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

The correct option is A First quadrant

Given, $z_1=\sqrt{3}+i\sqrt{3},z_2=\sqrt{3}+i$

Therefore, ${\left(\frac{z_1}{z_2}\right)}^{50}={\left(\frac{\sqrt{3}+i\sqrt{3}}{\sqrt{3}+i}\right)}^{50}$
$={\left[{\left(\frac{\sqrt{3}(1+i)}{\sqrt{3}+i}\right)}^2\right]}^{25}={\left[\frac{3\left(2i\right)}{3-1+2\sqrt{3}i}\right]}^{25}$
$={\left(\frac{3i}{1+\sqrt{3}i}\right)}^{25}=\frac{3^{25}{i}^{25}}{(-2{\omega }^2{)}^{25}}$
$=-i.\omega {\left(\frac{3}{2}\right)}^{25}=-i\left(\frac{-1+\sqrt{3}i}{2}\right){\left(\frac{3}{2}\right)}^{25}={\left(\frac{3}{2}\right)}^{25}(\frac{\sqrt{3}}{2}+\frac{1}{2}i)$
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.