha ={\tan }^{-1}(-\frac{2k}{1-k^2})=-2{\tan }^{-1}k$

Now, $\frac{z_1-z_2}{z_1+z_2}=\cos \alpha +i\sin \alpha$

where $\alpha$ is the angle between $(z_1-z_2)$ and $(z_1+z_2)$

$\Rightarrow \alpha =-2{\tan }^{-1}k$ is the angle between $(z_1-z_2)$ and $(z_1+z_2)$.