On this page
Many complex number problems look as if they need z = x + iy and a page of algebra. A lot of them collapse once you use one identity: a complex number times its conjugate is the square of its modulus. This post shows where the identity comes from, works four problems with it and lists the mistakes that cost marks.
Where is this in the syllabus?
The NCERT Class 11 chapter Complex Numbers and Quadratic Equations (2026-27 reprint) defines |z| and z̄ in section 4.4, shows that the multiplicative inverse is z̄/|z|², and lists properties such as |z₁z₂| = |z₁||z₂|. The CBSE 2026-27 Mathematics curriculum lists the algebraic properties of complex numbers and the Argand plane for Class 11; polar representation appears in its list of topics assessed only formatively this year. The JEE (Main) 2026 syllabus lists the algebra of complex numbers and modulus and argument. The JEE (Advanced) 2026 brochure lists conjugation, properties of modulus and the triangle inequality. Check the current year's documents.
Why is z z̄ always real?
Let z = a + ib, so z̄ = a − ib. Then
z z̄ = a² − (ib)² = a² + b² = |z|².
It is real, never negative, and zero only when z = 0. Three consequences follow at once:
1/z = z̄ ÷ |z|² for any nonzero z.
If |z| = 1, then z̄ = 1/z.
|z|² is generally not z². For z = i, z² = −1 but |z|² = 1.
Worked example 1: division
Find (3 + 4i) ÷ (1 − 2i).
Multiply top and bottom by 1 + 2i.
Top: (3 + 4i)(1 + 2i) = 3 + 6i + 4i + 8i² = −5 + 10i.
Bottom: |1 − 2i|² = 1 + 4 = 5.
Answer: −1 + 2i.
Check: (−1 + 2i)(1 − 2i) = −1 + 2i + 2i − 4i² = 3 + 4i. NCERT's Example 5 uses the same idea to find the inverse of 2 − 3i as (2 + 3i)/13.
Worked example 2: a |z| = 1 problem
If |z| = 1 and z ≠ −1, show that w = (z − 1)/(z + 1) is purely imaginary.
Because |z| = 1, z̄ = 1/z.
w̄ = (z̄ − 1)/(z̄ + 1) = (1/z − 1)/(1/z + 1). Multiply top and bottom by z: (1 − z)/(1 + z) = −w.
A number with w̄ = −w has zero real part, so w is purely imaginary.
Check with a number: z = (3 + 4i)/5 has |z| = 1. Then z − 1 = (−2 + 4i)/5 and z + 1 = (8 + 4i)/5, so w = (−2 + 4i)/(8 + 4i). Multiplying top and bottom by 8 − 4i gives 40i/80 = 0.5i. Purely imaginary, as claimed.
Worked example 3: expanding |z₁ ± z₂|²
Write the modulus squared as a product with the conjugate:
|z₁ + z₂|² = (z₁ + z₂)(z̄₁ + z̄₂) = |z₁|² + |z₂|² + z₁z̄₂ + z̄₁z₂.
The last two terms are conjugates of each other, so their sum is 2 Re(z₁z̄₂). Hence
|z₁ + z₂|² = |z₁|² + |z₂|² + 2 Re(z₁z̄₂), and |z₁ − z₂|² = |z₁|² + |z₂|² − 2 Re(z₁z̄₂).
Adding gives |z₁ + z₂|² + |z₁ − z₂|² = 2(|z₁|² + |z₂|²): the diagonals of a parallelogram, squared and added, equal the sum of the squares of its four sides.
Check with z₁ = 2 + i and z₂ = 1 − 3i: z₁ + z₂ = 3 − 2i (modulus squared 13) and z₁ − z₂ = 1 + 4i (17), total 30. And 2(5 + 10) = 30.
Worked example 4: a quick deduction
If |z₁| = 3, |z₂| = 4 and |z₁ + z₂| = 5, what is |z₁ − z₂|?
25 = 9 + 16 + 2 Re(z₁z̄₂), so Re(z₁z̄₂) = 0.
Then |z₁ − z₂|² = 9 + 16 − 0 = 25, so |z₁ − z₂| = 5.
On the Argand plane, z₁ and z₂ are perpendicular arrows, and the parallelogram is a rectangle with equal diagonals. Our post on complex numbers as geometry builds on this picture.
Which mistakes cost marks?
Writing |z|² = z². True only for real z.
Writing |z₁ + z₂| = |z₁| + |z₂|. That equality holds only when the two point the same way; in general the triangle inequality gives ≤.
Conjugating only part of a fraction. The conjugate of z₁/z₂ is z̄₁/z̄₂.
Using z̄ = 1/z without |z| = 1.
Dividing by zero. In example 2, z = −1 is excluded for a reason.
A practice routine
Divide (2 + 5i) by (3 − i) using the conjugate, then check by multiplying back. You should get 0.1 + 1.7i.
Prove that if |z| = 1 then (1 + z)/(1 + z̄) = z.
Verify the parallelogram identity for z₁ = 1 + 2i and z₂ = 3 − i.
Redo example 4 with |z₁| = 5, |z₂| = 12 and |z₁ + z₂| = 13.
For square roots of complex numbers, which use the same algebra, see our guide to finding the square root of a complex number. When a modulus problem refuses to simplify, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Today, take three complex-number problems from your last test and redo each one by multiplying by a conjugate first.
