## Addition of complex numbers

Given two complex numbers, $Z = a + bi$ and $W = u + vi$,

Their sum, $Z + W$ is calculated component wise: $$ Z + W = (a+ bi) + (u+vi) = (a+u) + (b+v)i $$

Their difference $Z-W$ is calculated component wise $$ Z - W = (a + bi) - (u + vi) = (a-u) + (b-v)i$$

These results extend all the way up to linearity, let $\lambda, \mu \in \mathbb{R}$, then

$$\lambda Z + \mu W = (\lambda a + \mu b) + ( \lambda u + \mu v)i$$

## multiplication of complex numbers

Given two complex numbers, $Z = a + bi$ and $W = u + vi$, then $$ \begin{align} ZW &= (a + bi)(u + vi) \\ &= au + ubi + avi + bvi^2 \\ &= au + ubi + avi + bv(-1) \\ &= au - bv + (ub + av)i \end{align} $$

*Note the minus sign that appears in the real part of the complex number.*

## division of complex numbers

Given two complex numbers, $Z = a + bi$ and $W = u + vi$, $\frac{Z}{W}$ is calculated as follows $$ \begin{align} \frac{Z}{W} &= \frac{a+bi}{u + vi} \\ &= \frac{(a+bi)(u-vi)}{(u+vi)(u-vi)} \\ &= \frac{(a+bi)(u-vi)}{u^2 + v^2} \\ &= \frac{au + vb - (bu + av)i}{u^2 + v^2} \end{align} $$

In terms of the complex conjugate, and modulus, we can express this operation somewhat more elegantly: $$ \frac{Z}{W} = \frac{Z\overline{W}}{W\overline{W}} = \frac{Z\overline{W}}{|W|^2} $$

## Integer powers of complex numbers

Powers of complex numbers can be calculated using repeated multiplication and by extension the binomial expansion, but in general this method is inefficient for high powers and more powerful techniques are used in practice - see the notes on modulus and argument. $$ \begin{align} Z^n &= (a + bi)^n \\ &= a^n + ^nC_1a^{n-1}bi + ^nC_2a^{n-2}(bi)^2 + \dots + (ib)^n \\ &= a^n + ^nC_1a^{n-1}bi - ^nC_2a^{n-2}b^2 + \dots + (i)^n(b)^n \end{align}$$ (As we've not specified the value of $n$, its a little difficult to evaluate $i^n$ as it can have four different values)

In the case of $n\lt 0$, we first evaluate $\frac{1}{Z}$ and then apply the binomial expansion, that is

$$ Z^{-n} = \left(\frac{1}{Z}\right)^n = (u + iv)^n = \dots $$