Mathematics · Ch 4 — Complex Numbers and Quadratic Equations

The Modulus and the Conjugate of a Complex Number

4.4

The Modulus and the Conjugate of a Complex Number

Every complex number $z = a + ib$ carries two fundamental real-valued companions: its modulus and its conjugate. These two tools let us measure the "size" of a complex number and reflect it across the real axis, and they turn out to be indispensable for division, for proving identities, and for understanding the geometry of the complex plane.

Definitions

Let $z = a + ib$ , where $a$ and $b$ are real numbers.

Modulus — denoted $|z|$ — is the non-negative real number

|z| = \sqrt{a^2 + b^2}.

It is the distance of the point $(a, b)$ from the origin in the complex plane.

Conjugate — denoted $\bar{z}$ — is the complex number

\bar{z} = a - ib.

Geometrically, $\bar{z}$ is the reflection of $z$ across the real axis.

For $z = 3 + i$ : $|z| = \sqrt{3^2 + 1^2} = \sqrt{10}$ , and $\bar{z} = 3 - i$ .

For $z = 2 - 5i$ : $|z| = \sqrt{2^2 + (-5)^2} = \sqrt{29}$ , and $\bar{z} = 2 + 5i$ .

For $z = -3 - 5i$ : $\bar{z} = -3 + 5i$ (notice the sign of both parts flips).

Watch out

The modulus is always a non-negative real number. It is never negative, and it is never imaginary. A common mistake is to write $|a+ib| = a^2 + b^2$ — the square root is essential.

The Multiplicative Inverse and the Conjugate

For a non-zero complex number $z = a + ib$ , its multiplicative inverse $z^{-1}$ is the complex number that satisfies $z \cdot z^{-1} = 1$ . We can find it using the conjugate:

z^{-1} = \frac{1}{a+ib} = \frac{a-ib}{(a+ib)(a-ib)} = \frac{a-ib}{a^2 + b^2} = \frac{\bar{z}}{|z|^2}.

This gives the compact formula

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

Tip

To find the multiplicative inverse of any non-zero complex number, just write $\frac{\bar{z}}{|z|^2}$ . No need to rationalise from scratch every time.

Properties of Modulus and Conjugate

For any two complex numbers $z_1$ and $z_2$ , the following five properties hold. Each one is proved directly from the definitions.

›Proof

Property (i): $\overline{z_1 z_2} = \bar{z}_1 \cdot \bar{z}_2$

Let $z_1 = a+ib$ , $z_2 = c+id$ . Then

$z_1 z_2 = (a+ib)(c+id) = (ac - bd) + i(ad + bc).$

Taking the conjugate:

$\overline{z_1 z_2} = (ac - bd) - i(ad + bc).$

Now compute $\bar{z}_1 \cdot \bar{z}_2 = (a-ib)(c-id) = (ac - bd) - i(ad + bc)$ .

The two expressions are identical, so the property holds.

›Proof

Property (ii): $\overline{\left(\frac{z_1}{z_2}\right)} = \frac{\bar{z}_1}{\bar{z}_2}$ , provided $z_2 \neq 0$ .

Using Property (i) and the fact that $\frac{z_1}{z_2} \cdot z_2 = z_1$ , take conjugates of both sides:

$\overline{\frac{z_1}{z_2}} \cdot \bar{z}_2 = \bar{z}_1.$

Since $\bar{z}_2 \neq 0$ (because $z_2 \neq 0$ ), divide through to get the result.

›Proof

Property (iii): $|z_1 z_2| = |z_1| \, |z_2|$

Let $z_1 = a+ib$ , $z_2 = c+id$ . Then

$|z_1 z_2|^2 = (ac - bd)^2 + (ad + bc)^2.$

Expand:

$= a^2 c^2 - 2abcd + b^2 d^2 + a^2 d^2 + 2abcd + b^2 c^2 = a^2(c^2+d^2) + b^2(c^2+d^2) = (a^2+b^2)(c^2+d^2).$

But $a^2+b^2 = |z_1|^2$ and $c^2+d^2 = |z_2|^2$ , so $|z_1 z_2|^2 = |z_1|^2 |z_2|^2$ . Taking square roots (all quantities are non-negative) gives $|z_1 z_2| = |z_1| \, |z_2|$ .

›Proof

Property (iv): $\overline{z_1 \pm z_2} = \bar{z}_1 \pm \bar{z}_2$

Let $z_1 = a+ib$ , $z_2 = c+id$ . Then

$z_1 \pm z_2 = (a \pm c) + i(b \pm d).$

Taking the conjugate:

$\overline{z_1 \pm z_2} = (a \pm c) - i(b \pm d) = (a-ib) \pm (c-id) = \bar{z}_1 \pm \bar{z}_2.$ …

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← Previous · 4.3.7Example 4 — Express (-√(3) + √(-2))(2√(3) - i) in the form of a + ib.Next · 4.4 →Example 5 — Find the multiplicative inverse of 2 - 3i.