Let $z$ be a purely imaginary number such that $\text{Im}(z) > 0$. Then $\text{arg}(z)$ is equal to
- A$\pi$
- B$\frac{\pi}{2}$
- C$0$
- D$-\frac{\pi}{2}$
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View Solution$\operatorname{Arg}\left(\sin \frac{6 \pi}{5}+i\left(1+\cos \frac{6 \pi}{5}\right)\right)=$
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