Complex Numbers

Complex Numbers

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

A complex number has the form z = a + bi, where a and b are real and i² = -1 is the imaginary unit. Re(z) = a is the real part; Im(z) = b is the imaginary part (note: Im(z) itself is a real number, not imaginary). The modulus is |z| = √(a² + b²) and the argument is arg(z) = tan⁻¹(b/a) measured from the positive real axis. The complex conjugate is z̄ = a − bi, satisfying z · z̄ = a² + b² = |z|², which rationalises denominators. ECAT usually asks 1–2 MCQs (≈2–3 marks) testing modulus/argument calculation, polar conversion, or De Moivre’s theorem. Remember that e^{iθ} = cos θ + i sin θ, so z = r e^{iθ} is the compact exponential form. Cube roots of unity 1, ω, ω² satisfy 1 + ω + ω² = 0, ω³ = 1 — a frequent shortcut.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Rectangular, Polar, and Exponential Forms

Every complex number z = a + bi corresponds to a point (a, b) on the Argand diagram. The modulus r = √(a² + b²) is the distance from the origin; the argument θ is the angle from the positive real axis, with the principal argument restricted to −π < Arg(z) ≤ π. Conversion identities:

• a = r cos θ, b = r sin θ
• Polar: z = r(cos θ + i sin θ)
• Exponential: z = r e^{iθ} (Euler’s formula)

Algebra and Conjugates

Addition and subtraction act componentwise on (a, b). Multiplication combines via (a+bi)(c+di) = (ac − bd) + (ad + bc)i — the cross terms from i² = −1 flip sign. Division rationalises the denominator by multiplying numerator and denominator by the conjugate of the divisor, since z · z̄ = |z|² is always real and non-negative. The conjugate obeys z̄̄ = z, (z₁ + z₂)̄ = z̄₁ + z̄₂, z₁z̄₂̄ = z̄₁ · z̄₂, and |z̄| = |z|, arg(z̄) = −arg(z).

De Moivre’s Theorem

For any real θ and integer n, (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ. Combined with the polar form, this means zⁿ = rⁿ(cos nθ + i sin nθ). ECAT questions exploit it to (i) compute powers like (1 + i)¹⁰ by first writing 1 + i = √2(cos π/4 + i sin π/4), or (ii) extract nth roots: the n solutions of zⁿ = w are r^{1/n} (cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)), k = 0, 1, …, n−1.

Cube Roots of Unity

The equation z³ = 1 yields 1, ω, ω² where ω = −½ + i(√3/2). Properties: ω³ = 1, 1 + ω + ω² = 0, ω² = ω̄. These appear in ECAT factorisation and in evaluating sums of roots of unity.

Common ECAT Patterns

• Compute |z| and Arg(z) for a given z, often with quadrant sign-checking.
• Convert z = −1 + i√3 into polar form: r = 2, θ = 2π/3.
• Apply De Moivre to evaluate (√3 + i)⁶ or simplify ((1 + i)/(1 − i))⁸.
• Solve z² + 1 = 0 → z = ±i; recognise discriminant-negative quadratics yield complex roots.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Quadratics with Negative Discriminants

When b² − 4ac < 0, roots are a complex conjugate pair α, ᾱ. Conjugate roots share the same modulus, opposite arguments, and a real product αᾱ = |α|². ECAT problems test sum = −b/a (real) and product = c/a (real) — never assume roots are real without checking the discriminant.

nth Roots — Getting All of Them

A frequent trap is returning only one root from zⁿ = w. The full solution set lies on a circle of radius r^{1/n} spaced evenly by 2π/n. For example, the cube roots of 8 are 2, 2ω, 2ω²; missing any is a guaranteed mark loss. Drawing the Argand diagram makes this geometry obvious and is recommended under exam pressure.

Edge Cases and Traps

• |z₁ · z₂| = |z₁||z₂| and |z₁ + z₂| ≤ |z₁| + |z₂| (triangle inequality) — useful for inequality questions.
• arg(z₁z₂) = arg(z₁) + arg(z₂) mod 2π; watch branch-cut jumps when arguments cross ±π.
• z is purely real iff z = z̄; purely imaginary iff z = −z̄. Useful for finding unknowns from conditions like z² = z̄².
• (1 + i)² = 2i, not 1 + i². Always expand using (a+bi)² = a² − b² + 2abi.

Worked Micro-Example

Find the modulus and argument of z = 1 − i and write it in polar form. |z| = √(1² + (−1)²) = √2. Since Re(z) > 0 and Im(z) < 0, the point lies in the fourth quadrant: Arg(z) = −π/4. Polar form: z = √2 (cos(−π/4) + i sin(−π/4)) = √2 e^{−iπ/4}.

ECAT Strategy

Complex numbers carry roughly 2–3 marks (1–2 MCQs). Prioritise: (1) fast polar conversion using r and quadrant signs, (2) De Moivre’s theorem applied to compact forms like (1 + i), √3 + i, and roots of unity, (3) recognising conjugate-root quadratics. Time budget: ≤ 90 seconds per MCQ.

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