Question

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

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

Answer 1

Answer: (D)

Let $\frac{z_1-z_2}{z_1+z_2}=cos\alpha +isin\alpha$

$\therefore \frac{2z_1}{-2z_2}=\frac{1+cos\alpha +isin\alpha }{1-cos\alpha -isin\alpha }$

$=\frac{2co{s}^2\frac{\alpha }{2}+2isin\frac{\alpha }{2}cos\frac{\alpha }{2}}{-2isin\frac{\alpha }{2}cos\frac{\alpha }{2}+2si{n}^2\frac{\alpha }{2}}$

$=\frac{2cos\frac{\alpha }{2}\left[cos\frac{\alpha }{2}+sin\frac{\alpha }{2}\right]}{-2isin\frac{\alpha }{2}i\left[cos\frac{\alpha }{2}+isin\frac{\alpha }{2}\right]}$

$\Rightarrow \frac{z_1}{z_2}=icot\frac{\alpha }{2}$
$\Rightarrow i{z}_1=-cot\frac{\alpha }{2}\cdot z_2$
But $i{z}_1=K{z}_2$
$\therefore K=-cot\frac{\alpha }{2}$
$\therefore tan\frac{\alpha }{2}=-\frac{1}{K}$.
Now $tan\alpha =\frac{2tan\alpha /2}{1-ta{n}^2\alpha /2}$
$=\frac{-\frac{2}{K}}{1-\frac{1}{K^2}}=\frac{-2K}{K^2-1}$
$\therefore \alpha =ta{n}^{-1}\left(\frac{2K}{1-K^2}\right)$
$=2ta{n}^{-1}\left(K\right)$
Now $\frac{z_1-z_2}{z_1+z_2}=cos\alpha +isin\alpha$
$\Rightarrow \alpha$ is the angle between $z_1-z_2$ and $z_1+z_2$.
$\Rightarrow \alpha =2ta{n}^{-1}K$ is the angle between $z_1-z_2$ and $z_1+z_2$