Question

If $3^{50}(x+iy)={\left(\frac{3}{2}+\frac{i\sqrt{3}}{2}\right)}^{100}\forall x,y\in R$, then ordered pair $(x,y)$ is given by

• A. $\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)$
• B. $\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$
• C. $\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$
• D. $\left(-\frac{3}{2},-\frac{1}{2}\right)$

Answer 1

Answer: (A)

The given equation can be written as
${\left(i\sqrt{3}\right)}^{100}{\left(\frac{1}{2}-\frac{i\sqrt{3}}{2}\right)}^{100}=3^{50}\left(x+iy\right)$
$\Rightarrow {\left(-\omega \right)}^{100}=x+iy$
$\Rightarrow \omega =x+iy$
$\Rightarrow \frac{-1+i\sqrt{3}}{2}=x+iy$
$\Rightarrow x=-1/2,y=\frac{\sqrt{3}}{2}$