Question

For any complex number w = a + bi, where a, b Ī R. If w =

cos 40⁰ + i sin 40⁰, then |w + 2w² + 3w³ + ……+ 9w⁹|^–1 equals:

• A. $\frac{1}{9}$sin 40⁰
• B. $\frac{2}{9}$sin 20⁰
• C. $\frac{1}{9}$cos 40⁰
• D. $\frac{9}{2}$cosec 20⁰

Answer 1

Answer: (B)

$\frac{2}{9}$sin 20⁰

Answer 2

Sol. S = W + 2W² + … + 9W⁹

WS = W² + …..+ 8W⁹ + 9W⁹

________________________

S (1–W) = W + … + W⁹ – 9W⁹

S (1 –W) = $\frac{W(1 - W^{9})}{1 - W}$– 9W⁹

1 –W⁹ = 1 – cos 360⁰ – i sin 360⁰ = 0

S = –$\frac{9W^{9}}{1 - W}$= –$\frac{9}{1 - W}$=

$\frac{- 9}{1 - \cos 40{^\circ} - i\sin 40{^\circ}}$

1– W = 1 – cos 40⁰–i sin 40⁰

= 2 sin² 20⁰ –2i sin 20⁰cos20⁰

= 2 sin 20⁰ (sin 20⁰–i cos 20⁰)

= –2i sin 20⁰ (cos20⁰–i sin 20⁰)

|1– w| = 2 sin 20⁰

|S| = $\frac{9}{2\sin 20{^\circ}}$ Ž |S|^–1 = $\frac{2\sin 20{^\circ}}{9}$