spec / complex.mdon commit refinements to calculus notes (6a1257b)
   1# Complex & Imaginary Numbers
   2
   3## Imaginary numbers
   4
   5$i^2 = -1$
   6
   7$\therefore i = \sqrt {-1}$
   8
   9### Simplifying negative surds
  10
  11$\sqrt{-2} = \sqrt{-1 \times 2}$
  12
  13          $= \sqrt{2}i$
  14
  15
  16## Complex numbers
  17
  18$\mathbb{C} = \{a+bi : a, b \in \mathbb{R} \}$
  19
  20General form: $z=a+bi$
  21- $\operatorname{Re}(z) = a$
  22- $\operatorname{Im}(z) = b$
  23
  24### Addition
  25
  26If $z_1 = a+bi$ and $z_2=c+di$, then
  27
  28            $z_1+z_2 = (a+c)+(b+d)i$
  29
  30### Subtraction
  31
  32If $z_1=a+bi$ and $z_2=c+di$, then
  33
  34           $z_1−z_2=(a−c)+(b−d)i$
  35
  36### Multiplication by a real constant
  37
  38If $z=a+bi$ and $k \in \mathbb{R}$, then
  39
  40           $kz=ka+kbi$
  41
  42### Powers of $i$
  43$i^0=1$
  44$i^1=i$
  45$i^2=-1$
  46$i^3=-i$
  47$i^4=1$
  48$\dots$
  49
  50Therefore..
  51- $i^{4n} = 1$
  52- $i^{4n+1} = i$
  53- $i^{4n+2} = -1$
  54- $i^{4n+3} = -i$
  55
  56### Multiplying complex expressions
  57
  58If $z_1 = a+bi$ and $z_2=c+di$, then
  59           $z_1 \times z_2 = (ac-bd)+(ad+bc)i$
  60
  61### Conjugates
  62
  63If $z=a+bi$, conjugate of $z$ is $\overline{z} = a-bi$ (flipped operator)
  64
  65Also, $z \overline{z} = (a+bi)(a-bi) = a^2+b^2$
  66
  67- Multiplication and addition are associative
  68
  69### Modulus
  70
  71Distance from origin.
  72$|{z}|=\sqrt{a^2+b^2}$
  73
  74$\therefore z \overline{z} = |z|^2$
  75
  76### Multiplicative inverse
  77
  78$z^{-1} = {1 \over z} = {{a-bi} \over {a^2+B^2}} = {\overline{z} \over {|z|^2}}$
  79
  80### Dividing complex numbers
  81
  82${{z_1} \over {z_2}} = {{z_1\ {z_2}^{-1}}} = {{z_1 \overline{z_2}} \over {{|z_2|}^2}}$
  83
  84(using multiplicative inverse)
  85
  86In practice, rationalise denominator:
  87${z_1 \over z_2} = {{(a+bi)(c-di)} \over {c^2+d^2}}$
  88
  89## Argand planes
  90
  91- Geometric representation of $\mathbb{C}$
  92- Horizontal $= \operatorname{Re}(z)$; vertical $= \operatorname{Im}(z)$
  93- Multiplication by $i$ results in an anticlockwise rotation of $\pi \over 2$
  94
  95## Solving complex quadratics
  96
  97To solve $z^2+a^2=0$ (sum of two squares):
  98
  99$z^2+a^2=z^2-(ai)^2$
 100              $=(z+ai)(z-ai)$ 
 101
 102## Polar form
 103
 104General form:
 105$z=r \operatorname{cis} \theta$
 106$= r\operatorname{cos}\theta+r\operatorname{sin}\theta i$
 107
 108where
 109- $z=a+bi$
 110- $r$ is the distance from origin, given by Pythagoras ($r=\sqrt{x^2+y^2}$)
 111- $\theta$ is the argument of $z$, CCW from origin
 112
 113Note each complex number has multiple polar representations:
 114$z=r \operatorname{cis} \theta = r \operatorname{cis} (\theta+2 n\pi$) where $n$ is integer number of revolutions
 115
 116### Multiplication and division in polar form
 117
 118$z_1z_2=r_1r_2\operatorname{cis}(\theta_1+\theta_2)$ (multiply moduli, add angles)
 119
 120${z_1 \over z_2} = {r_1 \over r_2} \operatorname{cis}(\theta_1-\theta_2)$ (divide moduli, subtract angles)
 121
 122## de Moivres' Theorum
 123
 124$(r\operatorname{cis}\theta)^n=r^n\operatorname{cis}(n\theta)$