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| 1 | +\providecommand{\main}{..} |
| 2 | +\documentclass[\main/notes.tex]{subfiles} |
| 3 | + |
| 4 | +\begin{document} |
| 5 | + \addtocontents{toc}{\protect\newpage} |
| 6 | + \setcounter{chapter}{3} |
| 7 | + \chapter{Complex Numbers} |
| 8 | + \section{Complex Numbers} |
| 9 | + \begin{definition}{Complex Number} |
| 10 | + The number $i$ is defined such that: |
| 11 | + \begin{align*} |
| 12 | + i &= \sqrt{-1}\\ |
| 13 | + i^{2} &= -1 |
| 14 | + \end{align*} |
| 15 | + |
| 16 | + Then, if $a$ and $b$ are real numbers, a \concept{complex number} can be written in the form: |
| 17 | + \begin{align*} |
| 18 | + z = a + bi |
| 19 | + \end{align*} |
| 20 | + |
| 21 | + The number $a$ is called the \concept{real part} of $z$, and is denoted $\Real(z)$, or $\Re(z)$. The number $b$ is called the \concept{imaginary part} of $z$, and is denoted $\Imaginary(z)$ or $\Im(z)$. |
| 22 | + \end{definition} |
| 23 | + \begin{definition}{The Complex Plane} |
| 24 | + A complex number $z = a + bi$ can be associated with an ordered pair of real numbers $(a, b)$. These numbers can then be represented geometrically by a point in the $xy$-plane. This plane is called the \concept{complex plane}. Points on the $x$-axis are real numbers, as they have an imaginary part of $0$, and points on the $y$-axis are \concept{pure imaginary numbers}, as they have a real part of $0$. |
| 25 | + \begin{center} |
| 26 | + \begin{tikzpicture} |
| 27 | + \begin{axis}[xmin=-1, xmax=1, ymin=-1, ymax=1, axis lines=center, xlabel={Real}, ylabel={Imaginary}, every axis y label/.style={at=(current axis.above origin),anchor=south}, every axis x label/.style={at=(current axis.right of origin),anchor=west}, ticks=none, scale=0.6] |
| 28 | + \end{axis} |
| 29 | + \end{tikzpicture} |
| 30 | + \end{center} |
| 31 | + \end{definition} |
| 32 | + \begin{definition}{Complex Conjugate} |
| 33 | + If $z = a + bi$ is a complex number, then the \concept{complex conjugate} of $z$ is denoted $\bar{z}$, and is defined by: |
| 34 | + \begin{align*} |
| 35 | + \bar{z} = a - bi |
| 36 | + \end{align*} |
| 37 | + |
| 38 | + You flip the sign of the imaginary part. This results in a vector that is reflected about the real axis. |
| 39 | + \end{definition} |
| 40 | + \begin{definition}{Modulus} |
| 41 | + The producr of a complex number $z = a + bi$ and its conjugate $\bar{z} = a - bi$ is a nonnegative neal number. |
| 42 | + \begin{align*} |
| 43 | + z\bar{z} = (a + bi)(a - bi) = a^{2} - abi + bai - b^{2}i^{2} = a^{2} + b^{2} |
| 44 | + \end{align*} |
| 45 | + |
| 46 | + The square root of the above value is the length of the vector corresponding to $z$. This is called the \concept{modulus}, or \concept{absolute value} of $z$, and is written $\left\lvert z \right\rvert$ |
| 47 | + \begin{align*} |
| 48 | + \left\lvert z\right\rvert = \sqrt{z \bar{z}} = \sqrt{a^{2} + b^{2}} |
| 49 | + \end{align*} |
| 50 | + \end{definition} |
| 51 | + \begin{definition}{Recipricols and Division} |
| 52 | + If $z \neq 0$, then the \concept{recipricol} (or \concept{multiplicative inverse}) of $z$ is denoted by $1/z$ or $z^{-1}$, and is defined such that: |
