Chapter 1: Complex Numbers
Exercise Solutions (1.1 to 1.5)
Exercise 1.1
Problem 1
Solve the following equations.
Solution
(a)
$\displaystyle \begin{aligned}
& x^2 - 6x + 10 = 0 \\
& x^2 - 6x = -10 \\
& x^2 - 6x + 9 = -1 \\
& (x - 3)^2 = -1 \\
& x - 3 = \pm i \\
& x = 3 \pm i
\end{aligned}$
(b)
$\displaystyle \begin{aligned}
& -2x^2 + 4x - 3 = 0 \\
& x^2 - 2x = -\frac{3}{2} \\
& x^2 - 2x + 1 = 1 - \frac{3}{2} \\
& (x - 1)^2 = -\frac{1}{2} \\
& x - 1 = \pm \frac{1}{\sqrt{2}}i \\
& x = 1 \pm \frac{\sqrt{2}}{2}i
\end{aligned}$
(c)
$\displaystyle \begin{aligned}
& 5x^2 - 2x + 1 = 0 \\
& x^2 - \frac{2}{5}x = -\frac{1}{5} \\
& x^2 - \frac{2}{5}x + \frac{1}{25} = \frac{1}{25} - \frac{5}{25} \\
& \left(x - \frac{1}{5}\right)^2 = -\frac{4}{25} \\
& x - \frac{1}{5} = \pm \frac{2}{5}i \\
& x = \frac{1}{5} \pm \frac{2}{5}i
\end{aligned}$
(d)
$\displaystyle \begin{aligned}
& 3x^2 + 7x + 5 = 0 \\
& x^2 + \frac{7}{3}x = -\frac{5}{3} \\
& x^2 + \frac{7}{3}x + \frac{49}{36} = \frac{49}{36} - \frac{60}{36} \\
& \left(x + \frac{7}{6}\right)^2 = -\frac{11}{36} \\
& x + \frac{7}{6} = \pm \frac{\sqrt{11}}{6}i \\
& x = -\frac{7}{6} \pm \frac{\sqrt{11}}{6}i
\end{aligned}$
Problem 2
Solve the following equations and check your answers.
Solution
(a)
$\displaystyle \begin{aligned}
& x^2 - 2x + 4 = 0 \\
& x^2 - 2x + 1 = -3 \\
& (x - 1)^2 = -3 \\
& x - 1 = \pm \sqrt{3}i \\
& x = 1 \pm \sqrt{3}i
\end{aligned}$
Check:
$\displaystyle \begin{aligned}
& \text{For } x = 1 + \sqrt{3}i: \\
& (1+\sqrt{3}i)^2 - 2(1+\sqrt{3}i) + 4 = (1 + 2\sqrt{3}i - 3) - 2 - 2\sqrt{3}i + 4 \\
& \phantom{(1+\sqrt{3}i)^2 - 2(1+\sqrt{3}i) + 4} = -2 + 2\sqrt{3}i - 2 - 2\sqrt{3}i + 4 \\
& \phantom{(1+\sqrt{3}i)^2 - 2(1+\sqrt{3}i) + 4} = 0
\end{aligned}$
(b)
$\displaystyle \begin{aligned}
& x^2 - 4x + 5 = 0 \\
& x^2 - 4x + 4 = -1 \\
& (x - 2)^2 = -1 \\
& x - 2 = \pm i \\
& x = 2 \pm i
\end{aligned}$
Check:
$\displaystyle \begin{aligned}
& \text{For } x = 2 + i: \\
& (2+i)^2 - 4(2+i) + 5 = (4 + 4i - 1) - 8 - 4i + 5 \\
& \phantom{(2+i)^2 - 4(2+i) + 5} = 3 + 4i - 8 - 4i + 5 \\
& \phantom{(2+i)^2 - 4(2+i) + 5} = 0
\end{aligned}$
Problem 3
Find the value of $\displaystyle i^n$ for every positive integer n.
Solution
$\displaystyle \begin{aligned}
& \text{Let } k \text{ be a positive integer or } 0. \\
& i^{4k} = 1 \\
& i^{4k+1} = i \\
& i^{4k+2} = -1 \\
& i^{4k+3} = -i
\end{aligned}$
Exercise 1.2
Problem 1
Compute:
Solution
(a)
$\displaystyle \begin{aligned}
& (2, 0)(3, 5) + (3, -2)(0, 1) = 2(3+5i) + (3-2i)(i) \\
& \phantom{(2, 0)(3, 5) + (3, -2)(0, 1)} = 6 + 10i + 3i - 2i^2 \\
& \phantom{(2, 0)(3, 5) + (3, -2)(0, 1)} = 6 + 13i + 2 \\
& \phantom{(2, 0)(3, 5) + (3, -2)(0, 1)} = 8 + 13i \\
& \phantom{(2, 0)(3, 5) + (3, -2)(0, 1)} \equiv (8, 13)
\end{aligned}$
(b)
$\displaystyle \begin{aligned}
& (2, -5)(-1, 0) + (1, 0)(5, 1) = (2-5i)(-1) + (1)(5+i) \\
& \phantom{(2, -5)(-1, 0) + (1, 0)(5, 1)} = -2 + 5i + 5 + i \\
& \phantom{(2, -5)(-1, 0) + (1, 0)(5, 1)} = 3 + 6i \\
& \phantom{(2, -5)(-1, 0) + (1, 0)(5, 1)} \equiv (3, 6)
\end{aligned}$
(c)
$\displaystyle \begin{aligned}
& (-3, -2)(-2, -3) + (-2, -3)(-3, -2) = 2[(-3-2i)(-2-3i)] \\
