Skip to main content
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

# 1.14: Representing Complex Multiplication as Matrix Multiplication

$$\newcommand{\vecs}{\overset { \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

Consider two complex number $$z_1 = a + bi$$ and $$z_2 = c + di$$ and their product

$z_1 z_2 = (a + bi) (c + id) = (ac - bd) + i(bc + ad) =: \omega \label{eq3}$

Now let's define two matrices

$Z_1 = \begin{bmatrix}a & -b \\ b & a \end{bmatrix}$

$Z_2 = \begin{bmatrix} c & -d \\ d & c \end{bmatrix}$

Note that these matrices store the same information as $$z_1$$ and $$z_2$$, respectively. Let’s compute their matrix product

$Z_1 Z_2 = \begin{bmatrix} a & -b \\ b & a \end{bmatrix} \begin{bmatrix} c & -d \\ d & c \end{bmatrix} = \begin{bmatrix} ac - bd & -(bc + ad) \\ bc + ad & ac - bd \end{bmatrix} := W.$

Comparing $$W$$ just above with $$w$$ in Equation \ref{eq3}, we see that $$W$$ is indeed the matrix corresponding to the complex number $$w = z_1 z_2$$. Thus, we can represent any complex number $$z$$ equivalently by the matrix

$Z = \begin{bmatrix} \text{Re} z & -\text{Im} z \\ \text{Im} z & \text{Re} z \end{bmatrix}$

and complex multiplication then simply becomes matrix multiplication. Further note that we can write

$Z = \text{Re} z \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \text{Im} z \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix},$

i.e., the imaginary unit $$i$$ corresponds to the matrix $$\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$$ and $$i^2 = -1$$ becomes

$\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = -\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.$

## Polar Form (Decomposition)

Writing $$z = re^{i \theta} = r(\cos \theta + i \sin \theta)$$, we find

\begin{align*} Z &= r \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \\[4pt] &= \begin{bmatrix} r & 0 \\ 0 & r \end{bmatrix} \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \end{align*}

corresponding to a stretch factor $$r$$ multiplied by a 2D rotation matrix. In particular, multiplication by $$i$$ corresponds to the rotation with angle $$\theta = \pi /2$$ and $$r = 1$$.

We will not make a lot of use of the matrix representation of complex numbers, but later it will help us remember certain formulas and facts.

• Was this article helpful?