<s>[INST] Find the roots of $z^4-3z^2+1=0$ in polar form. Question :
Prove that the solutions of $z^4-3z^2+1=0$ are given by :
$$z=2\cos{36^\circ},2\cos{72^\circ},2\cos{216^\circ},2\cos{252^\circ}$$
My work :
First of all, i want ro find the roots with quadratic formula
$\begin{align}
&(z^2)^2-3z^2+1=0\\
&z^2=\dfrac{3\pm \sqrt{5}}{2}\\
&z_{1,2}=\pm\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}\\
&z_{3,4}=\pm\dfrac{\sqrt{3-\sqrt{5}}}{\sqrt{2}}\\
\end{align}$
And i'm stuck. I don't know how to transform this complicated roots into polar form. Bcz, i'm not sure that the modulus for each number is exactly $2$?
Btw, i've found and read some possible duplicates
Here's one of the links :
Roots of $z^4 - 3z^2 + 1 = 0$.
But it seems doesn't answer my question...
Please give me a clear hint or another way to solve this without quadratic formula or something else.
 [/INST] Set $y = z^2$, your equation is then equivalent to
$$y^2 -3y +1=0.$$
This is a quadratic equation and can be solve with the quadratic formula that you mentioned:
$$y_{1,2} = \dfrac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1}}{2}=\dfrac{3 \pm \sqrt{5}}{2}.$$
Now observe that
$$y_1 = \dfrac{3+\sqrt{5}}{2}= \dfrac{\frac{1}{2}+\sqrt{5} + \frac{5}{2}}{2} = \dfrac{\frac{1+2\sqrt{5} + 5}{2}}{2} = \dfrac{(1 + \sqrt{5})^2}{2\cdot 2}= \dfrac{(1 + \sqrt{5})^2}{4}.$$
Similarly
$$y_2 = \dfrac{3-\sqrt{5}}{2}= \dfrac{\frac{1}{2}-\sqrt{5} + \frac{5}{2}}{2} = \dfrac{\frac{1-2\sqrt{5} + 5}{2}}{2} = \dfrac{(1 - \sqrt{5})^2}{2\cdot 2}= \dfrac{(1 - \sqrt{5})^2}{4}.$$
Therefore the four solutions $z_{1,2,3,4}$ of the original equations are given by
$$z_{1,2} = \pm\sqrt{y_1} = \pm \dfrac{1 + \sqrt{5}}{2}$$
and 
$$z_{3,4} = \pm\sqrt{y_2} = \pm \dfrac{-1+ \sqrt{5}}{2}.$$
Now you can use the well-known fact that $\cos(36°)=\dfrac{1+\sqrt{5}}{4}$ to get your result. To get a value of $\cos(72°)$ you can use the trigonometric relation
$$\cos(2x) = \cos^2(x)-1.$$
While for the others you can use reasoning on the unit circle.
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