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.

You can get a faster result by using a different way to complete the square: split the middle term
$$
z^4-3z^2+1=(z^2-1)^2-z^2=(z^2-z-1)(z^2+z-1)
$$
Now the remaining two quadratic equations are easy to solve.
$$
0=z^2-z-1\implies z=\frac12(1\pm\sqrt{5})\\
0=z^2+z-1\implies z=\frac12(-1\pm\sqrt{5})\\
$$

The fifth unit roots satisfy the equation $z^5-1=0$. The non-trivial ones  satisfy
$$
|z|=1\text{ and } z^2+\bar z^2 +z+\bar z + 1=0
$$
Eliminating the imaginary part leaves for the real part the equation
$$
0=2(x^2-(1-x^2))+2x+1=(2x)^2+(2x)-1
$$
so that $2x\in\{2\cos(72^\circ),2\cos(144^\circ)\}$ is one of the solutions of the second quadratic equation, and $(-2x)\in\{2\cos(36^\circ),2\cos(108^\circ)\}$ one of the solutions of the first quadratic equation.