| 53 | + \begin{align*} |
| 54 | + \frac{1}{z}z = 1 |
| 55 | + \end{align*} |
| 56 | + This equation has a unique solution for $1/z$: |
| 57 | + \begin{align*} |
| 58 | + \frac{1}{z} = \frac{\bar{z}}{\left\lvert z\right\rvert^{2}} |
| 59 | + \end{align*} |
| 60 | + \end{definition} |
| 61 | + \begin{definition}{Quotient} |
| 62 | + If $z_{2} \neq 0$, then the \concept{quotient} $z_{1}/z_{2}$ is defined to be the product of $z_{1}$ and $1/z_{2}$ |
| 63 | + \begin{align*} |
| 64 | + \frac{z_{1}}{z_{2}} = \frac{\overline{z_{2}}}{\left\lvert z_{2}\right\rvert^{2}}z_{1} = \frac{z_{1}\overline{z_{2}}}{\left\lvert z_{2}\right\rvert^{2}} |
| 65 | + \end{align*} |
| 66 | + |
| 67 | + The practical way to perform division of complex numbers is therefore to multiply both the top and bottom by the conjugate of the bottom. |
| 68 | + \end{definition} |
| 69 | + \pagebreak |
| 70 | + \begin{sidenote}{Conjugate Rules} |
| 71 | + For any complex numbers $z$, $z_{1}$ and $z_{2}$: |
| 72 | + \begin{enumerate}[label=(\alph*)] |
| 73 | + \item $\overline{z_{1} + z_{2}} = \overline{z_{1}} + \overline{z_{2}}$ |
| 74 | + \item $\overline{z_{1} - z_{2}} = \overline{z_{1}} - \overline{z_{2}}$ |
| 75 | + \item $\overline{z_{1}z_{2}} = \overline{z_{1}} \cdot \overline{z_{2}}$ |
| 76 | + \item $\overline{z_{1}/z_{2}} = \overline{z_{1}} / \overline{z_{2}}$ |
| 77 | + \item $\overline{\overline{z}} = z$ |
| 78 | + \item $z + \overline{z} = 2 \Real{z}$ |
| 79 | + \item $z - \overline{z} = 2 \Imaginary{z}$ |
| 80 | + \item $z\overline{z}$ is always real. |
| 81 | + \end{enumerate} |
| 82 | + \end{sidenote} |
| 83 | + \begin{sidenote}{Modulus Rules} |
| 84 | + For any complex numbers $z$, $z_{1}$ and $z_{2}$: |
| 85 | + \begin{enumerate}[label=(\alph*)] |
| 86 | + \item $\left\lvert \overline{z}\right\rvert = \left\lvert z\right\rvert $ |
| 87 | + \item $\left\lvert z_{1}z_{2}\right\rvert = \left\lvert z_{1}\right\rvert \left\lvert z_{2}\right\rvert $ |
| 88 | + \item $\left\lvert z_{1}/z_{2}\right\rvert = \left\lvert z_{1}\right\rvert /\left\lvert z_{2}\right\rvert $ |
| 89 | + \item $\left\lvert z_{1} + z_{2}\right\rvert \leq \left\lvert z_{1}\right\rvert + \left\lvert z_{2}\right\rvert $ |
| 90 | + \item $z\overline{z} = \left\lvert z\right\rvert^{2}$ |
| 91 | + \end{enumerate} |
| 92 | + \end{sidenote} |
| 93 | + |
| 94 | + \section{Polar Form} |
| 95 | + \begin{definition}{Polar Form} |
| 96 | + If $z = a + bi$ is a nonzero complex number, and $\phi$ is an angle from the real axis to the vector $z$, then the real and imaginary parts of $z$ can be expressed as: |
| 97 | + \begin{alignat*}{2} |
| 98 | + a &= \left\lvert z\right\rvert \cos \phi & \qquad b &= \left\lvert z\right\rvert \sin \phi |
| 99 | + \end{alignat*} |
| 100 | + |
| 101 | + The complex number $z = a + bi$ can therefore be expressed as: |
| 102 | + \begin{align*} |
| 103 | + z = \left\lvert z\right\rvert (\cos \phi + i \sin \phi) |
| 104 | + \end{align*} |
| 105 | + |
| 106 | + The angle $\phi$ is called an \concept{argument} of $z$. Is not unique, but only one argument (the \concept{principal argument}), which will satisfy in radians: |
