The Square Root of -1

What is the square root of $-1$ ? Whatever the quantity $\sqrt{-1}$ means, we want to denote $i = \sqrt{-1}$ , and then discuss calculations involving the set $\mathbb C$ of complex numbers, whose calculations are extended by usual algebraic techniques.

But having developed many ideas in linear algebra, let’s use some of them to properly define the complex numbers. Intuitively, we want to think of $i$ as a $90^{\circ}$ -counterclockwise rotation. Furthermore, we want this transformation to be linear.

Let $\mathbb K$ be a field. Not only will we develop the usual complex numbers, but we can define the complex numbers for basically any field we wish. Recall that $\mathbb K \cong \mathbb K \times \{\mathbf 0\} \subseteq \mathbb K^2$ as a subspace. This will be our starting point, and we will identify

$1 = \mathbf I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix},\quad c = c\cdot 1.$

We want the basis vector $\mathbf e_1$ to be transformed to $\mathbf e_2$ , and the basis vector $\mathbf e_2$ to be transformed to $-\mathbf e_1$ . Hence, our matrix that represents the rotation $i$ will be as follows:

Definition 1. $i := \begin{bmatrix}\mathbf e_2 & -\mathbf e_1\end{bmatrix} = \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}$ .

The complex numbers over $\mathbb K$ is then defined by

$\mathbb C_{\mathbb K} := \mathrm{span}\{1, i\} = \left\{\begin{bmatrix}a & -b \\ b & a\end{bmatrix} : a \in \mathbb K, b \in \mathbb K\right\} \subseteq \mathcal M_{2 \times 2}(\mathbb K).$

We abbreviate $\mathbb C = \mathbb C_{\mathbb R}$ for the usual complex numbers. Under this definition, rather remarkably, we recover the usual definitions and intuitions for the square root of $-1$ :

Lemma 1. $i^2 = -1$ .

Proof. By the definition of $i$ ,

$\begin{aligned}i^2 = \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}^2 &= \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}\begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix} \\ &= \begin{bmatrix}1\begin{bmatrix}-1 \\ 0 \end{bmatrix} & -1 \begin{bmatrix}0 \\ 1\end{bmatrix}\end{bmatrix} \\ &= \begin{bmatrix}-1 & 0 \\ 0 & -1\end{bmatrix} = -1 \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} \\ &= -1 \cdot 1 = -1.\end{aligned}$

Therefore, if there are no other elements $z \neq \pm i$ such that $z^2 = -1$ , we define $i := \sqrt{-1}$ by convention.

Interestingly, by this definition, $\sqrt{-1}$ can be defined from just the rational numbers $\mathbb Q$ , since $\mathbb Q$ is an ordered field $\mathbb K$ (in which for any $x \in \mathbb K$ , $x^2 \geq 0 > -1$ so that $x^2 \neq -1$ ).

If we limit ourselves a little more, as long as $\mathbb K$ is an ordered ring, we can define $\sqrt{-1}$ just as well. This is in fact the definition of the Gaussian integers, which we spell out more explicitly since $\mathbb Z$ is not a field:

Definition 2. The set of Gaussian integers is defined by

$\mathbb Z[i] := \mathbb Z + \mathbb Zi \equiv \{a + bi : a \in \mathbb Z, b \in \mathbb Z\}.$

By definition, $\mathbb C_{\mathbb K}$ forms a vector space over $\mathbb K$ , since it is a subspace of $\mathcal M_{2 \times 2}(\mathbb K)$ .

Lemma 2. $\mathbb C_{\mathbb K} \cong \mathbb K^2$ and $\dim(\mathbb C_{\mathbb K}) = 2$ .

This result means that addition in $\mathbb C_{\mathbb K}$ works just like addition in $\mathbb K^2$ . What about multiplication?

Lemma 3. Multiplication in $\mathbb C_{\mathbb K}\backslash \{\mathbf 0\}$ is closed, associative and commutative. Furthermore, $1$ is a multiplicative identity.

Proof. We can easily verify that $1 = 1 + 0i$ as the multiplicative identity. For the closure of multiplication, we observe that for $a+bi, c+di \in \mathbb C_{\mathbb K}$ , matrix multiplication yields

$\begin{aligned} (a+bi)(c+di) &= ac + (ad)i + (bc)i + (bd)i^2 \\ &= ac + (ad + bc)i + (bd)(-1) \\ &= (ac - bd) + (ad+bc)i \in \mathbb C_{\mathbb K}.\end{aligned}$

This computation also establishes that complex number multiplication is commutative. Associativity follows from the associativity of matrix multiplication.

In fact, we get something even more than a vector space: If $\mathbb K$ is an ordered field like $\mathbb Q$ or $\mathbb R$ , then $\mathbb C_{\mathbb K}$ forms a field, so that $\mathbb K \subseteq \mathbb C_{\mathbb K}$ as a sub-field.

Lemma 4. If $\mathbb K$ is an ordered field, then $\mathbb C_{\mathbb K}\backslash \{\mathbf 0\}$ forms an abelian group under multiplication. Furthermore, multiplication is distributive over addition.

Proof. By Lemma 3, we just need to check that every nonzero complex number $a+bi$ has an inverse. To that end, we need to find a complex number $c + di$ such that

$\begin{aligned} (ac - bd) + (ad+bc)i = (a+bi)(c+di) &= 0. \end{aligned}$

We leave it as an exercise to verify that $c = a/(a^2+b^2)$ and $d = -b/(a^2+b^2)$ yield the desired complex number, which is well-defined since $a^2 + b^2 > 0$ in an ordered field, so that

$\displaystyle (a+bi)^{-1} = \frac{a}{a^2+b^2} - \frac{b}{a^2+b^2}i.$

Distributivity of multiplication over addition in the complex numbers follows from the linearity property

$S \circ (T_1 + T_2) = S \circ T_1 + S \circ T_2.$

Finally, once we have completed our verifications, we have the canonical chain of number systems:

Theorem 1. We have the canonical chain of number systems:

$\mathbb N^+ \subsetneq \mathbb N_0 \subsetneq \mathbb Z \subsetneq \mathbb Q \subsetneq \mathbb R \subsetneq \mathbb C,$

with arithmetic being preserved at each step, except for order.

Interestingly, however, we cannot order $\mathbb C$ like how we usually order $\mathbb R$ .

Theorem 2. Suppose there exists a total order $\leq$ on $\mathbb C$ . Then $\leq$ is not the usual total order on $\mathbb R$ .

Proof. Suppose that $\leq$ is the usual total order on $\mathbb R$ . Since $\leq$ is a total order, either $1 \leq i$ or $i \leq 1$ . Suppose $1 \leq i$ for vibes. By the transitivity of $\leq$ , $0 \leq i$ . Hence, multiplying by $i$ yields

$i = i \cdot 1 \leq i \cdot i = -1 \leq 1.$

Since $\leq$ is antisymmetric, we must have $1 = i$ , a contradiction.

We’ll conclude by mentioning some other strange number systems. The dual numbers are defined by $\mathrm{span}\{1, \epsilon\}$ , where $\epsilon \neq 0$ but $\epsilon^2 = 0$ . The split-complex numbers are defined by $\mathrm{span}\{1, j\}$ , where $j \neq 1$ but $j^2 = 1$ . These definitions may seem absurd until we define

$\epsilon := \begin{bmatrix} \mathbf 0 & \mathbf e_1 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix},\quad j := -\begin{bmatrix} \mathbf e_2 & \mathbf e_1 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix}.$

The desired properties are then obtained by an exercise in matrix multiplication.

—Joel Kindiak, 2 Mar 25, 2308H

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The Square Root of -1

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