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Question: If \(- \frac{\pi}{6}\), then amp\(- 1 + i\sqrt{3}\)=...

If −π6- \frac{\pi}{6}, then amp−1+i3- 1 + i\sqrt{3}=

A

−60o- 60^{o}

B

60o60^{o}

C

120o120^{o}

D

−120o- 120^{o}

Answer

−120o- 120^{o}

Explanation

Solution

amp−zˉ\mathbf{-}\bar{\mathbf{z}}

1+2i1−(1−i)2=1+2i1−(1−1−2i)=1+2i1+2i=1+0i\frac{1 + 2i}{1 - (1 - i)^{2}} = \frac{1 + 2i}{1 - (1 - 1 - 2i)} = \frac{1 + 2i}{1 + 2i} = 1 + 0i.