& \phantom{(-3, -2)(-2, -3) + (-2, -3)(-3, -2)} = 2(6 + 9i + 4i + 6i^2) \\
& \phantom{(-3, -2)(-2, -3) + (-2, -3)(-3, -2)} = 2(6 + 13i - 6) \\
& \phantom{(-3, -2)(-2, -3) + (-2, -3)(-3, -2)} = 26i \\
& \phantom{(-3, -2)(-2, -3) + (-2, -3)(-3, -2)} \equiv (0, 26)
\end{aligned}$
(d)
$\displaystyle \begin{aligned}
& (1, 0)(0, 1) + (0, 1)(1, 0) = (1)(i) + (i)(1) \\
& \phantom{(1, 0)(0, 1) + (0, 1)(1, 0)} = 2i \\
& \phantom{(1, 0)(0, 1) + (0, 1)(1, 0)} \equiv (0, 2)
\end{aligned}$
Problem 2
Compute:
Solution
(a)
$\displaystyle \begin{aligned}
& (3+2i)(3-2i) + (-5+7i)(-1-i) = (9 - 4i^2) + (5 + 5i - 7i - 7i^2) \\
& \phantom{(3+2i)(3-2i) + (-5+7i)(-1-i)} = (9 + 4) + (5 - 2i + 7) \\
& \phantom{(3+2i)(3-2i) + (-5+7i)(-1-i)} = 13 + 12 - 2i \\
& \phantom{(3+2i)(3-2i) + (-5+7i)(-1-i)} = 25 - 2i
\end{aligned}$
(b)
$\displaystyle \begin{aligned}
& (-1+i)(1-i) + (2+3i) = (-1 + i + i - i^2) + 2 + 3i \\
& \phantom{(-1+i)(1-i) + (2+3i)} = (-1 + 2i + 1) + 2 + 3i \\
& \phantom{(-1+i)(1-i) + (2+3i)} = 2i + 2 + 3i \\
& \phantom{(-1+i)(1-i) + (2+3i)} = 2 + 5i
\end{aligned}$
(c)
$\displaystyle \begin{aligned}
& (1+i)(1-i) + (-2+i)(-2+i) = (1 - i^2) + (4 - 2i - 2i + i^2) \\
& \phantom{(1+i)(1-i) + (-2+i)(-2+i)} = 2 + (4 - 4i - 1) \\
& \phantom{(1+i)(1-i) + (-2+i)(-2+i)} = 2 + 3 - 4i \\
& \phantom{(1+i)(1-i) + (-2+i)(-2+i)} = 5 - 4i
\end{aligned}$
(d)
$\displaystyle \begin{aligned}
& (3+2i) + (7-i)(-3+3i) = 3+2i + (-21 + 21i + 3i - 3i^2) \\
& \phantom{(3+2i) + (7-i)(-3+3i)} = 3+2i - 21 + 24i + 3 \\
& \phantom{(3+2i) + (7-i)(-3+3i)} = -15 + 26i
\end{aligned}$
Exercise 1.3
Problem 1
Let $\displaystyle z_1 = -2+3i, z_2 = 5+2i$. Compute:
Solution
(a) $\displaystyle z_1^2 - 2z_1 + 1$
$\displaystyle \begin{aligned}
& z_1^2 - 2z_1 + 1 = (-2+3i)^2 - 2(-2+3i) + 1 \\
& \phantom{z_1^2 - 2z_1 + 1} = (4 - 12i - 9) + 4 - 6i + 1 \\
& \phantom{z_1^2 - 2z_1 + 1} = -5 - 12i + 5 - 6i \\
& \phantom{z_1^2 - 2z_1 + 1} = -18i
\end{aligned}$
(b) $\displaystyle 3z_2^2 + 2z_2 - 1$
$\displaystyle \begin{aligned}
& 3z_2^2 + 2z_2 - 1 = 3(5+2i)^2 + 2(5+2i) - 1 \\
& \phantom{3z_2^2 + 2z_2 - 1} = 3(25 + 20i - 4) + 10 + 4i - 1 \\
& \phantom{3z_2^2 + 2z_2 - 1} = 3(21 + 20i) + 9 + 4i \\
& \phantom{3z_2^2 + 2z_2 - 1} = 63 + 60i + 9 + 4i \\
& \phantom{3z_2^2 + 2z_2 - 1} = 72 + 64i
\end{aligned}$
(c) $\displaystyle z_1\overline{z_2} + z_2\overline{z_1}$
$\displaystyle \begin{aligned}
& z_1\overline{z_2} + z_2\overline{z_1} = (-2+3i)(5-2i) + (5+2i)(-2-3i) \\
& \phantom{z_1\overline{z_2} + z_2\overline{z_1}} = (-10 + 4i + 15i + 6) + (-10 - 15i - 4i + 6) \\
& \phantom{z_1\overline{z_2} + z_2\overline{z_1}} = (-4 + 19i) + (-4 - 19i) \\
& \phantom{z_1\overline{z_2} + z_2\overline{z_1}} = -8
\end{aligned}$
(d) $\displaystyle \frac{1}{z_1}$
$\displaystyle \begin{aligned}
& \frac{1}{z_1} = \frac{1}{-2+3i} \\
& \phantom{\frac{1}{z_1}} = \frac{-2-3i}{(-2+3i)(-2-3i)} \\
& \phantom{\frac{1}{z_1}} = \frac{-2-3i}{4+9} \\
& \phantom{\frac{1}{z_1}} = -\frac{2}{13} - \frac{3}{13}i
\end{aligned}$
(e) $\displaystyle \frac{1}{z_2}$
$\displaystyle \begin{aligned}
& \frac{1}{z_2} = \frac{1}{5+2i} \\
