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Question: If for complex numbers \(- 2\sqrt{3} + 2i\) and\(- \sqrt{3} + i\), \(z = \frac{1 - i\sqrt{3}}{1 + i\...

If for complex numbers −23+2i- 2\sqrt{3} + 2i and−3+i- \sqrt{3} + i, z=1−i31+i3,z = \frac{1 - i\sqrt{3}}{1 + i\sqrt{3}}, then arg(z)=arg(z) = is equal to.

A

60o60^{o}

B

120o120^{o}

C

240o240^{o}

D

0

Answer

240o240^{o}

Explanation

Solution

We have z=cos⁡π6+isin⁡π6=32+i2z = \cos\frac{\pi}{6} + i\sin\frac{\pi}{6} = \frac{\sqrt{3}}{2} + \frac{i}{2}

∴∣z∣=34+14=1\therefore|z| = \sqrt{\frac{3}{4} + \frac{1}{4}} = 1

where arg(z)=tan⁡−1(yx)=tan⁡−1(1/23/2)=tan⁡−1(13)arg(z) = \tan^{- 1}{}\left( \frac{y}{x} \right) = \tan^{- 1}\left( \frac{1/2}{\sqrt{3}/2} \right) = \tan^{- 1}\left( \frac{1}{\sqrt{3}} \right) and ⇒arg(z)=tan⁡−1(tan⁡π6)=π6\Rightarrow arg(z) = \tan^{- 1}\left( \tan\frac{\pi}{6} \right) = \frac{\pi}{6}

Since sin⁡π5+i(1−cos⁡π5)\sin\frac{\pi}{5} + i\left( 1 - \cos\frac{\pi}{5} \right)

∴=2sin⁡π10cos⁡π10+i2sin⁡2π10= 2\sin\frac{\pi}{10}\cos\frac{\pi}{10} + i2\sin^{2}\frac{\pi}{10}

=2sin⁡π10(cos⁡π10+isin⁡π10)= 2\sin\frac{\pi}{10}\left( \cos\frac{\pi}{10} + i\sin\frac{\pi}{10} \right)

⇒ tan⁡θ=sin⁡π10cos⁡π10=tan⁡π10\tan\theta = \frac{\sin\frac{\pi}{10}}{\cos\frac{\pi}{10}} = \tan\frac{\pi}{10}