Inverse of Complex Numbers

What is the Inverse of a Complex Number?

Every non-zero complex number $z = x + iy$ has a "reciprocal" friend called the multiplicative inverse (or just inverse), which we write as $z^{-1}$ or $1/z$ .

The defining characteristic of the multiplicative inverse is that if we multiply the complex number $z$ by its inverse $z^{-1}$ , the result is $1$ (the multiplicative identity element).

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}

Example Inverse Calculation

Let the complex number be $z = 1 - i$ . Find its inverse!

Solution:

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

Using the first formula:

x^2+y^2 = (1)^2 + (-1)^2 = 1 + 1 = 2

Using the conjugate and modulus formula:

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

The result is the same, namely:

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

Visualization of

z

and

z^{-1}

Visualization of

z = 1-i

and its inverse

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

. Notice their positions relative to the origin.

Exercise

Given the complex numbers $z_1 = 1-i$ and $z_2 = 2+3i$ . Find the inverse of $z_1 + z_2$ .

Answer Key

Step $1$ : Find $z_1 + z_2$ .

z = z_1 + z_2 = (1-i) + (2+3i) = (1+2) + (-1+3)i = 3+2i

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 $\frac{3}{13} - \frac{2}{13}i$ .

What is the Inverse of a Complex Number?

Every non-zero complex number $z = x + iy$ has a "reciprocal" friend called the multiplicative inverse (or just inverse), which we write as $z^{-1}$ or $1/z$ .

The defining characteristic of the multiplicative inverse is that if we multiply the complex number $z$ by its inverse $z^{-1}$ , the result is $1$ (the multiplicative identity element).

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}

Example Inverse Calculation

Let the complex number be $z = 1 - i$ . Find its inverse!

Solution:

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

Using the first formula:

x^2+y^2 = (1)^2 + (-1)^2 = 1 + 1 = 2

Using the conjugate and modulus formula:

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

The result is the same, namely:

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

Visualization of

z

and

z^{-1}

Visualization of

z = 1-i

and its inverse

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

. Notice their positions relative to the origin.

Exercise

Given the complex numbers $z_1 = 1-i$ and $z_2 = 2+3i$ . Find the inverse of $z_1 + z_2$ .

Answer Key

Step $1$ : Find $z_1 + z_2$ .

z = z_1 + z_2 = (1-i) + (2+3i) = (1+2) + (-1+3)i = 3+2i

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 $\frac{3}{13} - \frac{2}{13}i$ .