Multiplying by the conjugate makes a real number.

Drag z and w on the Argand plane; the product zw is drawn with |zw| = |z||w| shown as radii; when w is a real multiple of z̄ the product drops onto the real axis.

Place w so that zw is a positive real number for z = 3 + 4i (w along 3 − 4i), and read zz̄ = 25 = |z|².

Dividing part by part instead of multiplying by the conjugate.

• The modulus of z = a + ib is the non-negative real number |z| = √(a² + b²). For example |3 + 4i| = √25 = 5, |−5 + 12i| = 13 and |2 − 5i| = √29.
• The conjugate of z = a + ib is z̄ = a − ib: only the sign of the imaginary part changes. So the conjugate of 3 + 4i is 3 − 4i, of −2 − 7i is −2 + 7i, of 6i is −6i, and a real number is its own conjugate.
• z × z̄ = a² + b² = |z|². Hence for z ≠ 0 the inverse is z⁻¹ = z̄/|z|², which is the conjugate trick used in division.
• Modulus rules: |z₁z₂| = |z₁||z₂| and |z₁/z₂| = |z₁|/|z₂| when z₂ ≠ 0. For example |(3 + 4i)(5 − 12i)| = 5 × 13 = 65, without multiplying out; the product is 63 − 16i and √(63² + 16²) = √4225 = 65.
• Conjugate rules: the conjugate of z₁z₂ is z̄₁z̄₂, of z₁ ± z₂ is z̄₁ ± z̄₂, and of z₁/z₂ is z̄₁/z̄₂ when z₂ ≠ 0.
• Also z + z̄ = 2 Re z and z − z̄ = 2i Im z, so z is real exactly when z = z̄.
• Worked example: (1 + 7i)/(2 − i) = (1 + 7i)(2 + i)/5 = (2 + i + 14i − 7)/5 = (−5 + 15i)/5 = −1 + 3i, so its conjugate is −1 − 3i.

Teaching this idea? Put the simulation on your own page or LMS with this code.