Question

If $z_1 = \sqrt{3} + i \sqrt{3} $ and $z_2 = \sqrt{3} + i$ , then the complex number $\left(\frac{z_{1}}{z_{2}}\right)^{50} $ lies in the :

• A. first quadrant
• B. second quadrant
• C. third quadrant
• D. fourth quadrant

Answer 1

Answer: (A)

$\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} \left(1-i\right)}{\sqrt{3} +i}\right)^{2}\right]^{25} = \left[\frac{3\left(2i\right)}{3-1 +2 \sqrt{3i}}\right]^{25} $
$ = \left(\frac{3i}{1+\sqrt{3i}}\right)^{25} = \frac{3^{25} i^{25}}{\left(-2\omega^{2}\right)^{25}} $
$ = -i.\omega\left(\frac{3}{2}\right)^{25} = -i \left(\frac{-1 + \sqrt{3}i}{2}\right)\left(\frac{3}{2}\right)^{25} $
$ = \left(\frac{3}{2}\right)^{25} \left( \frac{\sqrt{3}}{2} + \frac{1}{2}i\right) $
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.