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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 + iy has a "reciprocal" friend called the multiplicative inverse (or just inverse), which we write as z1z^{-1} or 1/z1/z.

The defining characteristic of the multiplicative inverse is that if we multiply the complex number zz by its inverse z1z^{-1}, the result is 1 (the multiplicative identity element).

z×z1=1z \times 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 + iy, its inverse is:

z1=xx2+y2iyx2+y2z^{-1} = \frac{x}{x^2+y^2} - i\frac{y}{x^2+y^2}

This formula can also be written as an ordered pair:

z1=(xx2+y2,yx2+y2)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 (zˉ=xiy\bar{z} = x-iy) and the modulus squared (z2=x2+y2|z|^2 = x^2+y^2):

z1=zˉz2z^{-1} = \frac{\bar{z}}{|z|^2}

Example Inverse Calculation

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

Solution:

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

Using the first formula:

x2+y2=(1)2+(1)2=1+1=2x^2+y^2 = (1)^2 + (-1)^2 = 1 + 1 = 2
z1=xx2+y2iyx2+y2z^{-1} = \frac{x}{x^2+y^2} - i\frac{y}{x^2+y^2}
=12i12= \frac{1}{2} - i\frac{-1}{2}
=12+12i= \frac{1}{2} + \frac{1}{2}i

Using the conjugate and modulus formula:

zˉ=1(1)i=1+i\bar{z} = 1 - (-1)i = 1+i
z2=x2+y2=12+(1)2=2|z|^2 = x^2+y^2 = 1^2 + (-1)^2 = 2
z1=zˉz2=1+i2=12+12iz^{-1} = \frac{\bar{z}}{|z|^2} = \frac{1+i}{2} = \frac{1}{2} + \frac{1}{2}i

The result is the same, namely:

z1=12+12i or (12,12)z^{-1} = \frac{1}{2} + \frac{1}{2}i \text{ or } \left( \frac{1}{2}, \frac{1}{2} \right)
Visualization of zz and z1z^{-1}
Visualization of z=1iz = 1-i and its inverse z1=12+12iz^{-1} = \frac{1}{2} + \frac{1}{2}i. Notice their positions relative to the origin.

Exercise

Given the complex numbers z1=1iz_1 = 1-i and z2=2+3iz_2 = 2+3i. Find the inverse of z1+z2z_1 + z_2.

Answer Key

Step 1: Find z1+z2z_1 + z_2.

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

Step 2: Find the inverse of z=3+2iz = 3+2i. Here x=3x=3 and y=2y=2. We use the formula z1=zˉz2z^{-1} = \frac{\bar{z}}{|z|^2}.

zˉ=32i\bar{z} = 3-2i
z2=x2+y2=32+22=9+4=13|z|^2 = x^2+y^2 = 3^2 + 2^2 = 9 + 4 = 13
z1=32i13=313213iz^{-1} = \frac{3-2i}{13} = \frac{3}{13} - \frac{2}{13}i

So, the inverse of z1+z2z_1 + z_2 is 313213i\frac{3}{13} - \frac{2}{13}i.