Complex Numbers — Modulus, Argument and Operations
A complex number is written z = a + ib, where i² = −1. Students meet them first as "the answers you get when the discriminant is negative", which makes them feel like a trick. They are not. A complex number is a point in a plane, and its two most useful properties — modulus (how far from the origin) and argument (the angle) — turn hard multiplication problems into easy ones. This guide covers both, with every operation worked and checked.
The two forms of the same number
Modulus: |z| = √(a² + b²)
Argument: θ = the angle from the positive real axis, fixed by the quadrant
Polar form: z = |z| (cos θ + i sin θ) = |z| eiθ
Think of a + ib as the point (a, b) on a graph where the horizontal axis is real and the vertical axis is imaginary. The modulus is the distance from the origin — it is never negative. The argument is the angle you turn through to reach the point.
Getting the argument right — the quadrant table
Nearly every argument mistake comes from typing tan⁻¹(b/a) into a calculator and writing down whatever appears. A calculator's tan⁻¹ only returns angles between −90° and +90°, so it cannot tell 1 + i from −1 − i. Work out the reference angle α = tan⁻¹|b/a| first, then place it:
| Signs of (a, b) | Quadrant | Argument (principal value, −180° < θ ≤ 180°) |
|---|---|---|
| a > 0, b > 0 | First | θ = α |
| a < 0, b > 0 | Second | θ = 180° − α |
| a < 0, b < 0 | Third | θ = −(180° − α) = α − 180° |
| a > 0, b < 0 | Fourth | θ = −α |
Worked example 1 — modulus and argument of 3 + 4i
a = 3, b = 4, both positive → first quadrant.
|z| = √(3² + 4²) = √(9 + 16) = √25 = 5
α = tan⁻¹(4/3) = tan⁻¹(1.3333) = 53.13°, and since it is the first
quadrant, θ = 53.13°.
Polar form: z = 5 (cos 53.13° + i sin 53.13°)
Check: 5 cos 53.13° = 5 × 0.6 = 3 ✓ and 5 sin 53.13° = 5 × 0.8 = 4 ✓
Example 2 — a second-quadrant number: z = −1 + i√3
|z| = √(1 + 3) = √4 = 2
Reference angle α = tan⁻¹(√3 / 1) = 60°. The point is in the second quadrant (a negative,
b positive), so θ = 180° − 60° = 120°.
Polar form: z = 2 (cos 120° + i sin 120°).
Check: 2 cos 120° = 2 × (−0.5) = −1 ✓ and 2 sin 120° = 2 × 0.8660 = 1.732 = √3 ✓. A calculator that returned −60° would have given the point 1 − i√3 — the wrong number entirely.
The four operations
Multiplication: (a + ib)(c + id) = (ac − bd) + i(ad + bc)
Conjugate: z̄ = a − ib, and z z̄ = a² + b² = |z|²
Division: multiply numerator and denominator by the conjugate of the denominator
Addition and subtraction are easiest in Cartesian form. Multiplication and division are easiest in polar form, because moduli multiply and arguments add.
Example 3 — multiply (3 + 4i)(1 − 2i)
= 3(1) + 3(−2i) + 4i(1) + 4i(−2i)
= 3 − 6i + 4i − 8i²
= 3 − 2i + 8 (since −8i² = +8)
= 11 − 2i
Modulus check: |3 + 4i| = 5 and |1 − 2i| = √5, so the product must have modulus 5√5 = 11.180. And |11 − 2i| = √(121 + 4) = √125 = 11.180 ✓
Example 4 — divide (3 + 4i) ÷ (1 − 2i)
Multiply top and bottom by the conjugate of the bottom, 1 + 2i:
Numerator: (3 + 4i)(1 + 2i) = 3 + 6i + 4i + 8i² = 3 + 10i − 8 = −5 + 10i
Denominator: (1 − 2i)(1 + 2i) = 1² + 2² = 5 (a real number — that is the whole point
of the conjugate)
Result: (−5 + 10i)/5 = −1 + 2i
Check by multiplying back: (−1 + 2i)(1 − 2i) = −1 + 2i + 2i − 4i² = −1 + 4i + 4 = 3 + 4i ✓
De Moivre's theorem — powers made easy
Raise the modulus to the power; multiply the angle by the power. That is all.
Example 5 — find (1 + i)⁸
First convert: |1 + i| = √2, argument 45° (first quadrant, a = b = 1).
By De Moivre: (1 + i)⁸ = (√2)⁸ (cos 360° + i sin 360°).
(√2)⁸ = 2⁴ = 16, and cos 360° = 1, sin 360° = 0.
So (1 + i)⁸ = 16.
Check the slow way: (1 + i)² = 1 + 2i + i² = 2i. Then (2i)² = 4i² = −4, so (1 + i)⁴ = −4, and (−4)² = 16 ✓. Expanding the eighth power term by term would have taken a page.
Example 6 — a square root of i
i has modulus 1 and argument 90°. Halving the angle: √i = 1 × (cos 45° + i sin 45°) =
(1 + i)/√2.
Check: [(1 + i)/√2]² = (1 + 2i + i²)/2 = 2i/2 = i ✓
The cube roots of unity
Solving z³ = 1 gives three roots, not one: 1, ω = −½ + i(√3/2) and ω² = −½ − i(√3/2). Two properties are asked constantly:
Both follow directly from the polar form: the three roots sit at 0°, 120° and 240° on a circle of radius 1, spaced equally, so they cancel when added. Any question containing high powers of ω is solved by reducing the exponent modulo 3 — for instance ω¹⁰⁰ = ω99 × ω = (ω³)33 × ω = ω.
Common mistakes
- Taking tan⁻¹ without checking the quadrant. For z = −1 − i a calculator gives 45°, but the correct principal argument is −135°. Always sketch the point.
- Writing √(−4) × √(−9) = √36 = 6. It is 2i × 3i = 6i² = −6. The rule √a × √b = √(ab) does not survive negative numbers — convert to i form first.
- Calling the modulus negative. |z| is a distance; it is always ≥ 0.
- Dividing without the conjugate. A denominator of the form c + id is not a finished answer. Rationalise it so the result reads a + ib.
- Degrees and radians mixed. Physics and higher maths use radians (180° = π). Decide which you are working in and stay there.
- Forgetting that a real number is complex too, with b = 0. Questions that ask "for what value of k is z purely real" are asking you to set the imaginary part to zero.
Where this appears in exams
| Exam / class | Typical use |
|---|---|
| CBSE/ICSE Class 11 | Algebra of complex numbers, modulus, argument, polar form |
| JEE Main / Advanced | De Moivre's theorem, roots of unity, loci in the complex plane |
| Class 12 physics | AC circuits — impedance is handled as a complex quantity |
| IIT-JAM / GATE / CSIR-NET | Quantum chemistry wavefunctions and the eiθ form; |ψ|² = ψ*ψ |
In quantum chemistry the conjugate is not decoration: the probability density is ψ*ψ, which is exactly the "z z̄ = |z|²" rule above applied to a wavefunction. Check the current official notification for your exam's syllabus details.
Check your algebra. The complex number tool adds, multiplies, divides and converts between a + ib and polar form, so you can confirm a modulus, an argument or a messy division before you commit it to your answer sheet.
Open the Complex Number Calculator →Class 11–12 students who find the maths harder than the chemistry usually need the algebra fixed first. ABC Chemistry runs Class 11–12 chemistry coaching at the Gurugram centre plus online classes across India — abcchemistry.in.