The complex numbers are more than just a vector space; they are also an algebra, that is you can multiply them together. How do you compute the product of complex numbers? Simply multiply it out, that is \begin{align} (a+bi) (c+di) &= (a+bi) c + (a+bi) di \nonumber\\ &= (ac-bd) + (bc+ad)i \end{align}

What properties of the complex numbers have we used? First of all, we have distributed multiplication over addition. Second, $i$ is “the” square root of $-1$, that is \begin{equation} i^2 = -1 \end{equation} Third, we have used associativity, that is \begin{equation} (xy)z = x(yz) \end{equation} for any complex numbers $x$, $y$, $z$. Finally, we have used commutativity, i.e. \begin{equation} xy = yx \end{equation} to replace $bic$ with $bci$. 1)

We define the complex conjugate $\bar{z}$ of a complex number $z=a+bi$ by \begin{equation} \bar{z} = a - bi \end{equation} thus changing the sign of the imaginary part of $z$. Equivalently, complex conjugation is the (real) linear map which takes $1$ to $1$ and $i$ to $-i$. The norm $|z|$ of a complex number $z$ is defined by \begin{equation} |z|^2 = z\bar{z} = a^2 + b^2 \end{equation} The only complex number with norm $0$ is $0$. Furthermore, any nonzero complex number has a unique inverse, namely \begin{equation} z^{-1} = {\bar{z} \over  |z|^2} \end{equation} Since complex numbers are invertible, linear equations such as \begin{equation} c = a z + b \label{linear} \end{equation} can always be solved for $z$, so long as $a\ne0$.

The norms of complex numbers satisfy the following identity: \begin{equation} |yz| = |y| |z| \end{equation} Squaring both sides and expanding the result in terms of components yields \begin{equation} (ac-bd)^2 + (bc+ad)^2 = (a^2+b^2) (c^2+d^2) \end{equation} (where, say, $z=a+bi$ and $y=c+di$), which is called the 2-squares rule.