A) \[2\sin \frac{\alpha }{2}\]
B) \[\frac{\alpha }{2}\]
C) \[\alpha \]
D) None of these
Correct Answer: B
Solution :
\[z=\sin \alpha +i(1-\cos \alpha )\] Þ \[amp(z)={{\tan }^{-1}}\left( \frac{1-\cos \alpha }{\sin \alpha } \right)={{\tan }^{-1}}\left( \frac{2{{\sin }^{2}}\frac{\alpha }{2}}{2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}} \right)\] \[={{\tan }^{-1}}\tan \left( \frac{\alpha }{2} \right)=\frac{\alpha }{2}\].
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