Question

If $3^{49}(x+i y)=\left(\frac{3}{2}+\frac{\sqrt{3} i}{2}\right)^{100}$ and $x=k y$, then $k$ is

• A. $\sqrt{3}$
• B. $\frac{-1}{\sqrt{3}}$
• C. $\frac{-1}{3}$
• D. 3
• E. $\sqrt{3}$

Answer 1

Answer: (B)

$3^{49}(x+i y)=\left(\frac{3}{2}+\frac{\sqrt{3} i}{2}\right)^{100}$
$=\left[i \sqrt{3}\left(\frac{1-i \sqrt{3}}{2}\right)\right]^{100}$
$\Rightarrow 3^{49}(x+i y)=i^{100} \cdot 3^{50} \cdot(-\omega)^{100}$
$\Rightarrow 3^{49}(x+i y)=3^{50} \cdot \omega=3^{50}\left(\frac{-1}{2}+\frac{i \sqrt{3}}{2}\right)$
$\therefore x+i y=-\frac{3}{2}+\frac{3 \sqrt{3 i}}{2} \Rightarrow x=-\frac{1}{\sqrt{3}} y$
$\therefore k=-\frac{1}{\sqrt{3}}$