## Imaginary numbers
-$i^2 = -1$
-
-$\therefore i = \sqrt {-1}$
+$i^2 = -1 \quad \therefore i = \sqrt {-1}$
### Simplifying negative surds
-$\sqrt{-2} = \sqrt{-1 \times 2}$
-
- $= \sqrt{2}i$
+$\sqrt{-2} = \sqrt{-1 \times 2}$
+$= \sqrt{2}i$
## Complex numbers
$\mathbb{C} = \{a+bi : a, b \in \mathbb{R} \}$
-General form: $z=a+bi$
-- $\operatorname{Re}(z) = a$
-- $\operatorname{Im}(z) = b$
+General form: $z=a+bi$
+$\operatorname{Re}(z) = a, \quad \operatorname{Im}(z) = b$
### Addition
-If $z_1 = a+bi$ and $z_2=c+di$, then
-
- $z_1+z_2 = (a+c)+(b+d)i$
+If $z_1 = a+bi$ and $z_2=c+di$, then
+$z_1+z_2 = (a+c)+(b+d)i$
### Subtraction
-If $z_1=a+bi$ and $z_2=c+di$, then
-
- $z_1−z_2=(a−c)+(b−d)i$
+If $z_1=a+bi$ and $z_2=c+di$, then $z_1−z_2=(a−c)+(b−d)i$
### Multiplication by a real constant
-If $z=a+bi$ and $k \in \mathbb{R}$, then
-
- $kz=ka+kbi$
+If $z=a+bi$ and $k \in \mathbb{R}$, then $kz=ka+kbi$
### Powers of $i$
$i^0=1$
$\dots$
Therefore..
+
- $i^{4n} = 1$
- $i^{4n+1} = i$
- $i^{4n+2} = -1$
- $i^{4n+3} = -i$
-Divide by 4 and take remainder
+Divide by 4 and take remainder.
### Multiplying complex expressions
-If $z_1 = a+bi$ and $z_2=c+di$, then
- $z_1 \times z_2 = (ac-bd)+(ad+bc)i$
+If $z_1 = a+bi$ and $z_2=c+di$, then
+$z_1 \times z_2 = (ac-bd)+(ad+bc)i$
### Conjugates
If $z=a+bi$, conjugate of $z$ is $\overline{z} = a-bi$ (flipped operator)
-Also, $z \overline{z} = (a+bi)(a-bi) = a^2+b^2$
+Also, $z \overline{z} = (a+bi)(a-bi) = a^2+b^2 = |z|^2$
- Multiplication and addition are associative
+#### Properties
+
+- $\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}$
+- $\overline{z_1 z_2} = \overline{z_1} \cdot \overline{z_2}$
+- $\overline{kz} = k \overline{z}, \text{ for } k \in \mathbb{R}$
+- $z \overline{z} = |z|^2$
+- $z + \overline{z} = 2 \operatorname{Re}(z)$
+
+
### Modulus
Distance from origin.
$\therefore z \overline{z} = |z|^2$
+#### Properties
+
+- $|z_1 z_2| = |z_1| |z_2|$
+- $|{z_1 \over z_2}| = {|z_1| \over |z_2|}$
+- $|z_1 + z_2| \le |z_1 + |z_2|$
+
### Multiplicative inverse
$z^{-1} = {1 \over z} = {{a-bi} \over {a^2+B^2}} = {\overline{z} \over {|z|^2}}$
To solve $z^2+a^2=0$ (sum of two squares):
-$z^2+a^2=z^2-(ai)^2$
- $=(z+ai)(z-ai)$
+$z^2+a^2=z^2-(ai)^2=(z+ai)(z-ai)$
## Polar form