& \phantom{\frac{1}{z_2}} = \frac{5-2i}{(5+2i)(5-2i)} \\
& \phantom{\frac{1}{z_2}} = \frac{5-2i}{25+4} \\
& \phantom{\frac{1}{z_2}} = \frac{5}{29} - \frac{2}{29}i
\end{aligned}$
(f) $\displaystyle \frac{\overline{z_1}}{z_2}$
$\displaystyle \begin{aligned}
& \frac{\overline{z_1}}{z_2} = \frac{-2-3i}{5+2i} \\
& \phantom{\frac{\overline{z_1}}{z_2}} = \frac{(-2-3i)(5-2i)}{29} \\
& \phantom{\frac{\overline{z_1}}{z_2}} = \frac{-10 + 4i - 15i - 6}{29} \\
& \phantom{\frac{\overline{z_1}}{z_2}} = -\frac{16}{29} - \frac{11}{29}i
\end{aligned}$
(g) $\displaystyle \frac{z_1}{z_2}$
$\displaystyle \begin{aligned}
& \frac{z_1}{z_2} = \frac{-2+3i}{5+2i} \\
& \phantom{\frac{z_1}{z_2}} = \frac{(-2+3i)(5-2i)}{29} \\
& \phantom{\frac{z_1}{z_2}} = \frac{-10 + 4i + 15i + 6}{29} \\
& \phantom{\frac{z_1}{z_2}} = -\frac{4}{29} + \frac{19}{29}i
\end{aligned}$
(h) $\displaystyle \frac{1}{z_1z_2}$
$\displaystyle \begin{aligned}
& \frac{1}{z_1z_2} = \frac{1}{(-2+3i)(5+2i)} \\
& \phantom{\frac{1}{z_1z_2}} = \frac{1}{-10 - 4i + 15i - 6} \\
& \phantom{\frac{1}{z_1z_2}} = \frac{1}{-16 + 11i} \\
& \phantom{\frac{1}{z_1z_2}} = \frac{-16 - 11i}{(-16)^2 + 11^2} \\
& \phantom{\frac{1}{z_1z_2}} = -\frac{16}{377} - \frac{11}{377}i
\end{aligned}$
(i) $\displaystyle \frac{z_2}{z_1}$
$\displaystyle \begin{aligned}
& \frac{z_2}{z_1} = \frac{5+2i}{-2+3i} \\
& \phantom{\frac{z_2}{z_1}} = \frac{(5+2i)(-2-3i)}{13} \\
& \phantom{\frac{z_2}{z_1}} = \frac{-10 - 15i - 4i + 6}{13} \\
& \phantom{\frac{z_2}{z_1}} = -\frac{4}{13} - \frac{19}{13}i
\end{aligned}$
(k) $\displaystyle \frac{\overline{z_1}z_2}{z_1\overline{z_2}}$
$\displaystyle \begin{aligned}
& \frac{\overline{z_1}z_2}{z_1\overline{z_2}} = \frac{(-2-3i)(5+2i)}{(-2+3i)(5-2i)} \\
& \phantom{\frac{\overline{z_1}z_2}{z_1\overline{z_2}}} = \frac{-10 - 4i - 15i + 6}{-10 + 4i + 15i + 6} \\
& \phantom{\frac{\overline{z_1}z_2}{z_1\overline{z_2}}} = \frac{-4 - 19i}{-4 + 19i} \\
& \phantom{\frac{\overline{z_1}z_2}{z_1\overline{z_2}}} = \frac{(-4 - 19i)^2}{(-4)^2 + 19^2} \\
& \phantom{\frac{\overline{z_1}z_2}{z_1\overline{z_2}}} = \frac{16 + 152i + 361i^2}{16 + 361} \\
& \phantom{\frac{\overline{z_1}z_2}{z_1\overline{z_2}}} = \frac{16 + 152i - 361}{377} \\
& \phantom{\frac{\overline{z_1}z_2}{z_1\overline{z_2}}} = -\frac{345}{377} + \frac{152}{377}i
\end{aligned}$
(l) $\displaystyle \frac{z_2}{z_1} + \frac{z_1}{z_2}$
$\displaystyle \begin{aligned}
& \frac{z_2}{z_1} + \frac{z_1}{z_2} = \left(-\frac{4}{13} - \frac{19}{13}i\right) + \left(-\frac{4}{29} + \frac{19}{29}i\right) \\
& \phantom{\frac{z_2}{z_1} + \frac{z_1}{z_2}} = \frac{29(-4 - 19i) + 13(-4 + 19i)}{377} \\
& \phantom{\frac{z_2}{z_1} + \frac{z_1}{z_2}} = \frac{-116 - 551i - 52 + 247i}{377} \\
& \phantom{\frac{z_2}{z_1} + \frac{z_1}{z_2}} = -\frac{168}{377} - \frac{304}{377}i
\end{aligned}$
(m) $\displaystyle \left(\frac{z_2}{z_1}\right)^2$
$\displaystyle \begin{aligned}
& \left(\frac{z_2}{z_1}\right)^2 = \left(\frac{-4-19i}{13}\right)^2 \\
& \phantom{\left(\frac{z_2}{z_1}\right)^2} = \frac{16 + 152i - 361}{169} \\
& \phantom{\left(\frac{z_2}{z_1}\right)^2} = -\frac{345}{169} + \frac{152}{169}i
\end{aligned}$
Problem 2
