A) \[-\theta \]
B) \[\frac{\pi }{2}-\theta \]
C) \[\theta \]
D) \[\pi -\theta \]
Correct Answer: C
Solution :
Let \[z=\omega \] Now \[\frac{1+z}{1+\overline{z}}=\frac{1+\omega }{1+{{\omega }^{2}}}=\frac{-{{\omega }^{2}}}{-\omega }=\omega \] \[\therefore \] \[arg\frac{1+z}{1+z}=\arg \,\omega =\theta \] (put\[z=cos\text{ }\theta +i\text{ }sin\text{ }\theta \])
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