Conjugate Complex Numbers and Properties of Modulus

The conjugate of z = a + bi is z̄ = a − bi, and together with the modulus |z| = √(a² + b²) it satisfies z z̄ = |z|², which makes division of complex numbers and many proofs simple. In brief: conj(z + w) = z̄ + w̄, conj(zw) = z̄ w̄, |zw| = |z||w| and |z + w| ≤ |z| + |w|. For z = 3 + 4i: z̄ = 3 − 4i, z z̄ = 25 and |z| = 5. A number is real exactly when z = z̄.

Conjugation reflects a complex number in the real axis. It lets you express the real and imaginary parts, divide, find moduli and prove inequalities. The modulus behaves like an absolute value with an extra twist: it satisfies the triangle inequality, and equality conditions have a geometric meaning. This guide lists the properties of conjugation and modulus with proofs, shows how to use them in equations and inequalities and gives a table of the rules.

What are the properties of the conjugate?

For complex numbers z = a + bi and w:

• conj(z̄) = z (conjugating twice returns z).
• conj(z + w) = z̄ + w̄ and conj(z − w) = z̄ − w̄.
• conj(zw) = z̄ w̄ and conj(z/w) = z̄/w̄.
• conj(zⁿ) = (z̄)ⁿ for every integer n.
• z + z̄ = 2a = 2 Re z, and z − z̄ = 2bi = 2i Im z, so Re z = (z + z̄)/2 and Im z = (z − z̄)/(2i).
• z = z̄ if and only if z is real, and z = −z̄ if and only if z is purely imaginary (or 0).
• z z̄ = a² + b² = |z|².

Geometrically, conjugation is the reflection in the real axis. In polar form, the conjugate of r(cos θ + i sin θ) is r(cos θ − i sin θ) = r(cos(−θ) + i sin(−θ)), so the modulus is unchanged and the argument changes sign.

Proof of conj(zw) = z̄ w̄

With z = a + bi and w = c + di: zw = (ac − bd) + (ad + bc)i, so conj(zw) = (ac − bd) − (ad + bc)i. And z̄ w̄ = (a − bi)(c − di) = (ac − bd) − (ad + bc)i. They are equal ∎.

What are the properties of the modulus?

• |z| ≥ 0, with equality only for z = 0.
• |z̄| = |z| and |−z| = |z|.
• |zw| = |z||w| and |z/w| = |z|/|w|.
• |zⁿ| = |z|ⁿ.
• |Re z| ≤ |z| and |Im z| ≤ |z|.
• 1/z = z̄/|z|² for z ≠ 0.

Proof of |zw|² = |z|²|w|²: |zw|² = (zw)(conj(zw)) = z w z̄ w̄ = (z z̄)(w w̄) = |z|²|w|². The rule is also visible in polar form.

Example: |(3 + 4i)(5 − 12i)| = 5 · 13 = 65. Check: (3 + 4i)(5 − 12i) = 15 − 36i + 20i + 48 = 63 − 16i, and √(63² + 16²) = √(3969 + 256) = √4225 = 65 ✓.

What is the triangle inequality for complex numbers?

|z + w| ≤ |z| + |w|, with equality if and only if z and w point in the same direction (one is a nonnegative multiple of the other). Geometrically, in the triangle with vertices 0, z and z + w, the side from 0 to z + w is at most the sum of the other two. A consequence is the reverse inequality: ||z| − |w|| ≤ |z − w|.

Proof: |z + w|² = (z + w)(z̄ + w̄) = |z|² + |w|² + z w̄ + z̄ w = |z|² + |w|² + 2 Re(z w̄) ≤ |z|² + |w|² + 2|z||w| = (|z| + |w|)², since Re(z w̄) ≤ |z w̄| = |z||w|.

Example 1

z = 3 + 4i and w = 1 − 2i: |z + w| = |4 + 2i| = √20 ≈ 4.47, |z| + |w| = 5 + √5 ≈ 7.24. The inequality is strict. For z = 3 + 4i and w = 6 + 8i (in the same direction): |z + w| = 15 = 5 + 10, equality.

How do you use these properties?

Example 2: division

(2 + 3i)/(1 − i) = (2 + 3i)(1 + i)/|1 − i|² = (2 + 2i + 3i − 3)/2 = (−1 + 5i)/2.

Example 3: solving an equation with z and z̄

Solve z + 2z̄ = 6 − 3i. Let z = a + bi: (a + bi) + 2(a − bi) = 3a − bi. So 3a = 6 and −b = −3, giving a = 2 and b = 3, and z = 2 + 3i. Check: 2 + 3i + 2(2 − 3i) = 6 − 3i ✓.

Example 4: an equation with the modulus

Solve |z| = z + 2 for complex z. If z = x + yi, the right side z + 2 must be real and nonnegative, so y = 0. For real x: |x| = x + 2. If x ≥ 0: x = x + 2, impossible. If x < 0: −x = x + 2, so x = −1, but then x + 2 = 1 = |−1| ✓. The solution is z = −1.

Example 5: a real expression

For which z is (z − 1)/(z + 1) purely imaginary? Let the quotient be iλ: z − 1 = iλ(z + 1). Equivalently the real part of (z − 1)(z̄ + 1) must be 0: |z|² + z − z̄ − 1 has real part |z|² − 1. So |z| = 1 (z ≠ −1). Indeed, the points on the unit circle (other than −1) give purely imaginary values. For z = i: (i − 1)/(i + 1) = (i − 1)(1 − i)/2 = (i − i² − 1 + i)/2 = (2i)/2 = i ✓.

Example 6: an identity

Parallelogram law: |z + w|² + |z − w|² = 2|z|² + 2|w|². For z = 1 + i and w = 2 − i: |3|² + |−1 + 2i|² = 9 + 5 = 14, and 2 · 2 + 2 · 5 = 14 ✓. This relates the sides and the diagonals of a parallelogram.

Table of rules

Where do these ideas come from?

The conjugate pairs appeared with Cardano’s and Bombelli’s work on equations in the sixteenth century, because complex roots of real equations come in pairs. The modulus was introduced by Argand in 1806, and the triangle inequality has been known since Euclid for lengths. The use of the conjugate to divide complex numbers is the analogue of rationalizing the denominator.

Where are they used today?

In electrical engineering, the power in an AC circuit is S = V I*, where I* is the conjugate of the current phasor. In quantum mechanics, the probability density is |ψ|² = ψ ψ*, and observable quantities correspond to expressions that are equal to their conjugates (real). In signal processing, the Fourier transform of a real signal satisfies conj(F(ω)) = F(−ω).

Which mistakes are common?

Thinking |z|² = z²

|z|² = z z̄, which is real and nonnegative. z² is a complex number in general: for z = i, z² = −1, but |z|² = 1.

Assuming |z + w| = |z| + |w|

Only for numbers pointing in the same direction. In general, the inequality is strict.

Forgetting to conjugate the whole denominator

To divide by 2 + 3i, multiply by 2 − 3i, changing only the sign of the imaginary part.

Mixing up Im z and i Im z

Im z is a real number. For z = 3 + 4i, Im z = 4, not 4i.