Question

If 3⁴⁹ (x + iy) = $\left( \frac{3}{2} + \frac{\sqrt{}3}{2}i \right)^{100}$ and x = ky then k is

• A. $- \frac{1}{3}$
• B. Ö3
• C. –Ö3
• D. $- \frac{1}{\sqrt{}3}$

Answer 1

Answer: (D)

$- \frac{1}{\sqrt{}3}$

Answer 2

Sol. $\left( \frac{3}{2} + \frac{\sqrt{3}}{2}i \right)^{100}$ = $\sqrt { 3 }$ $\left( \frac{\sqrt{3}}{2} + \frac{1}{2}i \right)^{100}$

= ($\sqrt{3}$)¹⁰⁰ = (cos p/6 + i sinp/6)¹⁰⁰ = 3⁵⁰

$\left( \cos\frac{100\pi}{6} + i\sin\frac{100\pi}{6} \right)$3 cos $\frac{100\pi}{6}$= x,

3 sin $\frac{100\pi}{6}$ = Ky,k = cot $\frac{100\pi}{6}$ = cot $\frac{2\pi}{3}$ = $–\frac{1}{\sqrt{3}}$