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. $\sqrt{3}$
• C. –$\sqrt{3}$
• D. –$\frac{1}{\sqrt{3}}$

Answer 1

Answer: (D)

–$\frac{1}{\sqrt{3}}$

Answer 2

Sol. As, 3⁴⁹ (x + iy) = $\left( \frac{3}{2} + \frac{\sqrt{3}}{2}i \right)^{100}$

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

Ž 3⁴⁹ (x + iy) = i¹⁰⁰ . 3⁵⁰ . (–w)¹⁰⁰

Ž 3⁴⁹ (x + iy) = 3⁵⁰ . w = 3⁵⁰ . $\left( - \frac{1}{2} + \frac{i\sqrt{3}}{2} \right)$

\ x + iy = – $\frac{3}{2}$+ $\frac{3\sqrt{3}}{2}$i Ž x = –$\frac{1}{\sqrt{3}}$ y, \ k = –$\frac{1}{\sqrt{3}}$

Hence (4) is correct answer.