Division of Complex Numbers

Recall from the Addition and Multiplication of Complex Numbers page that if $z = a + bi, w = c + di \in \mathbb{C}$ then the sum $z + w$ by addition is defined as:

(1)

\begin{align} \quad z + w = (a + c) + (b + d)i \end{align}

Furthermore, the product $z \cdot w$ (or simply $zw$) by multiplication is defined as:

(2)

\begin{align} \quad zw = (ac - bd) + (ad + bc)i \end{align}

We will now look at the operation of division of complex numbers. The process of dividing two complex numbers is a tad more technical. For the set of real numbers, if $y \in \mathbb{R}$ and $y \neq 0$ then $\displaystyle{y^{-1} = \frac{1}{y}}$ is a well-defined and unique real number satisfying $\displaystyle{xx^{-1} = 1}$ and is called the multiplicative inverse of $y$ (or reciprocal of $y$). Then if $x \in \mathbb{R}$, the quotient $\displaystyle{\frac{x}{y} = xy^{-1}}$ is well defined and the operation is called division.

We wish to establish an analogous operation for dividing complex numbers.

Let $w = c + di \in \mathbb{C}$ and assume that $w \neq 0$. Then this means that $c$ and $d$ are not both zero. Using a trick from elementary algebra and we see that:

(3)

\begin{align} \quad \frac{1}{w} = \frac{1}{c + di} = \frac{1}{c + di} \cdot \frac{c - di}{c - di} = \frac{c - di}{c^2 - cdi + cdi - d^2i^2} = \frac{c - di}{c^2 - d^2i^2} = \frac{c - di}{c^2 + d^2} = \frac{c}{c^2 + d^2} - \frac{d}{c^2 + d^2} i \end{align}

This is a well defined complex number since the denominator $c^2 + d^2 \neq 0$. Similarly, if $z = a + bi \in \mathbb{C}$ then:

(4)

\begin{align} \quad \frac{a + bi}{c + di} = \frac{a + bi}{c + di} \frac{c - di}{c - di} = \frac{(a + bi)(c - di)}{c^2 + d^2} = \frac{ac - adi + bci - bdi^2}{c^2 + d^2} = \frac{ac + bd}{c^2 + d^2} - \frac{ad - bc}{c^2 + d^2} i \end{align}

Definition: If $z = a + bi, w = c + di \in \mathbb{C}$ and let $w \neq 0$ then the operation of Division denoted $\div$ between $z$ and $w$ yields the Quotient defined to be $\displaystyle{z \div w = \frac{z}{w} = \frac{ac + bd}{c^2 + d^2} - \frac{ad - bc}{c^2 + d^2}} i$.

For example:

(5)

\begin{align} \quad \frac{2 + 3i}{3 + 2i} = \frac{2 + 3i}{3 + 2i} \frac{3 - 2i}{3 - 2i} = \frac{(2 + 3i)(3 - 2i)}{9 + 4} = \frac{6 - 4i + 9i - 6i^2}{13} = \frac{12 + 5i}{13} = \frac{12}{13} + \frac{5}{13} i \end{align}