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प्रश्न
Show that `(sqrt(3)/2 + i/2)^5 + (sqrt(3)/2 - i/2)^5 = - sqrt(3)`
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उत्तर
LHS = `(sqrt3/2 + i/2)^5 + (sqrt3/2 - i/2)^5`
Polar form of `sqrt3/2 + i/2 = r (cos theta + i sin theta)`
`r = sqrt(x^2 + y^2) = sqrt(3/4 + 1/4) = 1`
`alpha = tan^-1 |y/x| = tan^-1 |1/sqrt3| = pi/6`
`theta = alpha = pi/6`
`sqrt3/2 + i/2 = 1 (cos pi/6 + i sin pi/6)`
Similarly `sqrt3/2 - i/2 = (cos pi/6 - i sin pi/6)`
LHS = `(cos pi/6 + i sin pi/6)^5 + (cos pi/6 - i sin pi/6)^5`
`= cos (5pi)/6 + i sin (5pi)/6 + cos (5pi)/6 - i sin (5pi)/6`
`= 2 cos (5pi)/6 = 2 cos (pi - pi/6) = -2 cos pi/6`
`= -2 xx sqrt3/2 = -sqrt3` = RHS
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