Andrew's git - notes.git/blob

   1# Complex & Imaginary Numbers
   2
   3## Imaginary numbers
   4
   5$i^2 = -1 \quad \therefore i = \sqrt {-1}$
   6
   7### Simplifying negative surds
   8
   9$\sqrt{-2} = \sqrt{-1 \times 2}$  
  10$= \sqrt{2}i$
  11
  12
  13## Complex numbers
  14
  15$\mathbb{C} = \{a+bi : a, b \in \mathbb{R} \}$
  16
  17General form: $z=a+bi$  
  18$\operatorname{Re}(z) = a, \quad \operatorname{Im}(z) = b$
  19
  20### Addition
  21
  22If $z_1 = a+bi$ and $z_2=c+di$, then  
  23$z_1+z_2 = (a+c)+(b+d)i$
  24
  25### Subtraction
  26
  27If $z_1=a+bi$ and $z_2=c+di$, then $z_1−z_2=(a−c)+(b−d)i$
  28
  29### Multiplication by a real constant
  30
  31If $z=a+bi$ and $k \in \mathbb{R}$, then $kz=ka+kbi$
  32
  33### Powers of $i$
  34$i^0=1$
  35$i^1=i$
  36$i^2=-1$
  37$i^3=-i$
  38$i^4=1$
  39$\dots$
  40
  41Therefore..
  42
  43- $i^{4n} = 1$
  44- $i^{4n+1} = i$
  45- $i^{4n+2} = -1$
  46- $i^{4n+3} = -i$
  47
  48For $i^n$, divide $n$ by 4 and let remainder $= r$. Then $i^n = i^r$.
  49
  50### Multiplying complex expressions
  51
  52If $z_1 = a+bi$ and $z_2=c+di$, then  
  53$z_1 \times z_2 = (ac-bd)+(ad+bc)i$
  54
  55### Conjugates
  56
  57If $z=a+bi$, conjugate of $z$ is $\overline{z} = a-bi$ (flipped operator)
  58
  59Also, $z \overline{z} = (a+bi)(a-bi) = a^2+b^2 = |z|^2$
  60
  61- Multiplication and addition are associative
  62
  63#### Properties
  64
  65- $\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}$
  66- $\overline{z_1 z_2} = \overline{z_1} \cdot \overline{z_2}$
  67- $\overline{kz} = k \overline{z}, \text{ for } k \in \mathbb{R}$
  68- $z \overline{z} = |z|^2$
  69- $z + \overline{z} = 2 \operatorname{Re}(z)$
  70
  71### Modulus
  72
  73Distance from origin.
  74$|{z}|=\sqrt{a^2+b^2}$
  75
  76$\therefore z \overline{z} = |z|^2$
  77
  78#### Properties
  79
  80- $|z_1 z_2| = |z_1| |z_2|$
  81- $|{z_1 \over z_2}| = {|z_1| \over |z_2|}$
  82- $|z_1 + z_2| \le |z_1 + |z_2|$
  83
  84### Multiplicative inverse
  85
  86$z^{-1} = {1 \over z} = {{a-bi} \over {a^2+B^2}} = {\overline{z} \over {|z|^2}}$
  87
  88### Dividing complex numbers
  89
  90${{z_1} \over {z_2}} = {{z_1\ {z_2}^{-1}}} = {{z_1 \overline{z_2}} \over {{|z_2|}^2}}$
  91
  92(using multiplicative inverse)
  93
  94In practice, rationalise denominator:
  95${z_1 \over z_2} = {{(a+bi)(c-di)} \over {c^2+d^2}}$
  96
  97## Argand planes
  98
  99- Geometric representation of $\mathbb{C}$
 100- Horizontal $= \operatorname{Re}(z)$; vertical $= \operatorname{Im}(z)$
 101- Multiplication by $i$ results in an anticlockwise rotation of $\pi \over 2$
 102
 103## Solving complex quadratics
 104
 105To solve $z^2+a^2=0$ (sum of two squares):
 106
 107$z^2+a^2=z^2-(ai)^2=(z+ai)(z-ai)$
 108
 109*Must include $\pm$ in solutions*
 110
 111## Solving complex polynomials
 112
 113Include $\pm$ for all solutions, including imaginary.
 114
 115#### Dividing complex polynomials
 116
 117Dividing $P(z)$ by $D(z)$ gives quotient $Q(z)$ and remainder $R(z)$ such that:
 118
 119$$P(z) = D(z)Q(z) + R(z)$$
 120
 121#### Remainder theorem
 122
 123Let $\alpha \in \mathbb{C}$. Remainder of $P(z) \div (z - \alpha)$ is $P(\alpha)$
 124
 125## Conjugate root theorem
 126
 127If $a+bi$ is a solution to $P(z)=0$, with $a, b \in \mathbb{R}$, the the conjugate $a-bi$ is also a solution.
 128
 129## Polar form
 130
 131$$\begin{equation}\begin{split}z & =r \operatorname{cis} \theta \\ & = r(\operatorname{cos}\theta+i \operatorname{sin}\theta) \\ & = a + bi \end{split}\end{equation}$$
 132
 133- $r=|z|$, given by Pythagoras ($r=\sqrt{\operatorname{Re}(z)^2 + \operatorname{Im}(z)^2}$)
 134- $\theta=\operatorname{arg}(z)$ (on CAS: `arg(a+bi)`)
 135- **principal argument** is $\operatorname{Arg}(z) \in (-\pi, \pi]$ (note capital $\operatorname{Arg}$)
 136
 137Note each complex number has multiple polar representations:
 138$z=r \operatorname{cis} \theta = r \operatorname{cis} (\theta+2 n\pi$) where $n$ is integer number of revolutions
 139
 140### Conjugate in polar form
 141
 142$$(r \operatorname{cis} \theta)^{-1} = r\operatorname{cis} (- \theta)$$
 143
 144Reflection of $z$ across horizontal axis.
 145
 146### Multiplication and division in polar form
 147
 148$z_1z_2=r_1r_2\operatorname{cis}(\theta_1+\theta_2)$ (multiply moduli, add angles)
 149
 150${z_1 \over z_2} = {r_1 \over r_2} \operatorname{cis}(\theta_1-\theta_2)$ (divide moduli, subtract angles)
 151
 152## de Moivres' Theorem
 153
 154$(r\operatorname{cis}\theta)^n=r^n\operatorname{cis}(n\theta)$ where $n \in \mathbb{Z}$
 155
 156## Roots of complex numbers
 157
 158$n$th roots of $r \operatorname{cis} \theta$ are:  
 159$z={r^{1 \over n}} \cdot (\cos ({{\theta + 2k \pi} \over n}) + i \sin ({{\theta + 2 k \pi} \over n}))$
 160
 161Same modulus for all solutions. Arguments are separated by ${2 \pi} \over n$