Let $\displaystyle z_1 = 3-2i, z_2 = 1+4i$. Show that:
Solution
(a)
$\displaystyle \begin{aligned}
& \text{LHS} = \overline{(3-2i + 1+4i)} \\
& \phantom{\text{LHS}} = \overline{4+2i} \\
& \phantom{\text{LHS}} = 4-2i \\
& \text{RHS} = (3+2i) + (1-4i) \\
& \phantom{\text{RHS}} = 4-2i \\
& \therefore \text{LHS} = \text{RHS}
\end{aligned}$
(b)
$\displaystyle \begin{aligned}
& \text{LHS} = \overline{(3-2i)(1+4i)} \\
& \phantom{\text{LHS}}= \overline{3 + 12i - 2i + 8} \\
& \phantom{\text{LHS}}= \overline{11+10i} \\
& \phantom{\text{LHS}}= 11-10i \\
& \text{RHS} = (3+2i)(1-4i) \\
& \phantom{\text{RHS}}= 3 - 12i + 2i + 8 \\
& \phantom{\text{RHS}}= 11-10i \\
& \therefore \text{LHS} = \text{RHS}
\end{aligned}$
(c)
$\displaystyle \begin{aligned}
& \text{LHS}= \overline{\left(\frac{3-2i}{1+4i}\right)} \\
& \phantom{\text{LHS}}= \overline{\left(\frac{(3-2i)(1-4i)}{17}\right)} \\
& \phantom{\text{LHS}}= \overline{\left(\frac{3 - 12i - 2i - 8}{17}\right)} \\
& \phantom{\text{LHS}}= \overline{\left(\frac{-5-14i}{17}\right)} \\
& \phantom{\text{LHS}}= -\frac{5}{17} + \frac{14}{17}i \\
& \text{RHS} = \frac{3+2i}{1-4i} \\
& \phantom{\text{RHS}}= \frac{(3+2i)(1+4i)}{17} \\
& \phantom{\text{RHS}}= \frac{3 + 12i + 2i - 8}{17} \\
& \phantom{\text{RHS}}= -\frac{5}{17} + \frac{14}{17}i \\
& \therefore \text{LHS} = \text{RHS}
\end{aligned}$
Exercise 1.4
Problem 1
Find the trigonometric form with $\displaystyle -\pi < \theta \le \pi$.
Solution
(a)
$\displaystyle \begin{aligned}
& z = 1 - \sqrt{3}i \\
& r = \sqrt{1^2 + (-\sqrt{3})^2} = 2 \\
& \tan\theta = -\sqrt{3} \quad (\text{4th Quadrant}) \implies \theta = -\frac{\pi}{3} \\
& z = 2\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right)
\end{aligned}$
(b)
$\displaystyle \begin{aligned}
& z = -\sqrt{2} + \sqrt{2}i \\
& r = \sqrt{2 + 2} = 2 \\
& \tan\theta = -1 \quad (\text{2nd Quadrant}) \implies \theta = \frac{3\pi}{4} \\
& z = 2\left(\cos\left(\frac{3\pi}{4}\right) + i\sin\left(\frac{3\pi}{4}\right)\right)
\end{aligned}$
(c)
$\displaystyle \begin{aligned}
& z = -2 - 2i \\
& r = \sqrt{4 + 4} = 2\sqrt{2} \\
& \tan\theta = 1 \quad (\text{3rd Quadrant}) \implies \theta = -\frac{3\pi}{4} \\
& z = 2\sqrt{2}\left(\cos\left(-\frac{3\pi}{4}\right) + i\sin\left(-\frac{3\pi}{4}\right)\right)
\end{aligned}$
(d)
$\displaystyle \begin{aligned}
& z = \sqrt{3} - i \\
& r = \sqrt{3 + 1} = 2 \\
& \tan\theta = -\frac{1}{\sqrt{3}} \quad (\text{4th Quadrant}) \implies \theta = -\frac{\pi}{6} \\
& z = 2\left(\cos\left(-\frac{\pi}{6}\right) + i\sin\left(-\frac{\pi}{6}\right)\right)
\end{aligned}$
(e)
$\displaystyle \begin{aligned}
& z = i \\
& r = 1, \quad \theta = \frac{\pi}{2} \\
& z = 1\left(\cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right)\right)
\end{aligned}$
(f)
$\displaystyle \begin{aligned}
& z = -3i \\
& r = 3, \quad \theta = -\frac{\pi}{2} \\
& z = 3\left(\cos\left(-\frac{\pi}{2}\right) + i\sin\left(-\frac{\pi}{2}\right)\right)
\end{aligned}$
Problem 2
Given that $\displaystyle z_1 = 2 - 2\sqrt{3}i, z_2 = -1 - i$, find the following complex numbers by using trigonometric forms. Check your answer by direct calculation.
