A complex number extends the reals by introducing the imaginary unit i, defined by i² = −1. Every complex number is written in the standard form z = a + ib, where a = Re(z) is the real part and b = Im(z) is the imaginary part — and crucially both a and b are real numbers (Im(z) is the coefficient b, not ib). This concept is the arithmetic of these numbers: adding, multiplying, dividing, and reading off when the result is real or imaginary.
1 — Powers of i. Because i² = −1, the powers cycle with period 4: i¹ = i, i² = −1, i³ = −i, i⁴ = 1, and then it repeats. To reduce any iⁿ, divide the exponent by 4 and keep the remainder r: iⁿ = iʳ. So i²⁰²⁴ = i⁰ = 1 (2024 is a multiple of 4) and i⁵⁷⁴ = i² = −1 (574 = 4·143 + 2). Negative powers use the same wheel: i⁻¹ = 1/i = −i, i⁻² = −1. A neat consequence: any four consecutive powers of i sum to 0 (iⁿ + iⁿ⁺¹ + iⁿ⁺² + iⁿ⁺³ = 0), which collapses long sums like i + i² + … + iⁿ to at most three surviving terms.
2 — Square roots of negative reals. For a > 0, √(−a) = i√a (so √(−16) = 4i). Danger: the rule √x · √y = √(xy) is valid only when at least one of x, y is non-negative. With two negatives you must convert to i-form first: √(−4) · √(−9) = (2i)(3i) = 6i² = −6, not √36 = 6. Mishandling this sign is one of the most common slips in the chapter.
3 — Addition, subtraction, multiplication. Add/subtract componentwise: (a + ib) ± (c + id) = (a ± c) + i(b ± d). Multiply like binomials and replace i² = −1: (a + ib)(c + id) = (ac − bd) + i(ad + bc). Useful special products: (1 + i)² = 2i, (1 − i)² = −2i, and a number times its conjugate gives a real number: (a + ib)(a − ib) = a² + b².
4 — The conjugate and division. The conjugate of z = a + ib is z̄ = a − ib (flip the sign of the imaginary part). To write a quotient in standard form, multiply top and bottom by the conjugate of the denominator — this makes the denominator real:
(a + ib)/(c + id) = (a + ib)(c − id) / (c² + d²).
For example (2 + 3i)/(1 − i) = (2 + 3i)(1 + i)/2 = (−1 + 5i)/2. The multiplicative inverse is the special case z⁻¹ = z̄ / (a² + b²) (dividing 1 by z). The additive inverse is simply −z = −a − ib.
5 — Powers of (1 ± i). These recur constantly. Build them from (1 + i)² = 2i: (1 + i)⁴ = (2i)² = −4, (1 + i)⁸ = 16, and (1 + i)/(1 − i) = i (so ((1 + i)/(1 − i))ⁿ = iⁿ). Reducing a high power to a power of 2i or of i beats multiplying it out — and stays inside pure algebra (no polar form needed here). …