# complex

#### bhitroofen01

Hi,

P(z)=$$\displaystyle z^6-2z^3cos(3\alpha)+1$$
How to show that the solutions of P(z) are: $$\displaystyle a_k=e^{i(-\alpha+\frac{2k\pi}{3}}$$ and $$\displaystyle b_k=e^{i(+\alpha+\frac{2k\pi}{3}}$$????

#### ques

Hi,

P(z)=$$\displaystyle z^6-2z^3cos(3\alpha)+1$$
How to show that the solutions of P(z) are: $$\displaystyle a_k=e^{i(-\alpha+\frac{2k\pi}{3}}$$ and $$\displaystyle b_k=e^{i(+\alpha+\frac{2k\pi}{3}}$$????
Simplest way is to substitute and check whether $$\displaystyle a_k and b_k give us a zero solution.$$

HallsofIvy

#### HallsofIvy

MHF Helper
Yes the best way to determine whether or not a given number satisfies an equation is, like ques said, to put the number into the equation and do the calculation!

But this equation isn't that difficult to solve. Let $$\displaystyle u= z^3$$ so that your equation becomes the quadratic equation $$\displaystyle u^2- 2ucos(3\alpha)+ 1= 0$$. Solve that using the quadratic formula and then Use DeMoivre's formula to solve $$\displaystyle z^3= u$$.

ques