Complex Numbers
What this domain tests
Arithmetic in ℂ, conjugate and modulus, the Argand diagram, polar (trigonometric) form, De Moivre's theorem, nth roots, quadratic equations with complex roots, and simple loci.
Core rules
- z = a + bi with i² = −1; z = w means equal real parts and equal imaginary parts.
- Conjugate z̄ = a − bi; z·z̄ = a² + b² = |z|²; 1/z = z̄/|z|², which is how division is done.
- |z₁z₂| = |z₁||z₂| and |z₁/z₂| = |z₁|/|z₂|, but |z₁ + z₂| ≤ |z₁| + |z₂| (triangle inequality).
- Polar form z = r(cos θ + i sin θ) with r = |z| and θ = arg z measured from the positive real axis in radians. Arg is defined modulo 2π; choose the principal value your textbook uses and stay consistent.
- De Moivre: [r(cos θ + i sin θ)]^n = r^n(cos nθ + i sin nθ).
- Roots: z^n = r(cos θ + i sin θ) has exactly n distinct roots r^(1/n)[cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)], k = 0, 1, …, n − 1.
- Loci: |z − z₀| = ρ is a circle centre z₀ radius ρ; |z − a| = |z − b| is the perpendicular bisector of the segment ab.
Worked example
Solve z² = 1 + i√3.
First write the right-hand side in polar form: |1 + i√3| = √(1 + 3) = 2, and its argument is π/3 (first quadrant). So z = √2·[cos((π/3 + 2kπ)/2) + i sin((π/3 + 2kπ)/2)] = √2·[cos(π/6 + kπ) + i sin(π/6 + kπ)] for k = 0, 1.
- k = 0: √2(cos π/6 + i sin π/6) = √2(√3/2 + i/2) = √6/2 + i√2/2.
- k = 1: the negative of this, −√6/2 − i√2/2.
The two roots are opposite points on the circle of radius √2, exactly as the theory predicts.
Traps that cost marks
- Reporting only the k = 0 root when the question asks for all roots.
- Mixing degrees and radians inside one calculation.
- Getting the quadrant of the argument wrong: arctan of the ratio only locates arguments in the first and fourth quadrants.
- Applying |z₁ + z₂| = |z₁| + |z₂|, which is false except in special cases.
- Forgetting to multiply numerator and denominator by the conjugate when dividing.