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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 12
α=cosA+isinA=cisA, where A=8π11
z=α+α2+α3+α4+α5=α(α51α1)
z=cisA((cisA)51cisA1)=cisA(1cos5Aisin5A1cosAisinA)
z=cisAsin5A2sinA2sin5A2icos5A2sinA2icosA2=cisAsin5A2sinA2isin5A2+cos5A2isinA2+cosA2
z=cisAsin5A2cis2AsinA2=cis3Asin5A2sinA2
Re(z)=cos3Asin5A2sinA2
Re(z)=cos3Asin5A22sinA4cosA4=cos(24π11)sin(20π11)2sin(2π11)cos(2π11)=cos(24π11)sin(20π11)2sin(2π2π11)cos(2π+2π11)
Re(z)=cos(24π11)sin(20π11)2sin(20π11)cos(24π11)=12
Hence, option B is correct.

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