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Factorise $z^5-1$ over the real field. Show that $\cos \frac{2\pi}{5}$ is a root of the equation $4x^2+2x-1=0$ and hence find its exact value.

I have worked out that $$ z^5-1=(z-1)(z^4+z^3+z^2+z+1)\\=(z-1)(z^2-2z\cos72^o+1) (z^2+2z\cos 72^o+1) $$

John Smith
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    Welcome to Mathematics Stack Exchange! Please tell us what you have tried so far. – Akiva Weinberger Nov 18 '14 at 11:01
  • By the way, an identity you should learn: $z^5-1=(z-1)(z^4+z^3+z^2+z+1)$, which you should verify. Similarly, for all $n$, we have $z^n-1=(z-1)(z^n+\dotsb+1)$. You should verify this for a few values of $n$. – Akiva Weinberger Nov 18 '14 at 11:03
  • I have previously answered a somewhat related question that might help: http://math.stackexchange.com/a/928762/131263. – barak manos Nov 18 '14 at 11:10
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    I have worked out that $z^5-1=(z-1)(z^4+z^3+z^2+z+1)=(z-1)(z^2-2zcos72+1)(z^2+2zcos72+1)=0$ but other than that I haven't made much progress. – John Smith Nov 18 '14 at 11:12

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Find the polynomials that cancel $\cos\left(\frac{2\pi}{5}\right)$ :

We know that the solution of $z^5=1$ are given by $\alpha^k=e^{\frac{2ik\pi}{5}}$, $k=0,1,2,3,4$. We can observe that $\alpha^{4}=\alpha^{-1}$ and $\alpha^{3}=\alpha^{-2}$. Moreover,

$$(*):\quad 0=1+\alpha+\alpha^2+\alpha^3+\alpha^4=1+\alpha+\alpha^{-1}+\alpha^2+\alpha^{-2}$$ and $\alpha+\alpha^{-1}=2\cos\left(\frac{2\pi}{5}\right)$.

Let set $X=\alpha+\alpha^{-1}$, then $$X^2=(\alpha+\alpha^{-1})^2=\alpha^2+\alpha^{-2}+2$$ and so $$\alpha^2+\alpha^{-2}=X^2-2$$

if we replace in $(*)$ we obtain $$0=X+X^2-2+1=X^2+X-1.$$ Now set $x=\frac{X}{2}=\cos\left(\frac{2\pi}{5}\right)$, therefore, $X=2x$ and so $$0=(2x)^2+2x-1=4x^4+2x-1$$ and thus, $\cos\left(\frac{2\pi}{5}\right)$ is solution of $$4x^2+2x-1=0$$

Find the value of $\cos\left(\frac{2\pi}{5}\right)$ :

$$4x^2+2x-1=0\iff x^2+\frac{1}{2}x-\frac{1}{4}=0\iff \left(x+\frac{1}{4}\right)^2-\frac{1}{16}-\frac{1}{4}=0\iff\left(x+\frac{1}{4}\right)^2=\frac{5}{16}\iff x+\frac{1}{4}=\pm\frac{\sqrt 5}{4}\iff x=\frac{-1\pm\sqrt 5}{4}$$

But $\cos\left(\frac{2\pi}{5}\right)>0$ because $\frac{2\pi}{5}<\frac{\pi}{2}$, then, $$\cos\left(\frac{2\pi}{5}\right)=\frac{-1+\sqrt 5}{4}$$

idm
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