Question

If $z_1,z_2$ are two complex number such that $\left|\frac{z_1-z_2}{z_1+z_2}\right|=1$ and $t{z}_1=k{z}_2$ where $k\in R$, then 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 $-2{\tan }^{-1}k$

Interpretating according to Coni's theorem.

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

Using componendo and dividendo

$\frac{2z_1}{-2z_2}=\frac{1+\cos \alpha +t\sin \alpha }{\cos \alpha -1+t\sin \alpha }$

$\Rightarrow \frac{z_1}{z_2}=\frac{2{\cos }^2\left(\frac{\alpha }{2}\right)+t2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}{-2{\sin }^2\left(\frac{\alpha }{2}\right)+t\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}$

$\Rightarrow \frac{z_1}{z_2}=t\cot \frac{\alpha }{2}\Rightarrow t{z}_1=-\cot \frac{\alpha }{2}{z}_2$

But $t{z}_1=k{z}_2$ ( where $k=-\cot \frac{\alpha }{2}$ (say) )

Now, $k=-\cot \frac{\alpha }{2}\Rightarrow \cot \frac{\alpha }{2}=-k$

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

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

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 $(z_1+z_2)$