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How Can You Simplify Complex Fractions

Here you will learn how to divide complex numbers both in Cartesian form and in polar form. Among other things, division can be used to find the inverses of complex numbers.

Cartesian Form

In Cartesian form, division of complex numbers is based on multiplication and conjugation. If you have a fraction with complex numbers in the numerator and the denominator, you need to write the fraction such that the imaginary unit is not in the denominator. This can be done by expanding the fraction using the conjugate of the denominator, in order to get a denominator consisting of only real numbers.

Formula

Division in Cartesian Form

Let $z_{1} = a + b i$ and $z_{2} = c + d i \neq 0$ be complex numbers, then

\begin{array}{l} \frac{z_{1}}{z_{2}} & = \frac{a + b i}{c + d i}, \\ = \frac{(a + b i) (c - d i)}{(c + d i) (c - d i)}, \\ = \frac{(a c + b d) + (b c - a d) i}{c^{2} + d^{2}}, \\ = \frac{a c + b d}{c^{2} + d^{2}} + \frac{b c - a d}{c^{2} + d^{2}} i . \end{array}

As with addition, subtraction, and multiplication, the set of complex numbers $ℂ$ is closed under division. This means that you get a new complex number when you divide two complex numbers. As with real numbers, it is not possible to divide by $0$ —so you must always be sure to not have $0$ in the denominator of a fraction at any point.

Example 1

Calculate $\frac{z_{1}}{z_{2}}$ with the complex numbers $z_{1} = 2 + i$ and $z_{2} = 3 - i$

In order to calculate $\frac{z_{1}}{z_{2}}$ , you find the conjugate of $z_{2}$ first:

\overset{}{z_{2}} = 3 + i .

Now you can simplify the fraction by expanding, multiplying by $\overset{}{z_{2}}$ in the numerator and denominator:

\begin{array}{l} \frac{z_{1}}{z_{2}} & = \frac{z_{1} \cdot \overset{}{z_{2}}}{z_{2} \cdot \overset{}{z_{2}}}, \\ = \frac{(2 + i) (3 + i)}{(3 - i) (3 + i)}, \\ = \frac{6 + 2 i + 3 i + i^{2}}{9 + 3 i - 3 i - i^{2}}, \\ = \frac{5 + 5 i}{10}, \\ = \frac{1}{2} + \frac{1}{2} i . \end{array}

Polar Form

If you write complex numbers with the complex exponential function, you can divide complex numbers by using normal power rules.

Formula

Division in Polar Form

Let $z_{1} = r_{1} e^{i 𝜃_{1}}$ and $z_{2} = r_{2} e^{i 𝜃_{2}}$ be complex numbers, then

\begin{array}{l} \frac{z_{1}}{z_{2}} & = \frac{r_{1} e^{i 𝜃_{1}}}{r_{2} e^{i 𝜃_{2}}} \\ = \frac{r_{1}}{r_{2}} e^{i (𝜃_{1} - 𝜃_{2})} . \end{array}

In polar form you divide complex numbers by dividing the norms and subtracting the arguments. This can be visualized in the complex plane:

Geometric visualization of division of complex numbers.

As with multiplication, division can be thought of as a combination of a scaling and a rotation in the complex plane. Unlike multiplication, the rotation is clockwise. This is because division with the number $z$ is the same as the multiplication with the inverse $z^{- 1}$ .

Inverse of Complex Numbers

For all complex numbers

z \neq 0

, there exists an inverse complex number denoted by

z^{- 1}

. The inverse number

z^{- 1}

has the property

z \cdot z^{- 1} = 1

. You find

z^{- 1}

by simplifying the fraction

\frac{1}{z}

Theory

Inverse of Complex Numbers

For every complex number

z \neq 0

, there exists an inverse

z^{- 1}

\begin{array}{l} z^{- 1} & = \frac{1}{z} \\ = \frac{1 \cdot \overset{}{z}}{z \cdot \overset{}{z}} \\ = \frac{\overset{}{z}}{{| z |}^{2}} . \end{array}

You can compute the inverse of a complex number $z$ by dividing the conjugate of $z$ by the square of the norm of $z$ . If you do not remember this formula, you can always find the inverse by solving $z^{- 1} = \frac{1}{z}$ .

Example 2

Calculate the inverse of the complex number $z = 4 - 3 i$ . Then check if you have found the correct inverse.

To find the inverse of $z$ , you need both the conjugate of $z$ and the norm of $z$ . Here you have the conjugate $\overset{}{z} = 4 + 3 i$ and the norm $| z | = 5$ . The inverse number is therefore:

\begin{array}{l} z^{- 1} & = \frac{\overset{}{z}}{{| z |}^{2}} \\ = \frac{4 + 3 i}{25} \\ = \frac{4}{25} + \frac{3}{25} i . \end{array}

To check if you found the correct inverse, you can see if $z^{- 1}$ fulfills $z \cdot z^{- 1} = 1$ : $\begin{array}{l} z \cdot z^{- 1} & = (4 - 3 i) (\frac{4}{25} + \frac{3}{25} i) \\ = \frac{16}{25} + \frac{12}{25} i - \frac{12}{25} i - \frac{9}{25} i^{2} \\ = \frac{16}{25} + \frac{9}{25} \\ = 1 . \end{array}$

The product of $z$ and $z^{- 1}$ is $1$ . Thus, you have found the correct inverse of $z$ .

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