Q47 / 160 · AP EAMCET 2024 (22 May Shift 2) · Mathematics
$\operatorname{Arg}\left[\frac{(1+i \sqrt{3})(-\sqrt{3}-i)}{(1-i)(-i)}\right]=$
- A$\frac{5 \pi}{6}$
- Correct answerB$\frac{\pi}{4}$
- C$\frac{2 \pi}{3}$
- D$\frac{-\pi}{2}$
Solution
$\begin{aligned}
& \text { Let } z=\left[\frac{(1+i \sqrt{3})(-\sqrt{3}-i)}{(1-i)(-i)}\right] \\
= & {\left[\frac{-\sqrt{3}-i-3 i+\sqrt{3}}{-i-1}\right]=\frac{4 i}{1+i} \times \frac{1-i}{1-i}=2+2 i }
\end{aligned}$
$\operatorname{Arg}(z)$ lies in $I^{\text {st }}$ quadrant
$\tan \theta=\frac{2}{2}=1 \Rightarrow \theta=n \pi+\frac{\pi}{4}$
For $I^{\text {st }}$ quadrant put $n=0, \therefore \arg (z)=\frac{\pi}{4}$
For $I^{\text {st }}$ quadrant put $n=0, \therefore \arg (z)=\frac{\pi}{4}$
Hence, the correct answer is option: 'B'.