202008181129 Homework 1 (Q1)

Solve for $z\in \mathbb{C}$ in the following equations:

(a) $z^4+z^3+z^2+z+1=0$ ,

(b) $3z^3+29z^2+497z-169=0$ .

Attempts.

(a) Take notice that $z\neq 1$ . Try and see having both sides of the equation multiplied by $(1-z)$ ,

$\begin{aligned} 0 & = (1-z)(z^4+z^3+z^2+z+1) \\ 0 & = 1-z^5 \\ z^5 & = 1 = 1(1+0\,\mathrm{i}) \\ z^5 & = e^{\mathrm{i}(2n\pi)}\qquad \qquad \textrm{where }n=0,1,2,3,4. \\ \end{aligned}$

$\therefore z=e^{\mathrm{i}(\frac{2n\pi}{5})}$ .

( $n=0$ is rejected for $z\neq 1=e^{\mathrm{i}(\frac{2(0)\pi}{5})}$ .)

In polar expression $z=e^{\mathrm{i\frac{2\pi}{5}}},\, e^{\mathrm{i\frac{4\pi}{5}}},\, e^{\mathrm{i\frac{6\pi}{5}}},\, e^{\mathrm{i\frac{8\pi}{5}}}$ .

In trigonometric expression

$z=\mathrm{cis}(\frac{2n\pi}{5})=\mathrm{cis}(\frac{2\pi}{5}),\, \mathrm{cis}(\frac{4\pi}{5}),\, \mathrm{cis}(\frac{6\pi}{5}),\, \mathrm{cis}(\frac{8\pi}{5})$

I.e.,

$\begin{aligned} z^1 & = \cos 72^\circ +\mathrm{i}\sin 72^\circ = \mathrm{cis\,} 72^\circ \\ z^2 & = \cos 144^\circ +\mathrm{i}\sin 144^\circ = \mathrm{cis\,} 144^\circ \\ z^3 & = \cos 216^\circ +\mathrm{i}\sin 216^\circ = \mathrm{cis\,} 216^\circ \\ z^4 & = \cos 288^\circ +\mathrm{i}\sin 288^\circ = \mathrm{cis\,} 288^\circ \\ \end{aligned}$

(b)Let $f(z)=3z^3+29z^2+497z-169=0$ . Then $(3z-1)$ is a factor, because $\frac{1}{3}$ is a zero (i.e., $f(\frac{1}{3}) = \frac{1}{9}+\frac{29}{9}+\frac{497}{3}-169=0$ ).