Solution
$\displaystyle \begin{aligned}
& z_1 = 4\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right), \quad z_2 = \sqrt{2}\left(\cos\left(-\frac{3\pi}{4}\right) + i\sin\left(-\frac{3\pi}{4}\right)\right)
\end{aligned}$
(a)
$\displaystyle \begin{aligned}
& z_1 z_2 = (4)(\sqrt{2})\left(\cos\left(-\frac{\pi}{3} - \frac{3\pi}{4}\right) + i\sin\left(-\frac{\pi}{3} - \frac{3\pi}{4}\right)\right) \\
& \phantom{z_1 z_2} = 4\sqrt{2}\left(\cos\left(-\frac{13\pi}{12}\right) + i\sin\left(-\frac{13\pi}{12}\right)\right) \\
& \text{Adding } 2\pi \text{ to find the principal argument: } -\frac{13\pi}{12} + 2\pi = \frac{11\pi}{12} \\
& \phantom{z_1 z_2} = 4\sqrt{2}\left(\cos\left(\frac{11\pi}{12}\right) + i\sin\left(\frac{11\pi}{12}\right)\right) \\
& \text{Direct Check:} \\
& z_1 z_2 = (2 - 2\sqrt{3}i)(-1 - i) \\
& \phantom{z_1 z_2} = -2 - 2i + 2\sqrt{3}i - 2\sqrt{3} \\
& \phantom{z_1 z_2} = (-2 - 2\sqrt{3}) + (2\sqrt{3} - 2)i
\end{aligned}$
(b)
$\displaystyle \begin{aligned}
& z_1^{-1} = \frac{1}{4}\left(\cos\left(\frac{\pi}{3}\right) + i\sin\left(\frac{\pi}{3}\right)\right) \\
& \text{Direct Check:} \\
& \frac{1}{2 - 2\sqrt{3}i} = \frac{2 + 2\sqrt{3}i}{4 + 12} = \frac{2 + 2\sqrt{3}i}{16} = \frac{1}{8} + \frac{\sqrt{3}}{8}i
\end{aligned}$
(c)
$\displaystyle \begin{aligned}
& z_2^{-1} = \frac{1}{\sqrt{2}}\left(\cos\left(\frac{3\pi}{4}\right) + i\sin\left(\frac{3\pi}{4}\right)\right) \\
& \text{Direct Check:} \\
& \frac{1}{-1 - i} = \frac{-1 + i}{1 + 1} = -\frac{1}{2} + \frac{1}{2}i
\end{aligned}$
(d)
$\displaystyle \begin{aligned}
& \frac{z_1}{z_2} = \frac{4}{\sqrt{2}}\left(\cos\left(-\frac{\pi}{3} - \left(-\frac{3\pi}{4}\right)\right) + i\sin\left(-\frac{\pi}{3} - \left(-\frac{3\pi}{4}\right)\right)\right) \\
& \phantom{\frac{z_1}{z_2}} = 2\sqrt{2}\left(\cos\left(\frac{5\pi}{12}\right) + i\sin\left(\frac{5\pi}{12}\right)\right) \\
& \text{Direct Check:} \\
& \frac{2 - 2\sqrt{3}i}{-1 - i} = \frac{(2 - 2\sqrt{3}i)(-1 + i)}{(-1 - i)(-1 + i)} \\
& \phantom{\frac{2 - 2\sqrt{3}i}{-1 - i}} = \frac{-2 + 2i + 2\sqrt{3}i + 2\sqrt{3}}{2} \\
& \phantom{\frac{2 - 2\sqrt{3}i}{-1 - i}} = (\sqrt{3} - 1) + (\sqrt{3} + 1)i \\
& \text{Verification of direct result: } z = (\sqrt{3} - 1) + (\sqrt{3} + 1)i \\
& \text{Magnitude: } |z| = \sqrt{(\sqrt{3}-1)^2 + (\sqrt{3}+1)^2} \\
& \phantom{\text{Magnitude: } |z|} = \sqrt{(3 - 2\sqrt{3} + 1) + (3 + 2\sqrt{3} + 1)} = \sqrt{8} = 2\sqrt{2} \\
& \text{Angle (1st Quadrant): } \theta = \tan^{-1}\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right) = \frac{5\pi}{12}
\end{aligned}$
(e)
$\displaystyle \begin{aligned}
& \frac{z_2}{z_1} = \frac{\sqrt{2}}{4}\left(\cos\left(-\frac{3\pi}{4} - \left(-\frac{\pi}{3}\right)\right) + i\sin\left(-\frac{3\pi}{4} - \left(-\frac{\pi}{3}\right)\right)\right) \\
& \phantom{\frac{z_2}{z_1}} = \frac{\sqrt{2}}{4}\left(\cos\left(-\frac{5\pi}{12}\right) + i\sin\left(-\frac{5\pi}{12}\right)\right) \\
& \text{Direct Check:} \\
& \frac{-1 - i}{2 - 2\sqrt{3}i} = \frac{(-1 - i)(2 + 2\sqrt{3}i)}{16} \\
& \phantom{\frac{-1 - i}{2 - 2\sqrt{3}i}} = \frac{-2 - 2\sqrt{3}i - 2i + 2\sqrt{3}}{16} \\
& \phantom{\frac{-1 - i}{2 - 2\sqrt{3}i}} = \frac{\sqrt{3} - 1}{8} - \frac{\sqrt{3} + 1}{8}i
\end{aligned}$
Problem 3
Given that $\displaystyle z = -2\sqrt{3} - 2i$, find (a) $\displaystyle z^5$ (b) $\displaystyle z^{-5}$.
