The quotient

Complex numbers: Calculating with complex numbers

The quotient

In our search to find all the known operations of the real numbers also when it comes to complex numbers, we have to consider division as well. For every complex number $w\neq 0$ an inverse exists $\frac{1}{w}$ . Therefore, we can work with fractions of complex numbers.

Quotient

If $z$ and $w$ are complex numbers with $w\ne0$ , then $\frac{z}{w}$ is the complex number defined by the following polar coordinates: $\begin{array}{rclcl} \left|\frac{z}{w}\right| & =&\frac{|z|}{|w|}&\phantom{xx}&\color{blue}{\text{absolute value}} \\ \arg\left (\frac{z}{w}\right) & =&\arg (z) -\arg (w) \, \pmod{2\pi}&\phantom{xx}&\color{blue}{\text{argument}} \end {array}$ This number is called the quotient of $z$ to $w$ .

As the notation as a fraction suggests, this number behaves like the usual quotient:

If $z$ and $w$ are complex numbers with $w\ne0$ , then we have $w\cdot\frac{z}{w} =z$ .

Proof: This follows directly from the laws of multiplication: $\begin{array}{rcl} \left|w\cdot \frac{z}{w}\right| &=&|w|\cdot \left| \frac{z}{w}\right|\\ &=&|w|\cdot\frac{|z|}{|w|}\\ &=&|z|\ ,\\ \arg \left(w\cdot \frac{z}{w}\right) &=&\arg(w)+\arg\left(\frac{z}{w}\right)\\ &=&\arg (w)+\arg (z) -\arg (w) \, \pmod{2\pi}\\ &=&\arg (z) \pmod{2\pi} \end {array}$ This shows that $w\cdot\frac{z}{w}$ and $z$ have the same polar coordinates.

For this quotient, the usual rules of calculation apply; for example:
$\begin{array}{rcl} \frac{z_1}{z_2} \cdot \frac{w_1}{w_2} &=& \frac{z_1\cdot w_1}{z_2 \cdot w_2}\\ \frac{z}{1}&=&z\end {array}$

This can be proven by calculating the absolute value and argument of the left and right hand side and checking they are equal (ie. modulo $2\pi$ for the argument).

We can also find the Cartesian coordinates of $\frac{w}{z}$ . Since $\frac{w}{z}=\frac{w}{1}\cdot\frac{1}{z}={w}\cdot\frac{1}{z}$ and the multiplication of two numbers in standard form is known, having the standard form for $\frac{1}{z}$ is sufficient.

Assume $a$ and $b$ are real numbers that are not both equal to zero. Then
$\frac{1}{a+b\ii}=\frac{a-b\ii}{a^2+b^2}\tiny.$

$\begin{array}{rcl} \frac{1}{a+b\ii}&=&\frac{1}{a+b\,\ii}\\ &=&\frac{a-b\,\ii}{(a+b\,\ii)\cdot(a-b\,\ii)}\\&&\phantom{xyzuvw}\color{blue}{\text{numerator and denominator multiplied by }a-b\,\ii\text{}} \\&=&\frac{a-b\,\ii}{a^2+b^2}\end {array}$

Just as with real numbers, we can now solve linear equations with an unknown that takes on complex values. Some examples are shown below.

Write the complex number $\frac{1}{-4-\ii}$ in standard form.

$\frac{1}{-4-\ii}=-{{4}\over{17}}+{{1}\over{17}}\,{\ii}$

We calculate this as follows: $\begin{array}{rclcl}\frac{1}{-4-\ii} &=& \frac{1}{-4-\ii}\cdot \frac{-4+\mathrm{i}}{-4+\mathrm{i}}& &\color{blue}{\text{multiplied by }1}\\ &=& \frac{-4+\mathrm{i}}{(-4-\ii)\cdot(-4+\mathrm{i})} &&\color{blue}{\text{fractions simplified}}\\ &=& \frac{-4+\mathrm{i}}{(-4)^2-(\mathrm{i})^2}&&\color{blue}{\text{notable product}}\\ &=&\frac{-4+\mathrm{i}}{(-4)^2+1^2}&&\color{blue}{{\ii}^2=-1}\\ &=&\frac{-4+\mathrm{i}}{17}& &\color{blue}{\text{denominator simplified}} \\ &=&-{{4}\over{17}}+{{1}\over{17}}\,{\ii}&& \color{blue}{\text{in standard form}} \end {array}$

New example