First slide

Algebra of complex numbers

Question

$For any complex number w = a + b i, where a, b \in R and i = \sqrt{- 1} . If w = \cos 40^{\circ} + i \sin 40^{\circ}, then$ ${|w + 2 w^{2} + 3 w^{3} + \dots + 9 w^{9}|}^{- 1} equals$

Difficult

A

$(\frac{1}{9} \sin 40^{\circ}) \sec (20^{\circ})$
B

$\frac{2}{9} \sin 20^{\circ}$
C

$\frac{1}{9} \cos 40^{\circ}$
D

$\frac{9}{2} cosec 20^{\circ}$

Solution

$\begin{array}{l} w = e^{i (\frac{2 π}{9})} \\ Let S = w + 2 w^{2} + 3 w^{3} + \dots + 9 w^{9} \to (1) \\ \Rightarrow w s = w^{2} + \dots \dots \dots . .9 w^{10} \to (2) \\ (1) - (2) \Rightarrow (1 - w) s = w + w^{2} + \dots . . + w^{9} - 9 w^{10} \\ = 0 - 9 w^{10} \\ \Rightarrow s = - \frac{9 w^{10}}{1 - w} \Rightarrow | s | = \frac{9}{2 \sin (20^{\circ})} \\ \Rightarrow | s |^{- 1} = \frac{2}{9} \sin (20^{\circ}) \end{array}$

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