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Question: If \(- 1 + \sqrt{- 3} = re^{i\theta},\) then θ is equal to...

If −1+−3=reiθ,- 1 + \sqrt{- 3} = re^{i\theta}, then θ is equal to

A

π3\frac{\pi}{3}

B

−π3- \frac{\pi}{3}

C

2π3\frac{2\pi}{3}

D

−2π3- \frac{2\pi}{3}

Answer

2π3\frac{2\pi}{3}

Explanation

Solution

Sol Here−1+−3=reiθ- 1 + \sqrt{- 3} = re^{i\theta} ⇒ −1+i3=reiθ=rcos⁡θ+irsin⁡θ- 1 + i\sqrt{3} = re^{i\theta} = r\cos\theta + ir\sin\theta

Equating real and imaginary part, we get rcos⁡θ=−1r\cos\theta = - 1 and rsin⁡θ=3r\sin\theta = \sqrt{3}

Hence tan⁡θ=−3\tan\theta = - \sqrt{3} ⇒ tan⁡θ=tan⁡2π3\tan\theta = \tan\frac{2\pi}{3}⇒ θ=2π3\theta = \frac{2\pi}{3}.