Division of Complex Numbers Examples 1

Division of Complex Numbers Examples 1

Recall from the Division of Complex Numbers page that if $z = a + bi, w = c + di \in \mathbb{C}$ then the quotient $\displaystyle{z \div w = \frac{z}{w}}$ is defined as:

(1)
\begin{align} \quad \frac{z}{w} = \frac{ac + bd}{c^2 + d^2} - \frac{ad - bc}{c^2 + d^2}i \end{align}

We will now look at some example problems regarding the division of complex numbers.

Example 1

Let $z = 2 + i$ and $w = -3 - 2i$. Find $\displaystyle{\frac{z}{w}}$.

We have that:

(2)
\begin{align} \frac{z}{w} &= \frac{(2)(-3) + (1)(-2)}{(-3)^2 + (-2)^2} - \frac{(2)(-2) - (1)(-3)}{(-3)^2 + (-2)^2} i \\ &= \frac{-6 - 2}{13} - \frac{-4 - (-3)}{13}i \\ &= \frac{-8}{13} + \frac{1}{13}i \end{align}

Example 2

Simplify $\displaystyle{\frac{1 + i}{1 - i}}$ and $\displaystyle{\frac{1 - i}{1 + i}}$.

We have that:

(3)
\begin{align} \quad \frac{1 + i}{1 - i} = \frac{1 + i}{1 - i} \cdot \frac{1 + i}{1 + i} = \frac{(1 + i)(1 + i)}{2} = \frac{1 +2i + i^2}{2} = i \end{align}

And also:

(4)
\begin{align} \quad \frac{1 - i}{1 + i} = \frac{1 - i}{1 + i} \cdot \frac{1 - i}{1 - i} = \frac{(1 - i)(1 - i)}{2} = \frac{1 -2i + i^2}{2} = -i \end{align}

Example 3

Prove or disprove the following statement: If $z = a + bi \in \mathbb{C}$, $z \neq 0$, then $\displaystyle{\frac{a + bi}{a - bi} = - \frac{a - bi}{a + bi}}$.

This statement is false in general. Take $z = a + bi \in \mathbb{R}$ with $z \neq 0$. Then $z = a + 0i$ for some nonzero $a \in \mathbb{R}$, and $z = a - 0i$ as well. So:

(5)
\begin{align} \quad 1 = \frac{a}{a} = \frac{a + 0i}{a - 0i} \neq - \frac{a - 0i}{a + 0i} = -\frac{a}{a} = -1 \end{align}