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Question: If z<sub>1</sub>, z<sub>2</sub> are two complex numbers such that \(\left| \frac{z_{1} - z_{2}}{z_{1...

If z1, z2 are two complex numbers such that ∣z1−z2z1+z2∣\left| \frac{z_{1} - z_{2}}{z_{1} + z_{2}} \right|= 1 and iz1 = kz2, where k Ī R, then the angle between z1 – z2 and z1 + z2 is –

A

tan–12k1+k2\frac{2k}{1 + k^{2}}

B

tan–12k1−k2\frac{2k}{1 - k^{2}}

C

–3 tan–1 k

D

3 tan–1 k

Answer

tan–12k1−k2\frac{2k}{1 - k^{2}}

Explanation

Solution

Sol. Let angle between z1 – z2 and z1 + z2 be a.

\ z1−z2z1+z2\frac{z_{1} - z_{2}}{z_{1} + z_{2}} = cos a + i sin a (∵∣z1−z2z1+z2∣=1)\left( \because\left| \frac{z_{1} - z_{2}}{z_{1} + z_{2}} \right| = 1 \right)

Ž 2z1−2z2\frac{2z_{1}}{- 2z_{2}} = cos⁡α+isin⁡α+1cos⁡α+isin⁡α−1\frac{\cos\alpha + i\sin\alpha + 1}{\cos\alpha + i\sin\alpha - 1}

Ž z1z2\frac{z_{1}}{z_{2}} = (1+cos⁡α)+isin⁡α(1−cos⁡α)−isin⁡α\frac{(1 + \cos\alpha) + i\sin\alpha}{(1 - \cos\alpha) - i\sin\alpha}

= 2cos⁡2α2+2isin⁡α2cos⁡α22sin⁡2α2−2isin⁡α2cos⁡α2\frac{2\cos^{2}\frac{\alpha}{2} + 2i\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}{2\sin^{2}\frac{\alpha}{2} - 2i\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}

=2cos⁡α2(cos⁡α2+isin⁡α2)2sin⁡α2(sin⁡α2−icos⁡α2)\frac{2\cos\frac{\alpha}{2}\left( \cos\frac{\alpha}{2} + i\sin\frac{\alpha}{2} \right)}{2\sin\frac{\alpha}{2}\left( \sin\frac{\alpha}{2} - i\cos\frac{\alpha}{2} \right)}= cos⁡α2−isin⁡α2\frac{\cos\frac{\alpha}{2}}{- i\sin\frac{\alpha}{2}}

= i cot α2\frac{\alpha}{2}.

\ iz1 =

(i2cot⁡α2)\left( i^{2}\cot\frac{\alpha}{2} \right)z2 = (−cot⁡α2)\left( - \cot\frac{\alpha}{2} \right)z2

But iz1 = kz2. \ k = – cot α2\frac{\alpha}{2}

\ tan α2\frac{\alpha}{2} = – 1k\frac{1}{k}

\ tan a = 2tan⁡α21−tan⁡2α2\frac{2\tan\frac{\alpha}{2}}{1 - \tan^{2}\frac{\alpha}{2}}= 2(−1k)1−(1k)2\frac{2\left( - \frac{1}{k} \right)}{1 - \left( \frac{1}{k} \right)^{2}}=−2kk2−1\frac{- 2k}{k^{2} - 1}=2k1−k2\frac{2k}{1 - k^{2}}

\ a = tan–1 2k1−k2\frac{2k}{1 - k^{2}}.