A) \[\frac{1}{2}\arg z\]
B) \[\arg z\]
C) \[-\arg z\]
D) \[-\frac{1}{2}\arg z\]
Correct Answer: A
Solution :
[a] \[|z|=1\Rightarrow z=\cos \theta +i\sin \theta ,\] where \[\theta =\arg z\] \[\bar{z}=\cos \theta -i\sin \theta \] \[\arg ({{z}^{2}}+\bar{z})=\arg \{\cos 2\theta +i\sin 2\theta +\cos \theta -i\sin \theta \}\]\[={{\tan }^{-1}}\frac{\sin 2\theta -\sin \theta }{\cos 2\theta +\cos \theta }\] \[={{\tan }^{-1}}\frac{2\cos \frac{3\theta }{2}\sin \frac{\theta }{2}}{2\cos \frac{3\theta }{2}\cos \frac{\theta }{2}}\] \[={{\tan }^{-1}}\tan \frac{\theta }{2}=\frac{\theta }{2}=\frac{1}{2}\arg z\]
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