Solution
$\displaystyle \begin{aligned}
& z = 4\left(\cos\left(-\frac{5\pi}{6}\right) + i\sin\left(-\frac{5\pi}{6}\right)\right)
\end{aligned}$
(a)
$\displaystyle \begin{aligned}
& z^5 = 4^5\left(\cos\left(5 \times -\frac{5\pi}{6}\right) + i\sin\left(5 \times -\frac{5\pi}{6}\right)\right) \\
& \phantom{z^5} = 1024\left(\cos\left(-\frac{25\pi}{6}\right) + i\sin\left(-\frac{25\pi}{6}\right)\right) \\
& \text{Adding } 4\pi \text{ to find the principal argument: } -\frac{25\pi}{6} + \frac{24\pi}{6} = -\frac{\pi}{6} \\
& \phantom{z^5} = 1024\left(\cos\left(-\frac{\pi}{6}\right) + i\sin\left(-\frac{\pi}{6}\right)\right) \\
& \phantom{z^5} = 1024\left(\frac{\sqrt{3}}{2} - \frac{1}{2}i\right) \\
& \phantom{z^5} = 512\sqrt{3} - 512i
\end{aligned}$
(b)
$\displaystyle \begin{aligned}
& z^{-5} = 4^{-5}\left(\cos\left(-5 \times -\frac{5\pi}{6}\right) + i\sin\left(-5 \times -\frac{5\pi}{6}\right)\right) \\
& \phantom{z^{-5}} = \frac{1}{1024}\left(\cos\left(\frac{25\pi}{6}\right) + i\sin\left(\frac{25\pi}{6}\right)\right) \\
& \text{Subtracting } 4\pi \text{ to find the principal argument: } \frac{25\pi}{6} - \frac{24\pi}{6} = \frac{\pi}{6} \\
& \phantom{z^{-5}} = \frac{1}{1024}\left(\cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right)\right) \\
& \phantom{z^{-5}} = \frac{1}{1024}\left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right) \\
& \phantom{z^{-5}} = \frac{\sqrt{3}}{2048} + \frac{1}{2048}i
\end{aligned}$
Exercise 1.5
Problem 1
Find the square roots of the following:
Solution
(a) $\displaystyle 1 + \sqrt{3}i$
$\displaystyle \begin{aligned}
& z = 2\left(\cos\left(\frac{\pi}{3}\right) + i\sin\left(\frac{\pi}{3}\right)\right) \\
& \text{Note: } r^{\frac{1}{2}} = \sqrt{2} \\
& w_k = \sqrt{2}\left(\cos\left(\frac{\frac{\pi}{3} + 2k\pi}{2}\right) + i\sin\left(\frac{\frac{\pi}{3} + 2k\pi}{2}\right)\right), \quad k = 0, 1 \\
& w_0 = \sqrt{2}\left(\cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right)\right) = \frac{\sqrt{6}}{2} + \frac{\sqrt{2}}{2}i \\
& w_1 = \sqrt{2}\left(\cos\left(\frac{7\pi}{6}\right) + i\sin\left(\frac{7\pi}{6}\right)\right) = -\frac{\sqrt{6}}{2} - \frac{\sqrt{2}}{2}i
\end{aligned}$
(b) $\displaystyle i$
$\displaystyle \begin{aligned}
& z = \cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right) \\
& w_k = \cos\left(\frac{\frac{\pi}{2} + 2k\pi}{2}\right) + i\sin\left(\frac{\frac{\pi}{2} + 2k\pi}{2}\right), \quad k = 0, 1 \\
& w_0 = \cos\left(\frac{\pi}{4}\right) + i\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i \\
& w_1 = \cos\left(\frac{5\pi}{4}\right) + i\sin\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i
\end{aligned}$
(c) $\displaystyle -\sqrt{3} + i$
$\displaystyle \begin{aligned}
& z = 2\left(\cos\left(\frac{5\pi}{6}\right) + i\sin\left(\frac{5\pi}{6}\right)\right) \\
& w_k = \sqrt{2}\left(\cos\left(\frac{\frac{5\pi}{6} + 2k\pi}{2}\right) + i\sin\left(\frac{\frac{5\pi}{6} + 2k\pi}{2}\right)\right), \quad k = 0, 1 \\
& w_0 = \sqrt{2}\left(\cos\left(\frac{5\pi}{12}\right) + i\sin\left(\frac{5\pi}{12}\right)\right) \\
& w_1 = \sqrt{2}\left(\cos\left(\frac{17\pi}{12}\right) + i\sin\left(\frac{17\pi}{12}\right)\right) = -\sqrt{2}\left(\cos\left(\frac{5\pi}{12}\right) + i\sin\left(\frac{5\pi}{12}\right)\right)
\end{aligned}$
(d) $\displaystyle -1 - \sqrt{3}i$
$\displaystyle \begin{aligned}
& z = 2\left(\cos\left(-\frac{2\pi}{3}\right) + i\sin\left(-\frac{2\pi}{3}\right)\right) \\
& w_k = \sqrt{2}\left(\cos\left(\frac{-\frac{2\pi}{3} + 2k\pi}{2}\right) + i\sin\left(\frac{-\frac{2\pi}{3} + 2k\pi}{2}\right)\right), \quad k = 0, 1 \\
& w_0 = \sqrt{2}\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right) = \frac{\sqrt{2}}{2} - \frac{\sqrt{6}}{2}i \\
& w_1 = \sqrt{2}\left(\cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right)\right) = -\frac{\sqrt{2}}{2} + \frac{\sqrt{6}}{2}i
\end{aligned}$
(e) $\displaystyle -i$
$\displaystyle \begin{aligned}
& z = \cos\left(-\frac{\pi}{2}\right) + i\sin\left(-\frac{\pi}{2}\right) \\
& w_k = \cos\left(\frac{-\frac{\pi}{2} + 2k\pi}{2}\right) + i\sin\left(\frac{-\frac{\pi}{2} + 2k\pi}{2}\right), \quad k = 0, 1 \\
& w_0 = \cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i \\
& w_1 = \cos\left(\frac{3\pi}{4}\right) + i\sin\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i
\end{aligned}$
(f) $\displaystyle \sqrt{3} - i$
$\displaystyle \begin{aligned}
& z = 2\left(\cos\left(-\frac{\pi}{6}\right) + i\sin\left(-\frac{\pi}{6}\right)\right) \\
& w_k = \sqrt{2}\left(\cos\left(\frac{-\frac{\pi}{6} + 2k\pi}{2}\right) + i\sin\left(\frac{-\frac{\pi}{6} + 2k\pi}{2}\right)\right), \quad k = 0, 1 \\
& w_0 = \sqrt{2}\left(\cos\left(-\frac{\pi}{12}\right) + i\sin\left(-\frac{\pi}{12}\right)\right) \\
& w_1 = \sqrt{2}\left(\cos\left(\frac{11\pi}{12}\right) + i\sin\left(\frac{11\pi}{12}\right)\right) = -\sqrt{2}\left(\cos\left(-\frac{\pi}{12}\right) + i\sin\left(-\frac{\pi}{12}\right)\right)
\end{aligned}$
Problem 2
Find the cube roots of the following complex numbers.
