Complex Numbers
Complex Numbers
Complex numbers are examined through arithmetic, the Argand diagram, modulus–argument form, De Moivre's theorem and simple loci. The paper usually includes a quadratic with complex roots and a geometric interpretation question.
Core rules
- i² = −1; for z = a + bi the conjugate is z* = a − bi and zz* = a² + b².
- Modulus |z| = √(a² + b²); argument arg z = arctan(b/a) adjusted for the quadrant, conventionally −π < θ ≤ π.
- Polar form z = r(cosθ + i sinθ) = r cisθ; multiplication multiplies moduli and adds arguments; division divides moduli and subtracts arguments.
- De Moivre: (r cisθ)^n = r^n cis(nθ). The nth roots are r^(1/n) cis[(θ + 2kπ)/n] for k = 0, 1, …, n − 1, equally spaced on a circle.
- Loci: |z − z₀| = r is a circle with centre z₀ and radius r; |z − z₀| = |z − z₁| is the perpendicular bisector; arg(z − z₀) = α is a half-line.
Worked example
Express z = −1 + i√3 in polar form and hence find z⁶. |z| = √(1 + 3) = 2. The point (−1, √3) lies in the second quadrant, so arg z = π − arctan(√3) = π − π/3 = 2π/3. Thus z = 2 cis(2π/3) and z⁶ = 2⁶ cis(6 × 2π/3) = 64 cis(4π) = 64, since 4π is a whole number of turns. The quadrant adjustment is essential: arctan(−√3) alone would give the wrong argument.
Traps
- Using arctan(b/a) without quadrant correction.
- Forgetting that a real-coefficient quadratic with one complex root has the conjugate as its other root.
- Mixing degrees and radians in the argument.
- Losing roots by taking only k = 0 in the nth-root formula.