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.