Inverse of Complex Numbers

Multiplying a Complex Number by Its Reciprocal

Every nonzero complex number $z = x + iy$ has a multiplicative inverse, written as $z^{-1}$ or $1/z$ .

Multiplying the complex number $z$ by its inverse $z^{-1}$ gives $1$ , the multiplicative identity.

z \times z^{-1} = 1

Finding the Inverse Formula

We already know from the material on properties of complex number multiplication that for $z = x + iy$ , its inverse is:

z^{-1} = \frac{x}{x^2+y^2} - i\frac{y}{x^2+y^2}

This formula can also be written as an ordered pair:

z^{-1} = \left( \frac{x}{x^2+y^2}, -\frac{y}{x^2+y^2} \right)

Remember also the other often useful form, using the conjugate ( $\bar{z} = x-iy$ ) and the modulus squared ( $|z|^2 = x^2+y^2$ ):

z^{-1} = \frac{\bar{z}}{|z|^2}

Finding the Inverse of a Complex Number

Find the inverse of $z = 1 - i$ .

Solution:

Here, $x=1$ and $y=-1$ .

Using the first formula:

\begin{aligned} x^2+y^2 &= (1)^2 + (-1)^2 \\ &= 1 + 1 \\ &= 2 \end{aligned}

Using the conjugate and modulus formula:

\bar{z} = 1 - (-1)i = 1+i

Both formulas give the same reciprocal:

z^{-1} = \frac{1}{2} + \frac{1}{2}i \text{ or} \left( \frac{1}{2}, \frac{1}{2} \right)

Verify $z^{-1}$ by multiplying it by the original number:

\begin{aligned} &(1-i)\left(\frac{1}{2}+\frac{1}{2}i\right) \\ &=\frac{1}{2}+\frac{1}{2}+\left(\frac{1}{2}-\frac{1}{2}\right)i \\ &=1 \end{aligned}

Visualization of

z

and

z^{-1}

The number

z = 1-i

and its inverse

z^{-1} = \frac{1}{2} + \frac{1}{2}i

lie on opposite sides of the real axis. The moduli of the two numbers are reciprocals.

Exercise

Given the complex numbers $z_1 = 1-i$ and $z_2 = 2+3i$ . Find the inverse of $z_1 + z_2$ . The result must satisfy $(z_1 + z_2) \cdot w = 1$ , which is the check you can run on your answer.

Worked Solutions

Step 1: Find $z_1 + z_2$ .

\begin{aligned} z &= z_1 + z_2 \\ &= (1-i) + (2+3i) \\ &= (1+2) + (-1+3)i \\ &= 3+2i \end{aligned}

Step 2: Find the inverse of $z = 3+2i$ . Here $x=3$ and $y=2$ . We use the formula $z^{-1} = \frac{\bar{z}}{|z|^2}$ .

\bar{z} = 3-2i

So, the inverse of $z_1 + z_2$ is $\frac{3}{13} - \frac{2}{13}i$ .