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Mathematics Question on Complex Numbers and Quadratic Equations

If z1,z2z_1,z_2 are two complex numbers and such that ∣z1−z2z1+z2∣=1\left|\frac{z_1-z_2}{z_1+z_2}\right|=1 and iz1=Kz2iz_1=Kz_2 where K ∈ RK\,\in\,R, then the angle between z1−z2z_1-z_2 and z1+z2z_1+z_2 is

A

tan−1(2KK2+1)tan^{-1}\left(\frac{2K}{K^2+1}\right)

B

tan−1(2K1−K2)tan^{-1}\left(\frac{2K}{1-K^2}\right)

C

−2 tan−1K-2 \,tan^{-1}K

D

2 tan−1K2 \,tan^{-1}K

Answer

2 tan−1K2 \,tan^{-1}K

Explanation

Solution

Let z1−z2z1+z2=cos α+i sin α\frac{z_{1}-z_{2}}{z_{1}+z_{2}}=cos\,\alpha+i\,sin\,\alpha ∴2z1−2z2=1+cos α+i sin α1−cos α−i sin α\therefore \frac{2z_{1}}{-2z_{2}}=\frac{1+cos\,\alpha+i\,sin\,\alpha}{1-cos\,\alpha-i\,sin\,\alpha} =2 cos2α2+2i sinα2cosα2−2i sinα2cosα2+2 sin2α2=\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}} =2 cosα2[cosα2+sinα2]−2i sinα2i[cosα2+i sinα2]=\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]} ⇒z1z2=i cotα2\Rightarrow \frac{z_{1}}{z_{2}}=i\,cot \frac{\alpha}{2} ⇒i z1=−cotα2⋅z2\Rightarrow i\,z_{1}=-cot \frac{\alpha}{2}\cdot z_{2} But i z1=K z2i\,z_{1}=K\,z_{2} ∴K=−cotα2\therefore K=-cot \frac{\alpha}{2} ∴tanα2=−1K\therefore tan \frac{\alpha}{2}=-\frac{1}{K}. Now tan α=2 tan α/21−tan2 α/2tan\,\alpha=\frac{2\,tan\,\alpha/2}{1-tan^{2}\,\alpha/2} =−2K1−1K2=−2KK2−1=\frac{-\frac{2}{K}}{1-\frac{1}{K^{2}}}=\frac{-2K}{K^{2}-1} ∴α=tan−1(2K1−K2)\therefore \alpha=tan^{-1}\left(\frac{2K}{1-K^{2}}\right) =2 tan−1(K)=2\,tan^{-1}\left(K\right) Now z1−z2z1+z2=cos α+i sin α\frac{z_{1}-z_{2}}{z_{1}+z_{2}}=cos\,\alpha+i\,sin\,\alpha ⇒α\Rightarrow \alpha is the angle between z1−z2z_{1}-z_{2} and z1+z2z_{1}+z_{2}. ⇒α=2 tan−1 K\Rightarrow \alpha=2\,tan^{-1}\,K is the angle between z1−z2z_{1}-z_{2} and z1+z2z_{1}+z_{2}