### Simplifying negative surds
-$\sqrt{-2} = \sqrt{-1 \times 2}$
-
- $= \sqrt{2}i$
+$\sqrt{-2} = \sqrt{-1 \times 2}$
+$= \sqrt{2}i$
## Complex numbers
### 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
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