Q54 / 125 · BITSAT 2016 · Mathematics
If $z_{1}=\sqrt{3}+i \sqrt{3}$ and $z_{2}=\sqrt{3}+i$, then the
complex number $\left(\frac{\mathrm{z}_{1}}{\mathrm{z}_{2}}\right)^{50}$ lies in the :
- Correct answerAfirst quadrant
- Bsecond quadrant
- Cthird quadrant
- Dfourth quadrant
Solution
$\left(\frac{z_{1}}{z_{2}}\right)^{50}=\left(\frac{\sqrt{3}+i \sqrt{3}}{\sqrt{3}+i}\right)^{50}$
$=\left[\left(\frac{\sqrt{3}(1+i)}{\sqrt{3}+i}\right)^{2}\right]^{25}=\left[\frac{3(2 i)}{3-1+2 \sqrt{3} i}\right]^{25}$
$=\left(\frac{3 i}{1+\sqrt{3} i}\right)^{25}=\frac{3^{25} i^{25}}{\left(-2 \omega^{2}\right)^{25}}$
$$
\begin{array}{l}
=-i . \omega\left(\frac{3}{2}\right)^{25}=-\mathrm{i}\left(\frac{-1+\sqrt{3} \mathrm{i}}{2}\right)\left(\frac{3}{2}\right)^{25} \\
=\left(\frac{3}{2}\right)^{25}\left(\frac{\sqrt{3}}{2}+\frac{1}{2} \mathrm{i}\right)
\end{array}
$$
Hence, $\left(\frac{z_{1}}{z_{2}}\right)^{50}$ lies in the first quadrant as both real and imaginary parts of this number are positive.
Hence, the correct answer is option: 'A'.