August 18, 2020 – Physicspupil

Express the following complex numbers in the form $a+b\,\mathrm{i}$ .

(a) $(1+\mathrm{i})^3$ ,

(b) $(2+3\mathrm{i})(3-4\mathrm{i})$ ,

(c) $\mathrm{i}^5+\mathrm{i}^{16}$ ,

(d) $\frac{1}{2}(1+\mathrm{i})(1+\mathrm{i}^{-8})$ .

Solution.

(a)

$\begin{aligned} & \quad\enspace (1+\mathrm{i})^3 \\ & = 1^3 + 3(1)^2(\mathrm{i}) +3(1)(\mathrm{i})^2 + \mathrm{i}^3 \\ & = 1 + 3\mathrm{i} - 3 -\mathrm{i} \\ & = -2+2\mathrm{i} \end{aligned}$

(b)

$\begin{aligned} & \quad\enspace \frac{2+3\mathrm{i}}{3-4\mathrm{i}} \\ & = \bigg(\frac{2+3\mathrm{i}}{3-4\mathrm{i}}\bigg) \bigg( \frac{3+4\mathrm{i}}{3+4\mathrm{i}} \bigg) \\ & = \frac{(2+3\mathrm{i})(3+4\mathrm{i})}{(3)^2-(4\mathrm{i})^2} \\ & = \frac{6+8\mathrm{i}+9\mathrm{i}+12\mathrm{i}^2}{9+16} \\ & = -\frac{6}{25} + \frac{17}{25}\mathrm{i} \end{aligned}$

(c)

$\begin{aligned} & \quad\enspace \mathrm{i}^5+\mathrm{i}^{16} \\ & = (\mathrm{i}^2)^2\mathrm{i} + (\mathrm{i}^2)^8 \\ & = (-1)^2\mathrm{i} + (-1)^8 \\ & = 1+\mathrm{i} \end{aligned}$

(d)

$\begin{aligned} & \quad\enspace \frac{1}{2}(1+\mathrm{i})(1+\mathrm{i}^{-8}) \\ & = \frac{1}{2}(1+\mathrm{i}^{-8}+\mathrm{i} + \mathrm{i}^{-7}) \\ & = \frac{1}{2} \bigg( 1+ (\mathrm{i}^{-2})^4 + \mathrm{i} + (\mathrm{i}^{-2})^4\mathrm{i} \bigg) \\ & (\, \dots \enspace \mathrm{i}^{-2} = \frac{1}{\mathrm{i}^2} = -1 \enspace \dots\, ) \\ & = \frac{1}{2} \bigg( 1+(-1)^4 + \mathrm{i} + (-1)^4\mathrm{i} \bigg) \\ & = \frac{1}{2} (2+2\mathrm{i}) \\ & = 1+\mathrm{i} \end{aligned}$

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., $f(\frac{1}{3}) = \frac{1}{9}+\frac{29}{9}+\frac{497}{3}-169=0$ ).