1 problem found
If $$ z^{4}+z^{3}+z^{2}+z+1=0\tag{*} $$ and \(u=z+z^{-1}\), find the possible values of \(u\). Hence find the possible values of \(z\). [Do not try to simplify your answers.] Show that, if \(z\) satisfies \((*)\), then \[z^{5}-1=0.\] Hence write the solutions of \((*)\) in the form \(z=r(\cos\theta+i\sin\theta)\) for suitable real \(r\) and \(\theta\). Deduce that \[\sin\frac{2\pi}{5}=\frac{\surd(10+2\surd 5)}{4} \ \ \hbox{and}\ \ \cos\frac{2\pi}{5}=\frac{-1+\surd 5}{4}.\]
Solution: \begin{align*} && 0 &= z^4+z^3+z^2+z+1 \\ \Rightarrow && 0 &= z^2+z+1+z^{-1}+z^{-2} \tag{\(z \neq 0\)} \\ &&&= \left ( z+z^{-1} \right)^2-2 + z+z^{-1} + 1 \\ &&&= u^2+u-1 \\ \Rightarrow && u &= \frac{-1 \pm \sqrt{5}}{2} \\ \Rightarrow && z+z^{-1} &= \frac{-1 \pm \sqrt{5}}{2} \\ \Rightarrow && 0 &= z^2-\left ( \frac{-1 \pm \sqrt{5}}{2}\right)z+1 \\ \Rightarrow && z &= \frac{\left ( \frac{-1 \pm \sqrt{5}}{2}\right) \pm \sqrt{\left ( \frac{-1 \pm \sqrt{5}}{2}\right)^2-4}}{2} \\ &&&= \frac{\left ( \frac{-1 \pm \sqrt{5}}{2}\right) \pm \sqrt{\frac{1+5\mp2\sqrt{5}-16}{4}}}{2} \\ &&&= \frac{\left ( \frac{-1 \pm \sqrt{5}}{2}\right) \pm \sqrt{\frac{-10\mp2\sqrt{5}-16}{4}}}{2} \\ &&&= \frac{-1\pm\sqrt{5}}{4} \pm i\frac{\sqrt{10\pm 2\sqrt{5}}}{4} \end{align*} Since \(z^4+z^3+z^2+z+1 = 0\) we can multiply both sides by \(z-1\) to obtain \(z^5-1 = 0\). Therefore if \(z = r(\cos \theta + i \sin \theta)\) we see that \(z^5 = 1 \Rightarrow r^5 (\cos 5 \theta + i \sin 5 \theta) = 1 \Rightarrow r = 1, 5 \theta = 2n \pi\) ie \(z = \cos \frac{2n\pi}{5} + i\sin \frac{2n \pi}{5}\). We are looking for a solution in the first quadrant, therefore \(\cos \frac{2\pi}{5} = \frac{-1 + \sqrt{5}}4\) and \(\sin \frac{2\pi}{5} = \frac{\sqrt{10+2\sqrt{5}}}{4}\)