Andrew's git - notes.git/blob - spec/spec-collated.tex

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  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}
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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}