Complex numbers

Among the real numbers the equation x² = −1 has no solution: no real number has a negative square. Complex numbers remove this “gap”: with them every quadratic equation (indeed every polynomial equation) can be solved. Complex numbers are widely used in electrical engineering, signal theory and the physics of oscillations. Exams ask about algebraic operations, modulus, conjugates, powers of i, the trigonometric form and quadratics with a negative discriminant.

§1. The imaginary unit and algebraic form

i is the imaginary unit. A complex number is z = a + bi, where a = Re z is the real part and b = Im z the imaginary part (a, b ∈ R). This is the algebraic form.

• If b = 0, z is real; if a = 0, b ≠ 0, it is purely imaginary. So the reals are a special case of the complex numbers: R ⊂ C.
• Two complex numbers are equal ⇔ their real parts and imaginary parts are equal separately. This is the main tool for equations: one complex equality becomes two real ones.
• Complex numbers cannot be ordered by “greater/less”; only their moduli can be compared.

Powers of `i` repeat with period 4:

Reason: i⁴ = (i²)² = (−1)² = 1, so the values come back every four powers.

§2. Operations

• Addition/subtraction: part by part: (3 + 2i) + (1 − 5i) = 4 − 3i.
• Multiplication: expand like polynomials and substitute i² = −1: (2 + 3i)(1 − i) = 2 − 2i + 3i − 3i² = 5 + i.
• Conjugate: z̄ = a − bi. z·z̄ = a² + b² is always real and non-negative; z + z̄ = 2a is real too.
• Division: multiply top and bottom by the conjugate of the denominator — the denominator becomes real. This resembles rationalising a denominator with roots.

Worked example

Calculate (5 + i)/(1 + i).

1. Multiply by the conjugate: (5 + i)(1 − i)/((1 + i)(1 − i)).
2. Numerator: 5 − 5i + i − i² = 6 − 4i; denominator: 1 + 1 = 2.
3. Result: 3 − 2i.

Worked example

Solve 2z + z̄ = 3 + 2i.

1. Let z = a + bi: 2a + 2bi + a − bi = 3a + bi.
2. Equate the parts: 3a = 3, b = 2.
3. Answer: z = 1 + 2i.

§3. Geometric meaning and modulus

z = a + bi is the point (a; b) of the coordinate plane (or the vector from the origin to it). The horizontal axis is real, the vertical axis imaginary. Addition corresponds to adding vectors, and the conjugate is the reflection in the real axis.

For example, |3 − 4i| = √(9 + 16) = 5. The points with |z − z₀| = R form a circle with centre z₀ and radius R; |z − z₀| ≤ R is a disc.

Worked example

Find |(3 + 4i)(1 − 2i)| without multiplying out.

1. |3 + 4i| = 5, |1 − 2i| = √5.
2. Moduli multiply: 5√5.

§4. Trigonometric form and De Moivre’s formula

A point can also be given by a distance and an angle:

When multiplying, moduli multiply and arguments add; when dividing, moduli divide and arguments subtract. This follows from the sine and cosine addition formulas.

Worked example

Calculate (1 + i)⁸.

1. r = √2, φ = π/4 (first quadrant).
2. (√2)⁸ = 16, 8·π/4 = 2π.
3. 16(cos 2π + i sin 2π) = 16.

§5. Quadratic equations and square roots

If the discriminant is negative, the roots are a conjugate complex pair:

For an equation with real coefficients, complex roots always come in conjugate pairs, and Vieta’s theorem still holds.

Worked example

Solve x² − 4x + 13 = 0.

1. D = 16 − 52 = −36, √(−D) = 6.
2. x = (4 ± 6i)/2 = 2 ± 3i.
3. Check (Vieta): x₁ + x₂ = 4, x₁x₂ = 4 + 9 = 13.

Worked example

Solve z² = −8 + 6i.

1. z = a + bi: a² − b² + 2abi = −8 + 6i.
2. a² − b² = −8, ab = 3.
3. Try an integer solution: a = 1, b = 3 (1 − 9 = −8).
4. Answer: z = ±(1 + 3i).

§6. Sets in the complex plane

Since the modulus is a distance, many equations and inequalities can be “read” geometrically:

In such problems a sketch often gives the answer at once instead of algebra. For example, |z − 1 − i| ≤ 2 is a disc with centre (1; 1) and radius 2; its point farthest from the origin is at distance √2 + 2.

Worked example

Find z such that |z| = |z − 2| = |z − 2i|.

1. |z| = |z − 2| — equidistant from 0 and 2, so Re z = 1.
2. |z| = |z − 2i| — equidistant from 0 and 2i, so Im z = 1.
3. Answer: z = 1 + i (the circumcentre of the triangle).

Check yourself

1. Calculate i²⁰²⁶.

2. Find |3 − 4i|.

3. Calculate (2 − i)².

4. Find z·z̄ for z = 1 + 2i.

5. Calculate (1 + i)/(1 − i).

6. Solve z² = −9.

7. Calculate (2 + 3i)(2 − 3i).

8. Solve x² + 2x + 5 = 0.

Common mistakes

• Writing i² = 1 — it is −1.
• Multiplying only the denominator by the conjugate and forgetting the numerator.
• Taking the modulus as a + b — it is √(a² + b²).
• Forgetting that remainder 0 gives 1 (i⁴ᵏ = 1).
• Ignoring the quadrant when finding the argument.
• Trying to compare complex numbers by “greater/less”.