Complex Numbers
Complex Numbers
The complex-numbers domain tests arithmetic in the form a + bi, where a is the real part and b is the imaginary part, with the defining relation i² = −1. Questions typically ask you to simplify powers of i, add and subtract complex numbers, multiply, divide using the conjugate, compute a modulus, and solve quadratic equations whose discriminant is negative.
Core rules
- Addition and subtraction: combine real parts and imaginary parts separately: (a + bi) + (c + di) = (a + c) + (b + d)i.
- Multiplication: distribute as with binomials, then replace i² with −1. In general, (a + bi)(c + di) = (ac − bd) + (ad + bc)i.
- Division: multiply numerator and denominator by the conjugate of the denominator, c − di. The new denominator is c² + d², which is real.
- Powers of i repeat every four steps: i¹ = i, i² = −1, i³ = −i, i⁴ = 1. Reduce the exponent modulo 4 before evaluating.
- Modulus: |a + bi| = √(a² + b²). It represents distance and is never negative.
- Quadratic formula: when b² − 4ac < 0, the roots are a conjugate pair written with i.
Worked example
Simplify (3 + 2i)/(1 − 4i). Multiply numerator and denominator by the conjugate 1 + 4i:
(3 + 2i)(1 + 4i) / [(1 − 4i)(1 + 4i)] = (3 + 12i + 2i + 8i²) / (1 + 16) = (3 + 14i − 8)/17 = (−5 + 14i)/17.
The quotient is −5/17 + (14/17)i. The denominator check is (1)² + (−4)² = 17, a positive real number.
Traps that cost marks
- Replacing i² with +1 instead of −1.
- Dividing only part of the numerator by the conjugate, or forgetting to multiply both terms by the conjugate.
- Writing the modulus as a² + b² instead of √(a² + b²).
- Mishandling i⁰ = 1 and remainders: remainder 1 gives i, remainder 2 gives −1, remainder 3 gives −i.
- Assuming a real-coefficient quadratic can have exactly one non-real root; non-real roots occur in conjugate pairs.
- Dropping the imaginary unit when adding: 2i + 3i = 5i, not 5.