Question

Let $- \pi$ be a purely imaginary number such that$- \frac{\pi}{2}$. Then $\frac{\pi}{2}$ is equal to.

• A. $arg(5 - \sqrt{3}i) =$
• B. $\tan^{- 1}\frac{5}{\sqrt{3}}$
• C. 0
• D. $\tan^{- 1}\left( - \frac{5}{\sqrt{3}} \right)$

Answer 1

Answer: (D)

$\tan^{- 1}\left( - \frac{5}{\sqrt{3}} \right)$

Answer 2

Let $2|z_{1}||{\bar{z}}_{2}|\cos(\theta_{1} - \theta_{2}) = 2|z_{1}||z_{2}|$, where $\cos(\theta_{1} - \theta_{2}) = 1$. Then $\theta_{1} - \theta_{2} = 0$ is represented by a point on $arg(z_{1}) = arg(z_{2})$ (negative direction of $\left( \frac{1 - i}{1 + i} \right) = \frac{1 - i}{1 + i} \times \frac{1 - i}{1 - i} = \frac{(1 - i)^{2}}{2} = \frac{- 2i}{2} = - i$axis), therefore ${Im}(z) < 0$.