Andrew's git - notes.git/blob

   1---
   2geometry: margin=2cm
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   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
  69If $z=a+bi$, conjugate is
  70
  71$$\overline{z} = a-bi$$
  72
  73##### Properties
  74
  75- $\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}$
  76- $\overline{z_1 z_2} = \overline{z_1} \cdot \overline{z_2}$
  77- $\overline{kz} = k \overline{z}, \text{ for } k \in \mathbb{R}$
  78- $z \overline{z} = = (a+bi)(a-bi) = a^2+b^2 = |z|^2$
  79- $z + \overline{z} = 2 \operatorname{Re}(z)$
  80
  81### Modulus
  82
  83Distance from origin.
  84
  85$$|{z}|=\sqrt{a^2+b^2} \quad  \therefore z \overline{z} = |z|^2$$
  86
  87###### Properties
  88
  89- $|z_1 z_2| = |z_1| |z_2|$
  90- $|{z_1 \over z_2}| = {|z_1| \over |z_2|}$
  91- $|z_1 + z_2| \le |z_1 + |z_2|$
  92
  93### Multiplicative inverse
  94
  95\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}
  96
  97### Dividing complex numbers
  98
  99$${{z_1} \over {z_2}} = {{z_1\ {z_2}^{-1}}} = {{z_1 \overline{z_2}} \over {{|z_2|}^2}} \quad \text{multiplicative inverse}$$
 100
 101(using multiplicative inverse)
 102
 103In practice, rationalise denominator:
 104${z_1 \over z_2} = {{(a+bi)(c-di)} \over {c^2+d^2}}$
 105
 106## Argand planes
 107
 108- Geometric representation of $\mathbb{C}$
 109- horizontal $= \operatorname{Re}(z)$; vertical $= \operatorname{Im}(z)$
 110- Multiplication by $i$ results in an anticlockwise rotation of $\pi \over 2$
 111
 112## Solving complex polynomials
 113
 114**Include $\pm$ for all solutions, including imaginary**
 115
 116## Solving complex quadratics
 117
 118To solve $z^2+a^2=0$ (sum of two squares):
 119
 120$$z^2+a^2=z^2-(ai)^2=(z+ai)(z-ai)$$
 121
 122#### Dividing complex polynomials
 123
 124Dividing $P(z)$ by $D(z)$ gives quotient $Q(z)$ and remainder $R(z)$ such that:
 125
 126$$P(z) = D(z)Q(z) + R(z)$$
 127
 128#### Remainder theorem
 129
 130Let $\alpha \in \mathbb{C}$. Remainder of $P(z) \div (z - \alpha)$ is $P(\alpha)$
 131
 132## Conjugate root theorem
 133
 134If $a+bi$ is a solution to $P(z)=0$, with $a, b \in \mathbb{R}$, then the conjugate $\overline{z}=a-bi$ is also a solution.
 135
 136## Polar form
 137
 138\begin{equation}\begin{split}z & =r \operatorname{cis} \theta \\ & = r(\operatorname{cos}\theta+i \operatorname{sin}\theta) \\ & = a + bi \end{split}\end{equation}
 139
 140- $r=|z|=\sqrt{\operatorname{Re}(z)^2 + \operatorname{Im}(z)^2}$
 141- $\theta=\operatorname{arg}(z)$ (on CAS: `arg(a+bi)`)
 142- **principal argument** is $\operatorname{Arg}(z) \in (-\pi, \pi]$ (note capital $\operatorname{Arg}$)
 143
 144Note each complex number has multiple polar representations:
 145$z=r \operatorname{cis} \theta = r \operatorname{cis} (\theta+2 n\pi$) where $n$ is integer number of revolutions
 146
 147### Conjugate in polar form
 148
 149$$(r \operatorname{cis} \theta)^{-1} = r\operatorname{cis} (- \theta)$$
 150
 151Reflection of $z$ across horizontal axis.
 152
 153### Multiplication and division in polar form
 154
 155$z_1z_2=r_1r_2\operatorname{cis}(\theta_1+\theta_2)$ (multiply moduli, add angles)
 156
 157${z_1 \over z_2} = {r_1 \over r_2} \operatorname{cis}(\theta_1-\theta_2)$ (divide moduli, subtract angles)
 158
 159## de Moivres' Theorem
 160
 161$$(r\operatorname{cis}\theta)^n=r^n\operatorname{cis}(n\theta) \text{ where } n \in \mathbb{Z}$$
 162
 163## Roots of complex numbers
 164
 165$n$th roots of $z = r \operatorname{cis} \theta$ are
 166
 167$$z={r^{1 \over n}} \operatorname{cis}({{\theta + 2 k \pi} \over n})$$
 168
 169Same modulus for all solutions. Arguments are separated by ${2 \pi} \over n$
 170
 171The solutions of $z^n=a \text{ where } a \in \mathbb{C}$ lie on circle
 172
 173$$x^2 + y^2 = (|a|^{1 \over n})^2$$
 174
 175## Sketching complex graphs
 176
 177- **Straight line:** $\operatorname{Re}(z) = c$ or $\operatorname{Im}(z) = c$ (perpendicular bisector) or $\operatorname{Arg}(z) = \theta$
 178- **Circle:** $|z-z_1|^2 = c^2 |z_2+2|^2$ or $|z-(a + bi)| = c$
 179- **Locus:** $\operatorname{Arg}(z) < \theta$