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Question

If z=cosθ+isinθ is a root of xn+p1xn1+p2xn2+...+pn=0, where p1,p2,...,pnR, Find the value of 22+7(p1sinθ+p2sin2θ+...+pnsin(nθ))

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Solution

xn+p1xn1+p2xn2+.......+pn1x+pn=0
1+p1x1+p2x2+.......+pn1xn+1+pnxn=0
Since, cosθ+isinθ is the root
Therefore, 1+p1(cosθ+isinθ)1+p2(cosθ+isinθ)2+.......+pn1(cosθ+isinθ)n+1+pn(cosθ+isinθ)n=0
1+p1(cosθisinθ)+p2(cos2θisin2θ)+.......+pn1(cos(n1)θisin(n1)θ)+pn(cosnθisinnθ)=0
On comparing imaginary parts, we get
p1(sinθ)+p2(sin2θ)+.......+pn1(sin(n1)θ)+pn(sinnθ)=0
Therefore, 22+7(p1sinθ+p2sin2θ+...+pnsin(nθ))=22
Ans: 22

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