Imaginary Numbers

Define the 2 × 2 matrices $\mathbf I$ and $\mathbf J$ as follows:

$\mathbf I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \quad \mathbf J = \begin{bmatrix} 0 & -1 \\ 1 & 0\end{bmatrix}.$

Problem 1. Evaluate $\mathbf J^2$ .

(Click for Solution)

Solution. Using matrix multiplication,

$\begin{aligned} \mathbf J^2 &= \begin{bmatrix} 0 & -1 \\ 1 & 0\end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0\end{bmatrix} \\ &= \begin{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0\end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} & \begin{bmatrix} 0 & -1 \\ 1 & 0\end{bmatrix} \begin{bmatrix} -1 \\ 0 \end{bmatrix} \end{bmatrix} \\ &= \begin{bmatrix} 0\begin{bmatrix} 0 \\ 1\end{bmatrix} + 1\begin{bmatrix} -1 \\ 0\end{bmatrix} & (-1)\begin{bmatrix} 0 \\ 1\end{bmatrix} + 0\begin{bmatrix} -1 \\ 0\end{bmatrix} \end{bmatrix} \\ &= \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = (-1) \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} \\ & = (-1)\mathbf I = -\mathbf I. \end{aligned}$

Let $a,b,c,d$ be real numbers.

Problem 2. Show that

$a \mathbf I + b \mathbf J = c \mathbf I + d \mathbf J$

implies that $a = c$ and $b = d$ .

(Click for Solution)

Solution. Using the definition of $\mathbf I$ and $\mathbf J$ ,

$\begin{aligned} a\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} + b \begin{bmatrix}0 & -1 \\ 1 & 0 \end{bmatrix} &= c \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} + d \begin{bmatrix}0 & -1 \\ 1 & 0 \end{bmatrix} \\ \begin{bmatrix}a & 0 \\ 0 & a \end{bmatrix} + \begin{bmatrix}0 & -b \\ b & 0 \end{bmatrix} &= \begin{bmatrix}c & 0 \\ 0 & c \end{bmatrix} + \begin{bmatrix}0 & -d \\ d & 0 \end{bmatrix} \\ \begin{bmatrix}a & -b \\ b & a \end{bmatrix} &= \begin{bmatrix}c & -d \\ d & c \end{bmatrix}.\end{aligned}$

Comparing the entries of the matrices on both sides yields $a = c$ and $b = d$ .

Define $\mathbf Z := a \mathbf I + b \mathbf J$ and $\mathbf W := c \mathbf I + d \mathbf J$ .

Problem 3. Show that $\mathbf Z \mathbf W = \mathbf W \mathbf Z$ .

(Click for Solution)

Solution. We observe that $\mathbf I \mathbf J = \mathbf J = \mathbf J \mathbf I$ . By Problem 1,

$\begin{aligned}\mathbf Z \mathbf W &= (a \mathbf I + b \mathbf J)(c \mathbf I + d \mathbf J) \\ &= (ac) \mathbf I^2 + (ad) \mathbf I \mathbf J + (bc) \mathbf J \mathbf I + (bd) \mathbf J^2 \\ &= (ac) \mathbf I +(ad)\mathbf J + (bc)\mathbf J + (bd)(-1)\mathbf I \\ &= (ac - bd) \mathbf I + (ad + bc) \mathbf J.\end{aligned}$

Similarly, $\mathbf W \mathbf Z = (ca - db) \mathbf I + (da + cb) \mathbf J$ . Therefore, $\mathbf Z \mathbf W = \mathbf W \mathbf Z$ .

Define $\mathbf Z^* := a \mathbf I - b \mathbf J$ .

