Roots of $z^4 - 3z^2 + 1 = 0$.

Question

Show that the solutions of $z^4 - 3z^2 + 1 = 0$ are given by $$z = 2\cos 36^{\def\o{\mathrm{o}}\o}, 2\cos 72^\o, 2\cos 216^\o, 2\cos 252^\o.$$ Hence show that

• $\cos 36^\o = \frac 14(\sqrt{5}+1)$
• $\cos 72^\o = \frac 14(\sqrt{5}-1)$

I can solve for the roots, however I could not express them as the question wanted. Is there a specific way to express the root in kinda polar form without the imaginary part.

Answer 1

I would solve the equation $$z^4-3z^2+1=0$$ by substituting $$z^2=t$$ we get $$t^2-3t+1=0$$ and by the quadratic formula we obtain the solutions as $$t_{1,2}=\frac{3}{2}\pm\sqrt{\frac{9}{4}-1}$$ Can you finish?

Answer 2

Let $w=e^{i\theta}$. Try to search for $z=w+\bar w=w+\frac{1}{w}$. Setting it in the equation gives $$ w^8+w^6+w^4+w^2+1=\frac{w^{10}-1}{w^2-1}=0. $$ Now you need to pick those $w$ that make $w^{10}=1$ and $w^2\ne 1$.