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Complex Number

Inverse of Complex Numbers

What is the Inverse of a Complex Number?

Every non-zero complex number z=x+iyz = x + iyz=x+iy has a "reciprocal" friend called the multiplicative inverse (or just inverse), which we write as z−1z^{-1}z−1 or 1/z1/z1/z.

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

z×z−1=1z \times z^{-1} = 1z×z−1=1

Finding the Inverse Formula

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

z−1=xx2+y2−iyx2+y2z^{-1} = \frac{x}{x^2+y^2} - i\frac{y}{x^2+y^2}z−1=x2+y2x​−ix2+y2y​

This formula can also be written as an ordered pair:

z−1=(xx2+y2,−yx2+y2)z^{-1} = \left( \frac{x}{x^2+y^2}, -\frac{y}{x^2+y^2} \right)z−1=(x2+y2x​,−x2+y2y​)

Remember also the other often useful form, using the conjugate (zˉ=x−iy\bar{z} = x-iyzˉ=x−iy) and the modulus squared (∣z∣2=x2+y2|z|^2 = x^2+y^2∣z∣2=x2+y2):

z−1=zˉ∣z∣2z^{-1} = \frac{\bar{z}}{|z|^2}z−1=∣z∣2zˉ​

Example Inverse Calculation

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

Solution:

Here, x=1x=1x=1 and y=−1y=-1y=−1.

Using the first formula:

x2+y2=(1)2+(−1)2=1+1=2x^2+y^2 = (1)^2 + (-1)^2 = 1 + 1 = 2x2+y2=(1)2+(−1)2=1+1=2
z−1=xx2+y2−iyx2+y2z^{-1} = \frac{x}{x^2+y^2} - i\frac{y}{x^2+y^2}z−1=x2+y2x​−ix2+y2y​
=12−i−12= \frac{1}{2} - i\frac{-1}{2}=21​−i2−1​
=12+12i= \frac{1}{2} + \frac{1}{2}i=21​+21​i

Using the conjugate and modulus formula:

zˉ=1−(−1)i=1+i\bar{z} = 1 - (-1)i = 1+izˉ=1−(−1)i=1+i
∣z∣2=x2+y2=12+(−1)2=2|z|^2 = x^2+y^2 = 1^2 + (-1)^2 = 2∣z∣2=x2+y2=12+(−1)2=2
z−1=zˉ∣z∣2=1+i2=12+12iz^{-1} = \frac{\bar{z}}{|z|^2} = \frac{1+i}{2} = \frac{1}{2} + \frac{1}{2}iz−1=∣z∣2zˉ​=21+i​=21​+21​i

The result is the same, namely:

z−1=12+12i or (12,12)z^{-1} = \frac{1}{2} + \frac{1}{2}i \text{ or } \left( \frac{1}{2}, \frac{1}{2} \right)z−1=21​+21​i or (21​,21​)
Visualization of zzz and z−1z^{-1}z−1
Visualization of z=1−iz = 1-iz=1−i and its inverse z−1=12+12iz^{-1} = \frac{1}{2} + \frac{1}{2}iz−1=21​+21​i. Notice their positions relative to the origin.

Exercise

Given the complex numbers z1=1−iz_1 = 1-iz1​=1−i and z2=2+3iz_2 = 2+3iz2​=2+3i. Find the inverse of z1+z2z_1 + z_2z1​+z2​.

Answer Key

Step 1: Find z1+z2z_1 + z_2z1​+z2​.

z=z1+z2=(1−i)+(2+3i)=(1+2)+(−1+3)i=3+2iz = z_1 + z_2 = (1-i) + (2+3i) = (1+2) + (-1+3)i = 3+2iz=z1​+z2​=(1−i)+(2+3i)=(1+2)+(−1+3)i=3+2i

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

zˉ=3−2i\bar{z} = 3-2izˉ=3−2i
∣z∣2=x2+y2=32+22=9+4=13|z|^2 = x^2+y^2 = 3^2 + 2^2 = 9 + 4 = 13∣z∣2=x2+y2=32+22=9+4=13
z−1=3−2i13=313−213iz^{-1} = \frac{3-2i}{13} = \frac{3}{13} - \frac{2}{13}iz−1=133−2i​=133​−132​i

So, the inverse of z1+z2z_1 + z_2z1​+z2​ is 313−213i\frac{3}{13} - \frac{2}{13}i133​−132​i.

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Properties of Multiplication of Complex Numbers

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Complex Number Conjugate

  • Inverse of Complex NumbersCalculate complex number inverses using conjugate and modulus formulas. Master z⁻¹ = z̄/|z|² for division and reciprocal operations with examples.
On this page
  • What is the Inverse of a Complex Number?
  • Finding the Inverse Formula
  • Example Inverse Calculation
  • Exercise
    • Answer Key
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