1---
2geometry: margin=1.9cm
3<!-- columns: 2 -->
4graphics: yes
5tables: yes
6author: Andrew Lorimer
7classoption: twocolumn
8header-includes:
9- \usepackage{harpoon}
10- \usepackage{amsmath}
11- \pagenumbering{gobble}
12
13---
14
15
16# Complex & Imaginary Numbers
17
18## Imaginary numbers
19
20$$i^2 = -1 \quad \therefore i = \sqrt {-1}$$
21
22### Simplifying negative surds
23
24\begin{equation}\begin{split}\sqrt{-2} & = \sqrt{-1 \times 2} \\ & = \sqrt{2}i\end{split}\end{equation}
25
26
27## Complex numbers
28
29$$\mathbb{C} = \{a+bi : a, b \in \mathbb{R} \}$$
30
31General form: $z=a+bi$
32$\operatorname{Re}(z) = a, \quad \operatorname{Im}(z) = b$
33
34### Addition
35
36If $z_1 = a+bi$ and $z_2=c+di$, then
37
38$$z_1+z_2 = (a+c)+(b+d)i$$
39
40### Subtraction
41
42If $z_1=a+bi$ and $z_2=c+di$, then
43
44$$z_1−z_2=(a−c)+(b−d)i$$
45
46### Multiplication by a real constant
47
48If $z=a+bi$ and $k \in \mathbb{R}$, then
49
50$$kz=ka+kbi$$
51
52### Powers of $i$
53
54- $i^{4n} = 1$
55- $i^{4n+1} = i$
56- $i^{4n+2} = -1$
57- $i^{4n+3} = -i$
58
59For $i^n$, find remainder $r$ when $n \div 4$. Then $i^n = i^r$.
60
61### Multiplying complex expressions
62
63If $z_1 = a+bi$ and $z_2=c+di$, then
64
65$$z_1 \times z_2 = (ac-bd)+(ad+bc)i$$
66
67### Conjugates
68
69$$\overline{z} = a \mp bi$$
70
71##### Properties
72
73- $\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}$
74- $\overline{z_1 z_2} = \overline{z_1} \cdot \overline{z_2}$
75- $\overline{kz} = k \overline{z}, \text{ for } k \in \mathbb{R}$
76- $z \overline{z} = = (a+bi)(a-bi) = a^2+b^2 = |z|^2$
77- $z + \overline{z} = 2 \operatorname{Re}(z)$
78
79### Modulus
80
81Distance from origin.
82
83$$|{z}|=\sqrt{a^2+b^2} \quad \therefore z \overline{z} = |z|^2$$
84
85###### Properties
86
87- $|z_1 z_2| = |z_1| |z_2|$
88- $|{z_1 \over z_2}| = {|z_1| \over |z_2|}$
89- $|z_1 + z_2| \le |z_1 + |z_2|$
90
91### Multiplicative inverse
92
93\begin{equation}\begin{split}z^{-1} & = {1 \over z} \\ & = {{a-bi} \over {a^2+B^2}} \\ & = {\overline{z} \over {|z|^2}}\end{split}\end{equation}
94
95### Dividing complex numbers
96
97$${{z_1} \over {z_2}} = {{z_1\ {z_2}^{-1}}} = {{z_1 \overline{z_2}} \over {{|z_2|}^2}} \quad \text{(multiplicative inverse)}$$
98
99In practice, rationalise denominator:
100
101$${z_1 \over z_2} = {{(a+bi)(c-di)} \over {c^2+d^2}}$$
102
103## Argand planes
104
105- Geometric representation of $\mathbb{C}$
106- horizontal $= \operatorname{Re}(z)$; vertical $= \operatorname{Im}(z)$
107- Multiplication by $i$ results in an anticlockwise rotation of $\pi \over 2$
108
109\vfil \break
110
111## Complex polynomials
112
113**Include $\pm$ for all solutions, including imaginary**
114
115### Sum of two squares (quadratics)
116
117$$z^2+a^2=z^2-(ai)^2=(z+ai)(z-ai)$$
118
119Complete the square to get to this point.
120
121#### Dividing complex polynomials
122
123$P(z) \div D(z)$ gives quotient $Q(z)$ and remainder $R(z)$:
124
125$$P(z) = D(z)Q(z) + R(z)$$
126
127#### Remainder theorem
128
129Let $\alpha \in \mathbb{C}$. Remainder of $P(z) \div (z - \alpha)$ is $P(\alpha)$
130
131#### Factor theorem
132
133If $a+bi$ is a solution to $P(z)=0$, then:
134
135- $P(a+bi)=0$
136- $z-(a+bi)$ is a factor of $P(z)$
137
138#### Sum of two cubes
139
140$$a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)$$
141
142## Conjugate root theorem
143
144If $a+bi$ is a solution to $P(z)=0$, then the conjugate $\overline{z}=a-bi$ is also a solution.
145
146## Polar form
147
148\begin{equation}\begin{split}z & =r \operatorname{cis} \theta \\ & = r(\operatorname{cos}\theta+i \operatorname{sin}\theta) \\ & = a + bi \end{split}\end{equation}
149
150- $r=|z|=\sqrt{\operatorname{Re}(z)^2 + \operatorname{Im}(z)^2}$
151- $\theta=\operatorname{arg}(z)$ (on CAS: `arg(a+bi)`)
152- **principal argument** is $\operatorname{Arg}(z) \in (-\pi, \pi]$ (note capital $\operatorname{Arg}$)
153
154Each complex number has multiple polar representations:
155$z=r \operatorname{cis} \theta = r \operatorname{cis} (\theta+2 n\pi$) with $n \in \mathbb{Z}$ revolutions
156
157### Conjugate in polar form
158
159$$(r \operatorname{cis} \theta)^{-1} = r\operatorname{cis} (- \theta)$$
160
161Reflection of $z$ across horizontal axis.
162
163### Multiplication and division in polar form
164
165$$z_1z_2=r_1r_2\operatorname{cis}(\theta_1+\theta_2)$$
166
167$${z_1 \over z_2} = {r_1 \over r_2} \operatorname{cis}(\theta_1-\theta_2)$$
168
169## de Moivres' Theorem
170
171$$(r\operatorname{cis}\theta)^n=r^n\operatorname{cis}(n\theta) \text{ where } n \in \mathbb{Z}$$
172
173## Roots of complex numbers
174
175$n$th roots of $z = r \operatorname{cis} \theta$ are
176
177$$z={r^{1 \over n}} \operatorname{cis}({{\theta + 2 k \pi} \over n})$$
178
179Same modulus for all solutions. Arguments are separated by ${2 \pi} \over n$
180
181The solutions of $z^n=a \text{ where } a \in \mathbb{C}$ lie on circle
182
183$$x^2 + y^2 = (|a|^{1 \over n})^2$$
184
185## Sketching complex graphs
186
187### Straight line
188
189- $\operatorname{Re}(z) = c$ or $\operatorname{Im}(z) = c$ (perpendicular bisector)
190- $\operatorname{Arg}(z) = \theta$
191- $|z+a|=|z+bi|$ where $m={a \over b}$
192- $|z+a|=|z+b| \longrightarrow 2(a-b)x=b^2-a^2$
193
194### Circle
195
196$|z-z_1|^2 = c^2 |z_2+2|^2$ or $|z-(a + bi)| = c$
197
198### Locus
199
200$\operatorname{Arg}(z) < \theta$