Solution
(a) $\displaystyle 1 + i$
$\displaystyle \begin{aligned}
& z = \sqrt{2}\left(\cos\left(\frac{\pi}{4}\right) + i\sin\left(\frac{\pi}{4}\right)\right) \\
& \text{Note: } r^{\frac{1}{3}} = (\sqrt{2})^{\frac{1}{3}} = (2^{\frac{1}{2}})^{\frac{1}{3}} = 2^{\frac{1}{6}} \\
& w_k = 2^{\frac{1}{6}}\left(\cos\left(\frac{\frac{\pi}{4} + 2k\pi}{3}\right) + i\sin\left(\frac{\frac{\pi}{4} + 2k\pi}{3}\right)\right), \quad k = 0, 1, 2 \\
& w_0 = 2^{\frac{1}{6}}\left(\cos\left(\frac{\pi}{12}\right) + i\sin\left(\frac{\pi}{12}\right)\right) \\
& w_1 = 2^{\frac{1}{6}}\left(\cos\left(\frac{3\pi}{4}\right) + i\sin\left(\frac{3\pi}{4}\right)\right) \\
& w_2 = 2^{\frac{1}{6}}\left(\cos\left(\frac{17\pi}{12}\right) + i\sin\left(\frac{17\pi}{12}\right)\right)
\end{aligned}$
(b) $\displaystyle i$
$\displaystyle \begin{aligned}
& z = \cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right) \\
& w_k = \cos\left(\frac{\frac{\pi}{2} + 2k\pi}{3}\right) + i\sin\left(\frac{\frac{\pi}{2} + 2k\pi}{3}\right), \quad k = 0, 1, 2 \\
& w_0 = \cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} + \frac{1}{2}i \\
& w_1 = \cos\left(\frac{5\pi}{6}\right) + i\sin\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2} + \frac{1}{2}i \\
& w_2 = \cos\left(\frac{3\pi}{2}\right) + i\sin\left(\frac{3\pi}{2}\right) = -i
\end{aligned}$
(c) $\displaystyle -1 + i$
$\displaystyle \begin{aligned}
& z = \sqrt{2}\left(\cos\left(\frac{3\pi}{4}\right) + i\sin\left(\frac{3\pi}{4}\right)\right) \\
& \text{Note: } r^{\frac{1}{3}} = (\sqrt{2})^{\frac{1}{3}} = (2^{\frac{1}{2}})^{\frac{1}{3}} = 2^{\frac{1}{6}} \\
& w_k = 2^{\frac{1}{6}}\left(\cos\left(\frac{\frac{3\pi}{4} + 2k\pi}{3}\right) + i\sin\left(\frac{\frac{3\pi}{4} + 2k\pi}{3}\right)\right), \quad k = 0, 1, 2 \\
& w_0 = 2^{\frac{1}{6}}\left(\cos\left(\frac{\pi}{4}\right) + i\sin\left(\frac{\pi}{4}\right)\right) \\
& w_1 = 2^{\frac{1}{6}}\left(\cos\left(\frac{11\pi}{12}\right) + i\sin\left(\frac{11\pi}{12}\right)\right) \\
& w_2 = 2^{\frac{1}{6}}\left(\cos\left(\frac{19\pi}{12}\right) + i\sin\left(\frac{19\pi}{12}\right)\right)
\end{aligned}$
(d) $\displaystyle -1 - i$
$\displaystyle \begin{aligned}
& z = \sqrt{2}\left(\cos\left(-\frac{3\pi}{4}\right) + i\sin\left(-\frac{3\pi}{4}\right)\right) \\
& \text{Note: } r^{\frac{1}{3}} = (\sqrt{2})^{\frac{1}{3}} = (2^{\frac{1}{2}})^{\frac{1}{3}} = 2^{\frac{1}{6}} \\
& w_k = 2^{\frac{1}{6}}\left(\cos\left(\frac{-\frac{3\pi}{4} + 2k\pi}{3}\right) + i\sin\left(\frac{-\frac{3\pi}{4} + 2k\pi}{3}\right)\right), \quad k = 0, 1, 2 \\
& w_0 = 2^{\frac{1}{6}}\left(\cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right)\right) \\
& w_1 = 2^{\frac{1}{6}}\left(\cos\left(\frac{5\pi}{12}\right) + i\sin\left(\frac{5\pi}{12}\right)\right) \\
& w_2 = 2^{\frac{1}{6}}\left(\cos\left(\frac{13\pi}{12}\right) + i\sin\left(\frac{13\pi}{12}\right)\right)
\end{aligned}$
(e) $\displaystyle -i$
$\displaystyle \begin{aligned}
& z = \cos\left(-\frac{\pi}{2}\right) + i\sin\left(-\frac{\pi}{2}\right) \\
& w_k = \cos\left(\frac{-\frac{\pi}{2} + 2k\pi}{3}\right) + i\sin\left(\frac{-\frac{\pi}{2} + 2k\pi}{3}\right), \quad k = 0, 1, 2 \\
& w_0 = \cos\left(-\frac{\pi}{6}\right) + i\sin\left(-\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} - \frac{1}{2}i \\
& w_1 = \cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right) = i \\
& w_2 = \cos\left(\frac{7\pi}{6}\right) + i\sin\left(\frac{7\pi}{6}\right) = -\frac{\sqrt{3}}{2} - \frac{1}{2}i
\end{aligned}$
(f) $\displaystyle 1 - i$
$\displaystyle \begin{aligned}
& z = \sqrt{2}\left(\cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right)\right) \\
