Question

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

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

Answer 1

Answer: (B)

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