InterviewSolution
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InterviewSolution
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If `z=costheta+isintheta`is a root of the equation `a_0z^n+a_2z^(n-2)++a_(n-1)z^+a_n=0,`then prove that`a_0+a_1costheta+a_2^cos2theta++a_ncosntheta=0``a_1"sin"theta+a_2^sin2theta++a_nsinntheta=0` |
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Answer» Dividing the given equation by `z^(n),` we get `a_(0)+a_(1)z^(-1)+a_(2)z^(-2)+…+a_(n-1)z^(1-n)+a_(n)z^(-n)=0` Now, `z=costheta+isintheta=e^(itheta)` satisfies the above equation. Hence, `a_(0)+a_(1)e^(-itheta)+a_(2)e^(-2itheta)+…+a_(n-1)e^(-i(n-1)theta)+a_(n)e^(-intheta)=0` `implies(a_(0)+a_(1)costheta+a_(2)cos2theta+...+a_(n)cosntheta)` `-i(a_(1)sintheta+a_(2)sin2theta+...+a_(n)sinntheta)=0` `impliesa_(0)+a_(1)costheta+a_(2)cos2theta+...+a_(n)cosntheta=0` and `a_(1)sintheta+theta_(2)sin2theta+...+a_(n)sinntheta=0` |
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