& \text{Note: } r^{\frac{1}{3}} = (\sqrt{2})^{\frac{1}{3}} = (2^{\frac{1}{2}})^{\frac{1}{3}} = 2^{\frac{1}{6}} \\
& w_k = 2^{\frac{1}{6}}\left(\cos\left(\frac{-\frac{\pi}{4} + 2k\pi}{3}\right) + i\sin\left(\frac{-\frac{\pi}{4} + 2k\pi}{3}\right)\right), \quad k = 0, 1, 2 \\
& w_0 = 2^{\frac{1}{6}}\left(\cos\left(-\frac{\pi}{12}\right) + i\sin\left(-\frac{\pi}{12}\right)\right) \\
& w_1 = 2^{\frac{1}{6}}\left(\cos\left(\frac{7\pi}{12}\right) + i\sin\left(\frac{7\pi}{12}\right)\right) \\
& w_2 = 2^{\frac{1}{6}}\left(\cos\left(\frac{5\pi}{4}\right) + i\sin\left(\frac{5\pi}{4}\right)\right)
\end{aligned}$
Problem 3
Solve the following equations:
Solution
(a) $\displaystyle z^4 = -i$
$\displaystyle \begin{aligned}
& z^4 = \cos\left(-\frac{\pi}{2}\right) + i\sin\left(-\frac{\pi}{2}\right) \\
& z_k = \cos\left(\frac{-\frac{\pi}{2} + 2k\pi}{4}\right) + i\sin\left(\frac{-\frac{\pi}{2} + 2k\pi}{4}\right), \quad k = 0, 1, 2, 3 \\
& z_0 = \cos\left(-\frac{\pi}{8}\right) + i\sin\left(-\frac{\pi}{8}\right), \quad z_1 = \cos\left(\frac{3\pi}{8}\right) + i\sin\left(\frac{3\pi}{8}\right) \\
& z_2 = \cos\left(\frac{7\pi}{8}\right) + i\sin\left(\frac{7\pi}{8}\right), \quad z_3 = \cos\left(\frac{11\pi}{8}\right) + i\sin\left(\frac{11\pi}{8}\right)
\end{aligned}$
(b) $\displaystyle z^4 = -1$
$\displaystyle \begin{aligned}
& z^4 = \cos(\pi) + i\sin(\pi) \\
& z_k = \cos\left(\frac{\pi + 2k\pi}{4}\right) + i\sin\left(\frac{\pi + 2k\pi}{4}\right), \quad k = 0, 1, 2, 3 \\
& z_0 = \cos\left(\frac{\pi}{4}\right) + i\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i \\
& z_1 = \cos\left(\frac{3\pi}{4}\right) + i\sin\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i \\
& z_2 = \cos\left(\frac{5\pi}{4}\right) + i\sin\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i \\
& z_3 = \cos\left(\frac{7\pi}{4}\right) + i\sin\left(\frac{7\pi}{4}\right) = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i
\end{aligned}$
(c) $\displaystyle z^4 = -8 - 8\sqrt{3}i$
$\displaystyle \begin{aligned}
& z^4 = 16\left(\cos\left(-\frac{2\pi}{3}\right) + i\sin\left(-\frac{2\pi}{3}\right)\right) \\
& \text{Note: } r^{\frac{1}{4}} = 16^{\frac{1}{4}} = 2 \\
& z_k = 2\left(\cos\left(\frac{-\frac{2\pi}{3} + 2k\pi}{4}\right) + i\sin\left(\frac{-\frac{2\pi}{3} + 2k\pi}{4}\right)\right), \quad k = 0, 1, 2, 3 \\
& z_0 = 2\left(\cos\left(-\frac{\pi}{6}\right) + i\sin\left(-\frac{\pi}{6}\right)\right) = \sqrt{3} - i \\
& z_1 = 2\left(\cos\left(\frac{\pi}{3}\right) + i\sin\left(\frac{\pi}{3}\right)\right) = 1 + \sqrt{3}i \\
& z_2 = 2\left(\cos\left(\frac{5\pi}{6}\right) + i\sin\left(\frac{5\pi}{6}\right)\right) = -\sqrt{3} + i \\
& z_3 = 2\left(\cos\left(\frac{4\pi}{3}\right) + i\sin\left(\frac{4\pi}{3}\right)\right) = -1 - \sqrt{3}i
\end{aligned}$
(d) $\displaystyle z^6 = -1$
$\displaystyle \begin{aligned}
& z^6 = \cos(\pi) + i\sin(\pi) \\
& z_k = \cos\left(\frac{\pi + 2k\pi}{6}\right) + i\sin\left(\frac{\pi + 2k\pi}{6}\right), \quad k = 0, 1, 2, 3, 4, 5 \\
& z_0 = \cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} + \frac{1}{2}i \\
& z_1 = \cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right) = i \\
& z_2 = \cos\left(\frac{5\pi}{6}\right) + i\sin\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2} + \frac{1}{2}i \\
& z_3 = \cos\left(\frac{7\pi}{6}\right) + i\sin\left(\frac{7\pi}{6}\right) = -\frac{\sqrt{3}}{2} - \frac{1}{2}i \\
& z_4 = \cos\left(\frac{3\pi}{2}\right) + i\sin\left(\frac{3\pi}{2}\right) = -i \\
& z_5 = \cos\left(\frac{11\pi}{6}\right) + i\sin\left(\frac{11\pi}{6}\right) = \frac{\sqrt{3}}{2} - \frac{1}{2}i
\end{aligned}$