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 $iz_1=Kz_2$ 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 D

Let $\frac{z_{1}-z_{2}}{z_{1}+z_{2}}=cos\,\alpha+i\,sin\,\alpha$ $\therefore \frac{2z_{1}}{-2z_{2}}=\frac{1+cos\,\alpha+i\,sin\,\alpha}{1-cos\,\alpha-i\,sin\,\alpha}$ $=\frac{2\,cos^{2} \frac{\alpha}{2}+2i\,sin \frac{\alpha}{2} cos \frac{\alpha}{2}}{-2i\,sin \frac{\alpha}{2} cos \frac{\alpha}{2}+2\,sin^{2} \frac{\alpha}{2}}$ $=\frac{2\,cos \frac{\alpha}{2}\left[cos \frac{\alpha}{2}+sin \frac{\alpha}{2}\right]}{-2i\,sin \frac{\alpha}{2} i\left[cos \frac{\alpha}{2}+i\,sin \frac{\alpha}{2}\right]}$ $\Rightarrow \frac{z_{1}}{z_{2}}=i\,cot \frac{\alpha}{2}$ $\Rightarrow i\,z_{1}=-cot \frac{\alpha}{2}\cdot z_{2}$ But $i\,z_{1}=K\,z_{2}$ $\therefore K=-cot \frac{\alpha}{2}$ $\therefore tan \frac{\alpha}{2}=-\frac{1}{K}$. Now $tan\,\alpha=\frac{2\,tan\,\alpha/2}{1-tan^{2}\,\alpha/2}$ $=\frac{-\frac{2}{K}}{1-\frac{1}{K^{2}}}=\frac{-2K}{K^{2}-1}$ $\therefore \alpha=tan^{-1}\left(\frac{2K}{1-K^{2}}\right)$ $=2\,tan^{-1}\left(K\right)$ Now $\frac{z_{1}-z_{2}}{z_{1}+z_{2}}=cos\,\alpha+i\,sin\,\alpha$ $\Rightarrow \alpha$ 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 $z_{1}+z_{2}$