Complex poly

A

aycaramba

Member

May 13, 2017

#1

How to solve z⁵ - 4z =0 over complex field??? Cambridges theory notes were confusing

J

$z^5-4z=0 \\ \Rightarrow z(z^4-4)=0 \\ \Rightarrow z(z^2-2)(z^2+2)=0 \\\Rightarrow z(z+\sqrt{2})(z-\sqrt{2}) (z+\sqrt{2}i)(z-\sqrt{2}i)=0 \\ z=0, \pm \sqrt{2}, \pm \sqrt{2}i\\ \\ \\ $another way to solve $ z^4=4: \\ z=\sqrt[\frac{1}{4}]{4} \\= \left ( 4cis( 2n \pi) \right )^{\frac{1}{4}} \\ = 4^{\frac{1}{4}}cis\left ( \frac{2n \pi}{4} \right ) \\ =\sqrt{2}cis\left ( \frac{n \pi}{2}\right ) \\$ When $n= -1, 0, 1, 2$ you get $z=-\sqrt{2}i, \sqrt{2}, \sqrt{2}i, -\sqrt{2} $ respectively$