<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] 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 $\dfrac{z^5+1}{z+1}=0$
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