Question

For two complex numbers $z_1$ and $z_2$. It is given that $\left|\frac{z_1{-z}_2}{{}_1+z_2}\right|=1$. Prove that $i\frac{z_1}{z_2}=\lambda$, where $\lambda$ is real. Also determine the angle between the lines drawn from origin to points $z_1+z_2$ and $z_1-z_2$.

Answer 1

Here $\frac{z_1-z_2}{z_1+z_2}=e^{i\theta }=\frac{e^{i\theta /2}}{e^{-i\theta /2}}$
Apply comp. and divid.
$\frac{2z_1}{-2z_2}=\frac{2\cos \left(\theta /2\right)}{2i\sin \left(\theta /2\right)}=-i\cot \frac{\theta }{2}$
$\because i\frac{z_1}{z_2}=-\cot \frac{\theta }{2}\lambda$ (Real), say ....(1)
Now angle between the lines joining origin to the points $z_1+z_2$ and $z_1-z_2$ is
$\arg \frac{z_1+z_2}{z_1-z_2}=\arg Z$
$Z=\frac{z_1+z_2}{z_1-z_2}=\frac{\frac{z_1}{z_2}+1}{\frac{z_1}{z_2}-1}=\frac{\lambda /i+1}{\lambda /i-1}=\frac{\lambda +i}{\lambda -i}$
$\frac{{(\lambda +i)}^2}{{\lambda }^2+1}=\frac{{\lambda }^2-1+2\lambda i}{{\lambda }^2+1}=X+iY$
$\therefore \arg Z={\tan }^{-1}\frac{Y}{X}={\tan }^{-1}\frac{2\lambda }{{\lambda }^2-1}$
$\theta ={\tan }^{-1}\frac{2\lambda }{{\lambda }^2-1}=-{\tan }^{-1}\frac{2\lambda }{1-{\lambda }^2}$
$=-2{\tan }^{-1}\lambda .$