Question

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`

Answer 1

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`