| 107 | + \begin{align*} |
| 108 | + - \pi < \phi \leq \pi |
| 109 | + \end{align*} |
| 110 | + \end{definition} |
| 111 | + \begin{sidenote}{Geometric Interpretation} |
| 112 | + \begin{itemize} |
| 113 | + \item Multiplying two complex numbers has the geometric effect of multiplying their moduli, and adding their arguments. |
| 114 | + \item Dividing two complex numbers has the geometric effect of dividing their moduli, and subtracting their arguments. |
| 115 | + \end{itemize} |
| 116 | + \end{sidenote} |
| 117 | + \begin{theorem}{De Moivre's Theorem} |
| 118 | + If $n$ is a positive integer, and if $z$ is a nonzero complex number with polar form: |
| 119 | + \begin{align*} |
| 120 | + z = \left\lvert z\right\rvert (\cos \phi + i \sin \phi) |
| 121 | + \end{align*} |
| 122 | + Then: |
| 123 | + \begin{align*} |
| 124 | + z^{n} = \left\lvert z\right\rvert^{n} (\cos (n \phi) + i \sin (n \phi)) |
| 125 | + \end{align*} |
| 126 | + \end{theorem} |
| 127 | + \begin{theorem}{Binomial Theorem and Pascal's Triangle} |
| 128 | + A method to expand any expression that has been raised to a power. Used to determine the coefficents of a term when expanded. |
| 129 | + |
| 130 | + \concept{Pascal's Triangle} is a way to determine the terms: |
| 131 | + \begin{center} |
| 132 | + \begin{tikzpicture}[rotate=-90] |
| 133 | + \foreach \x in {0,1,...,5} |
| 134 | + { |
| 135 | + \foreach \y in {0,...,\x} |
| 136 | + { |
| 137 | + \pgfmathsetmacro\binom{factorial(\x)/(factorial(\y)*factorial(\x-\y))} |
| 138 | + \pgfmathsetmacro\shift{\x/2} |
| 139 | + \node[xshift=-\shift cm] at (\x,\y) {\pgfmathprintnumber\binom}; |
| 140 | + } |
| 141 | + } |
| 142 | + \end{tikzpicture} |
| 143 | + \end{center} |
| 144 | + \end{theorem} |
| 145 | + \pagebreak |
| 146 | + \begin{example} |
| 147 | + \question{Express $\cos(4\theta)$ and $\sin(4\theta)$ in terms of powers of $\sin\theta$ and $\cos\theta$} |
| 148 | + \begin{enumerate} |
| 149 | + \item Find $(\cos\theta + i\sin\theta)^{4}$ using De Moivre: |
| 150 | + \begin{align*} |
| 151 | + (\cos\theta + i\sin\theta)^{4} = \cos(4\theta) + i\sin(4\theta) |
| 152 | + \end{align*} |
| 153 | + \item Use binomial expansion of $(\cos\theta + i\sin\theta)^{4}$ |
| 154 | + \begin{align*} |
| 155 | + (cos\theta + i\sin\theta)^{4} &= \cos^{4}\theta + 4i\cos^{3}\theta\sin\theta - 6cos^{2}\theta\sin^{2}\theta - 4i\cos\theta\sin^{3}\theta + \sin^{4}\theta\\ |
| 156 | + &= (cos^{4}\theta - 6\cos^{2}\theta\sin^{2}\theta + \sin^{4}\theta) + i(4\cos^{3}\theta\sin\theta - 4\cos\theta\sin^{3}\theta) |
| 157 | + \end{align*} |
| 158 | + \item Equate the real and nonreal parts. |
| 159 | + \begin{align*} |
| 160 | + \cos(4\theta) &= cos^{4}\theta - 6\cos^{2}\theta\sin^{2}\theta + \sin^{4}\theta\\ |
| 161 | + \sin(4\theta) &= 4\cos^{3}\theta\sin\theta - 4\cos\theta\sin^{3}\theta |
| 162 | + \end{align*} |
| 163 | + \end{enumerate} |
| 164 | + \end{example} |
| 165 | + \ifSubfilesClassLoaded{% |
| 166 | + \vbox{\rulechapterend}}{\vspace*{\parskip}\rulebookend} |
| 167 | +\end{document} |
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