Problem 4. Evaluate $\mathbf Z \mathbf Z^*$ . Hence, if $\mathbf Z \neq \mathbf 0$ , construct a matrix $\mathbf Z^{-1}$ with the property that

$\mathbf Z \mathbf Z^{-1} = \mathbf Z^{-1} \mathbf Z = \mathbf I.$

(Click for Solution)

Solution. Using the multiplication in Problem 3 but setting $c = a$ and $d = -b$ ,

$\mathbf Z \mathbf Z^* = (a^2 - b(-b))\mathbf I + (a(-b) + ba)\mathbf J = (a^2 + b^2) \mathbf I.$

If $\mathbf Z \neq \mathbf 0$ , then either $a \neq 0$ or $b \neq 0$ , so that $a^2 + b^2 \neq 0$ . Therefore,

$\begin{aligned} \mathbf Z\left( \frac 1{a^2 + b^2} \mathbf Z^* \right) &= \frac 1{a^2 + b^2} \mathbf Z \mathbf Z^* \\ &= \frac 1{a^2 + b^2} (a^2 + b^2) \mathbf I \\ &= \mathbf I. \end{aligned}$

Denoting $\alpha := (\alpha^2 + \beta^2)^{-1}$ , define $\mathbf Z^{-1} := \alpha \mathbf Z^*$ , so that

$\mathbf Z \mathbf Z^{-1} = \mathbf Z (\alpha \mathbf Z^*) = \mathbf I.$

By Problem 3, $\mathbf Z^{-1} \mathbf Z = \mathbf Z \mathbf Z^{-1} = \mathbf I$ .

Problem 5. Determine the two possible matrices $\mathbf Z = a\mathbf I + b \mathbf J$ such that

$\mathbf Z^2 - 4\mathbf Z + 13\mathbf I = \mathbf 0.$

(Click for Solution)

Solution. Write $\mathbf Z = a\mathbf I + b \mathbf J$ . Using the multiplication in Problem 3 with $c = a$ and $d = b$ ,

$\mathbf Z^2 = (a^2 - b^2) \mathbf I + (2ab) \mathbf J.$

Therefore,

$((a^2 - b^2) \mathbf I + (2ab) \mathbf J) - 4(a\mathbf I + b\mathbf J) + 13\mathbf I = \mathbf 0.$

Grouping the terms together,

$(a^2 - 4a - b^2 + 13) \mathbf I + 2b (a - 2) \mathbf J = \mathbf 0 = 0 \mathbf I + 0\mathbf J.$

Using Problem 2,

$a^2 - 4a - b^2 + 13 = 0\quad \mathbf{and} \quad 2b (a - 2) = 0.$

Using the second equation, either $b = 0$ or $a = 2$ . If $b = 0$ , then substituting into the first equation,

$a^2 - 4a+ 13 = 0.$

However, this equation has discriminant $(-4)^2 - 4 \cdot 1 \cdot 13 = - 36 < 0$ , and so there are no real roots to the equation, a contradiction.

Therefore, we must have $a = 2$ . Substituting into the first equation again,

$2^2 - 4 \cdot 2 - b^2 + 13 = 0 \quad \Rightarrow \quad b^2 = 9.$

Therefore, $b = \pm 3$ . Hence, the two possible matrices for $\mathbf Z$ are

$\mathbf Z = 2\mathbf I +3\mathbf J \quad \text{or} \quad \mathbf Z = 2\mathbf I - 3 \mathbf J.$

We can condense them to the expression $\mathbf Z = 2\mathbf I \pm 3 \mathbf J$ .

Remark 1. By denoting $a \mathbf I \equiv a$ and $b\mathbf J \equiv bi$ , we have created a model for the complex numbers $a + bi$ , where $i^2 = -1$ by Problem 1. In particular, numbers of the form $bi$ are called purely imaginary. The solution to Problem 5 would then look like

$z^2 - 2z + 5 = 0 \quad \iff \quad z = 2 \pm 3i.$

The letter ‘z‘ is used to denote a complex number by convention. Furthermore, the calculation $i^2 = -1$ motivates the (somewhat debatable) notation $\sqrt{-1} := i$ . For more information, see this post.

—Joel Kindiak, 25 Mar 26, 0056H

KindiakMath

Imaginary Numbers

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Imaginary Numbers

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