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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.

Abec
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  • I think these numbers refer to the real part of the root $z$. If a real part is $$ Re(z) = 2\cos{36^{\circ}} \qquad \Rightarrow \qquad Im(z) = 2\sin{36^{\circ}} $$ – Matti P. Dec 20 '18 at 08:20
  • The question clearly states that all roots are real. Why would you need a polar form ? –  Dec 20 '18 at 08:23
  • Is your problem about showing $z=2\cos 36^\circ$ etc are roots of the quartic, or about expressing roots of the quartic as $\pm(1\pm\sqrt5)/4$ (independent $\pm$s)? – user10354138 Dec 20 '18 at 08:33
  • Do you need to solve the equation? Maybe you can set $s = z^2$ . – Matti P. Dec 20 '18 at 08:38

2 Answers2

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?

Dr. Sonnhard Graubner
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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 $w^2\ne 1$.

A.Γ.
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