Question

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

A
12
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B
12
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C
0
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D
None of these
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Solution

## The correct option is D −12Givenα=cos8π11+isin8π11=ei8π11α2=ei16π11=cos16π11+isin16π11α3=ei24π11=cos24π11+isin24π11α4=ei32π11=cos32π11+isin32π11α5=ei40π11=cos40π11+isin40π11Now,Re(α+α2+α3+α4+α5)=cos8π11+cos16π11+cos24π11+cos32π11+cos40π11Say x=8π11,thenRe(α+α2+α3+α4+α5)=cosx+cos2x+cos3x+cos4x+cos5xMultiply and divide by 2sinx2Re(α+α2+α3+α4+α5)=12sinx2(2sinx2cosx+2sinx2.cos2x+2sinx2.cos3x+2sinx2.cos4x+2sinx2.cos5x)=12sinx2[sin3x2−sinx2+sin5x2−sin3x2+sin7x2−sin5x2+sin9x2−sin7x2+sin11x2−sin9x2]=12sinx2[sin11x2−sinx2]Putting x=8π11Re(α+α2+α3+α4+α5)=12sin4π11[sin4π−sin4π11]=−12

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