A) \[\frac{\pi }{4}\]
B) \[\frac{3\pi }{4}\]
C) \[\frac{\pi }{12}\]
D) \[\frac{\pi }{2}\]
E) \[\frac{3\pi }{2}\]
Correct Answer: D
Solution :
Let \[z=\frac{i}{2}-\frac{2}{i}=\frac{i}{2}-\frac{2i}{{{i}^{2}}}\] \[Z=\frac{i}{2}+2i=\frac{5}{2}i=0+\frac{5}{2}i\] \[\arg (z)={{\tan }^{-1}}\left( \frac{\operatorname{Im}(z)}{\operatorname{Re}(z)} \right)\] \[={{\tan }^{-1}}\left( \frac{5/2}{0} \right)={{\tan }^{-1}}(\infty )\] \[={{\tan }^{-1}}\left( \tan \frac{\pi }{2} \right)=\frac{\pi }{2}\]
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