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
113#### Dividing complex polynomials
114
115Dividing $P(z)$ by $D(z)$ gives quotient $Q(z)$ and remainder $R(z)$ such that:
116
117$$P(z) = D(z)Q(z) + R(z)$$
118
119#### Remainder theorem
120
121Let $\alpha \in \mathbb{C}$. Remainder of $P(z) \div (z - \alpha)$ is $P(\alpha)$
122
123## Conjugate root theorem
124
125Let $P(z)$ be a polynomial with real coefficients. If $a+bi$ is a solution to $P(z)=0$, with $a, b \in \mathbb{R}$, the the conjugate $a-bi$ is also a solution.
126
127## Polar form
128
129$$\begin{equation}\begin{split}z & =r \operatorname{cis} \theta \\ & = r(\operatorname{cos}\theta+i \operatorname{sin}\theta) \\ & = a + bi \end{split}\end{equation}$$
130
131- $r=|z|$, given by Pythagoras ($r=\sqrt{\operatorname{Re}(z)^2 + \operatorname{Im}(z)^2}$)
132- $\theta=\operatorname{arg}(z)$ (on CAS: `arg(a+bi)`)
133- **principal argument** is $\operatorname{Arg}(z) \in (-\pi, \pi]$ (note capital $\operatorname{Arg}$)
134
135Note each complex number has multiple polar representations:
136$z=r \operatorname{cis} \theta = r \operatorname{cis} (\theta+2 n\pi$) where $n$ is integer number of revolutions
137
138### Conjugate in polar form
139
140$$(r \operatorname{cis} \theta)^{-1} = r\operatorname{cis} (- \theta)$$
141
142Reflection of $z$ across horizontal axis.
143
144### Multiplication and division in polar form
145
146$z_1z_2=r_1r_2\operatorname{cis}(\theta_1+\theta_2)$ (multiply moduli, add angles)
147
148${z_1 \over z_2} = {r_1 \over r_2} \operatorname{cis}(\theta_1-\theta_2)$ (divide moduli, subtract angles)
149
150## de Moivres' Theorem
151
152$(r\operatorname{cis}\theta)^n=r^n\operatorname{cis}(n\theta)$ where $n \in \mathbb{Z}$