72544ce67b07af595aa53677ba971bd98b9218e2
1\documentclass[a4paper]{article}
2\usepackage[a4paper,margin=2cm]{geometry}
3\usepackage{multicol}
4\usepackage{amsmath}
5\usepackage{amssymb}
6\usepackage{harpoon}
7\usepackage{tabularx}
8\usepackage{graphicx}
9\usepackage{wrapfig}
10\usepackage{tikz}
11\usepackage{fancyhdr}
12\pagestyle{fancy}
13\fancyhead[LO,LE]{Year 12 Specialist}
14\fancyhead[CO,CE]{Andrew Lorimer}
15
16\usepackage{mathtools}
17\usepackage{xcolor} % used only to show the phantomed stuff
18\renewcommand\hphantom[1]{{\color[gray]{.6}#1}} % comment out!
19\setlength\fboxsep{0pt} \setlength\fboxrule{.2pt} % for the \fboxes
20\newcommand*\leftlap[3][\,]{#1\hphantom{#2}\mathllap{#3}}
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22
23\begin{document}
24
25\begin{multicols}{2}
26
27 \section{Complex numbers}
28
29 \[\mathbb{C}=\{a+bi:a,b\in\mathbb{R}\}\]
30
31 \subsection*{Operations}
32
33 \begin{align*}
34 z_1 \pm z_2&=(a \pm c)(b \pm d)i\\
35 k \times z &= ka + kbi\\
36 z_1 \cdot z_2 &= ac-bd+(ad+bc)i\\
37 z_1 \div z_2 &= (z_1 \overline{z_2}) \div |z_2|^2
38 \end{align*}
39
40 \subsection*{Conjugate}
41
42 \[\overline{z} = a \pm bi\]
43
44 \subsubsection*{Properties}
45
46 \begin{align*}
47 \overline{z_1 \pm z_2} &= \overline{z_1}\pm\overline{z_2}\\
48 \overline{z_1 \cdot z_2} &= \overline{z_1}\cdot\overline{z_2}\\
49 \overline{kz} &= k\overline{z} \quad | \quad k \in \mathbb{R}\\
50 z\overline{z} &= (a+bi)(a-bi)\\
51 &= a^2 + b^2\\
52 &= |z|^2
53 \end{align*}
54
55 \subsection*{Modulus}
56
57 \[|z|=|\vec{Oz}|=\sqrt{a^2 + b^2}\]
58
59 \subsubsection*{Properties}
60
61 \begin{align*}
62 |z_1z_2|&=|z_1||z_2|\\
63 \left|\frac{z_1}{z_2}\right|&=\frac{|z_1|}{|z_2|}\\
64 |z_1+z_2|&\le|z_1|+|z_2|
65 \end{align*}
66
67 \subsection*{Multiplicative inverse}
68
69 \begin{align*}
70 z^{-1}&=\frac{a-bi}{a^2+b^2}\\
71 &=\frac{\overline{z}}{|z|^2}
72 a
73 \end{align*}
74
75 \subsection*{Dividing over \(\mathbb{C}\)}
76
77 \begin{align*}
78 \frac{z_1}{z_2}&=z_1z_2^{-1}\\
79 &=\frac{z_1\overline{z_2}}{|z_2|^2}\\
80 &=\frac{(a+bi)(c-di)}{c^2+d^2}\\
81 & \qquad \text{(rationalise denominator)}
82 \end{align*}
83
84 \subsection*{Argand planes}
85
86 \begin{tikzpicture}\begin{scope}[thick,font=\scriptsize]
87 \draw [->] (-1.5,0) -- (1.5,0) node [above left] {$\operatorname{Re}(z)$};
88 \draw [->] (0,-1.5) -- (0,1.5) node [below right] {$\operatorname{Im}(z)$};
89
90 % If you only want a single label per axis side:
91 \draw (1,-3pt) -- (1,0pt) node [below] {$1$};
92 \draw (-1,-3pt) -- (-1,0pt) node [below] {$-1$};
93 \draw (-3pt,1) -- (0pt,1) node [left] {$i$};
94 \draw (-3pt,-1) -- (0pt,-1) node [left] {$-i$};
95 \end{scope}\end{tikzpicture}
96
97 Multiplication by \(i \implies\) anticlockwise rotation of \(\frac{\pi}{2}\)
98
99 \subsection*{de Moivres' theorem}
100
101 \[(r \operatorname{cis} \theta)^n = r^n \operatorname{cis}(n\theta) \text{ where } n \in \mathbb{Z}\]
102
103 \subsection*{Complex polynomials}
104
105 Include \(\pm\) for all solutions, incl. imaginary
106
107\newcolumntype{R}{>{\raggedleft\arraybackslash}X}
108\newcolumntype{L}{>{\raggedright\arraybackslash}X}
109 \begin{tabularx}{\columnwidth}{rX}
110 \hline
111 Sum of squares & \(\begin{aligned}
112 z^2 + a^2 &= z^2-(ai)^2\\
113 &= (z+ai)(z-ai) \end{aligned}\) \\
114 \hline
115 Sum of cubes & \(a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)\)\\
116 \hline
117 Division & \(P(z)=D(z)Q(z)+R(z)\) \\
118 \hline
119 \parbox[t]{2cm}{Remainder} & Let \(\alpha \in \mathbb{C}\). Remainder of \(P(z) \div (z-\alpha)\) is \(P(\alpha)\)\\
120 \hline
121\end{tabularx}
122
123\subsection*{Roots}
124
125\(n\)th roots of \(z=r\operatorname{cis}\theta\) are:
126
127\[z = r^{\frac{1}{n}} \operatorname{cis}\left(\frac{\theta+2k\pi}{n}\right)\]
128
129\begin{itemize}
130
131 \item{Same modulus for all solutions}
132 \item{Arguments are separated by \(\frac{2\pi}{n}\)}
133
134\item{Solutions of \(z^n=a\) where \(a \in \mathbb{C}\) lie on the circle \(x^2+y^2=\left(|a|^{\frac{1}{n}}\right)^2\)}
135\end{itemize}
136
137\subsubsection*{Conjugate root theorem}
138
139If \(a+bi\) is a solution to \(P(z)=0\), then the conjugate \(\overline{z}=a-bi\) is also a solution.
140
141\end{multicols}
142\end{document}