Proving $a_1\cos t + a_2\cos 2t+\dots+a_n\cos nt=-1$ Given: $f(z)=z^n+a_1z^{n-1}+a_2z^{n-2}+...+a_{n-1}z^{1}+a_nz^{0}=0$, $z=cist$, $a_n \in R$. Need to prove: $a_1\cos t + a_2\cos 2t+\dots+a_n\cos nt=-1$ So i see...--prophetes.ai

Proving $a_1\cos t + a_2\cos 2t+\dots+a_n\cos nt=-1$ Given: $f(z)=z^n+a_1z^{n-1}+a_2z^{n-2}+...+a_{n-1}z^{1}+a_nz^{0}=0$, $z=cist$, $a_n \in R$. Need to prove: $a_1\cos t + a_2\cos 2t+\dots+a_n\cos nt=-1$ So i see that: $Re(f(z))=cos(nt)+a_1cos((n-1)t)+...+a_ncos((n-n)t)=0$ can i just put n=0 and using $cos(-x)=cos(x)$ to prove? Because i can't see how to continue. Thanks.

**Hint.** Divide the expression $f(z)=0$ throughout by $z^n.$ Use $z=\cos \theta + \sin \theta$ with De Moivre's theorem and take the real part.