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.
 A: Note that $$z=\sqrt{\frac{3 \pm \sqrt{5}}{2}}= \pm \left(\frac{1\pm \sqrt{5}}{2} \right).$$ So $$z_1=2\frac{1+\sqrt{5}}{4} =2 \cos 36^0$$ Hence the other three roots.
A: 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.
A: Consider $z^5-1=0$
So the roots of $$0=\dfrac{z^5-1}{z-1}=z^4+z^3+z^2+z+1$$ are $e^{2i\pi r/5},r=1,2,3,4$
As $z\ne0,$ like Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$ , divide both sides by $z^2$
Replace $z+\dfrac1z=w\implies w^2=?$ to find $$w^2-2+w+1=0$$ whose roots are $$2\cos\dfrac{2r\pi}5;r=1,2$$
Similarly consider $\dfrac{z^5+1}{z+1}=0$
A: 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.
A: Note that 
$$\cos36 = -\cos216 = \frac{1+\sqrt5}{4}$$
$$\cos72 = -\cos252 = \frac{1-\sqrt5}{4}$$
Then,
$$(x-2\cos36)(x-2\cos216) = x^2 - \frac{3+\sqrt5}{2}=0$$
$$(x-2\cos72)(x-2\cos252) = x^2 - \frac{3-\sqrt5}{2}=0$$
Together,
$$(x-2\cos36)(x-2\cos216)(x-2\cos72)(x-2\cos252)=x^4-3z^2+1=0$$
