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Question

Question: If \(\frac{\alpha}{2}\)then amp\(\alpha\)– amp\(\frac{1 + \sqrt{3}i}{\sqrt{3} + 1}\)...

If α2\frac{\alpha}{2}then ampα\alpha– amp1+3i3+1\frac{1 + \sqrt{3}i}{\sqrt{3} + 1}

A

0

B

π3\frac{\pi}{3}

C

π3- \frac{\pi}{3}

D

π6\frac{\pi}{6}

Answer

π3- \frac{\pi}{3}

Explanation

Solution

amp(z) z2=π+[arg6mu(zˉ)]z_{2} = \pi + \lbrack ar ⥂ g\mspace{6mu}(\bar{z})\rbrack