Question

If z₁, z₂ are two complex numbers such that $\left| \frac{z_{1} - z_{2}}{z_{1} + z_{2}} \right|$ = 1 and i z₁ = Kz₂, where K Ī R, then the angle between z₁ – z₂ and z₁ + z₂ 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

Answer: (D)

2 tan^–1 K

Answer 2

Sol. $\frac{z_{1} - z_{2}}{z_{1} + z_{2}}$ = cos a + i sin a

Ž $\frac{2z_{1}}{- 2z_{2}}$ = $\frac{\cos\alpha + i\sin\alpha + 1}{\cos\alpha + 1 + i\sin\alpha}$

$\frac{2\cos^{2}\alpha/2 + 2i\sin\alpha/2\cos\alpha/2}{2i\sin\alpha/2\cos\alpha/2 - 2\sin^{2}\alpha/2}$

Ž $\frac{2\cos\alpha/2\lbrack\cos\alpha/2 + i\sin\alpha/2\rbrack}{2i\sin\alpha/2\lbrack\cos\alpha/2 + i\sin\alpha/2\rbrack}$

$\frac{z_{1}}{z_{2}}$ = i cot $\frac{\alpha}{2}$

Ž iz₁ = – cos $\frac{\alpha}{2}$

i z₁ = k z₂ \ k = – cot a/2

tan a/2 = – 1/k

tan a = $\frac{2\tan\alpha/2}{1 - \tan^{2}\alpha/2}$ Ž $\frac{- 2/k}{1 - 1/k^{2}}$ Ž $\frac{- 2k}{k^{2} - 1}$

a = tan^–1 $\left( \frac{2k}{1 - k^{2}} \right)$ Ž 2 tan^–1 (k)