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Let z=acosπ5+isinπ5,aR,|a|<1, then S=z2015+z2016+z2017+equals

a
a2015z−1
b
a20151-z
c
a20151-a
d
a2015a-1

detailed solution

Correct option is A

We have z=a<1, thus            S=a2015z−1But z2015=a2015cos(403π)+i sin (403π)=a2015∴        S=a2015z−1

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