Complex number and quadratic in combained

[Balaviswanth]

I don't know how to approach this problem
Pls solve and show me the full question.

i have attached the question below.

 

[Dr.Peterson]

We aren't here to do the work for you; we ask you to show your own thinking, so we can help you go further. See our submission guidelines.

The first thing I see here is the cosines of multiple angles, which suggests a connection to the fact that [MATH](\cos\theta + i \sin\theta)^n = \cos(n\theta) + i \sin(n\theta)[/MATH]. Is that something you are familiar with?

Try writing the polynomial, with x replaced by [MATH]\cos\theta + i \sin\theta[/MATH], with that in mind.

 

[pka]

I don't know how to approach this problem

You are given the polynomial $\displaystyle P(z) = \sum\limits_{k = 0}^n {{a_k}{z^k}} = 0$ with $\displaystyle z_0=\cos(\theta)+i\sin(\theta)$ is a root.
Notation: $\displaystyle z_0=\cos(\theta)+i\sin(\theta)=cis(\theta)=exp(i\theta)$
To do this one needs to know these facts $\displaystyle (z_0)^k=\cos(k\theta)+i\sin(k\theta)=cis(k\theta)=exp(ik\theta)$
By definition of a root we know that means $\displaystyle P(z_0) = \sum\limits_{k = 0}^n {{a_k}{(z_0)^k}} = 0$
OR $\displaystyle \sum\limits_{k = 0}^n {{a_k}\cos (k\theta )} + i\sum\limits_{k = 0}^n {{a_k}\sin (k\theta )} = 0$
If a complex number $\displaystyle P(z_0)$ is equal to zero then $\displaystyle \Re(P(z_0))=?$