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Question

# If α=cos(8π11)+isin(8π11), then Re(α+α2+α3+α4+α5) is

A
12
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B
12
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C
0
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D
None of the above
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Solution

## The correct option is B −12α=cosA+isinA=cisA, where A=8π11z=α+α2+α3+α4+α5=α(α5−1α−1)⇒z=cisA((cisA)5−1cisA−1)=cisA(1−cos5A−isin5A1−cosA−isinA)⇒z=cisAsin5A2sinA2⎛⎝sin5A2−icos5A2sinA2−icosA2⎞⎠=cisAsin5A2sinA2⎛⎝isin5A2+cos5A2isinA2+cosA2⎞⎠⇒z=cisAsin5A2cis2AsinA2=cis3Asin5A2sinA2⇒Re(z)=cos3Asin5A2sinA2⇒Re(z)=cos3Asin5A22sinA4cosA4=cos(24π11)sin(20π11)2sin(2π11)cos(2π11)=cos(24π11)sin(20π11)−2sin(2π−2π11)cos(2π+2π11)⇒Re(z)=cos(24π11)sin(20π11)−2sin(20π11)cos(24π11)=−12Hence, option B is correct.

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