Complex numbers and their algebra

x + 1 = 0 and x² – 1 = 0 have real solutions, but x² + 2 = 0 has none: no real number squared gives –2. In Class 10 this was the case 'discriminant negative, no real roots'. To solve such equations we need new numbers.

Notice that √–4 is not –2, because (–2)² = +4. So we invent a number i = √–1, with i² = –1. Then √–4 = √4 × √–1 = 2i, √–9 = 3i and √–5 = √5 i.

A complex number is z = a + ib, where a and b are real.

• a = Re z is the real part and b = Im z is the imaginary part. Note that Im(3 – 4i) is –4, not –4i.
• If b = 0, z is purely real: every real number is a complex number, so R ⊂ C. If a = 0 (and b ≠ 0), z is purely imaginary, like 7i.

Equal complex numbers: a + ib = c + id exactly when a = c and b = d. One complex equation gives two real equations. If (2x – y) + i(x + y) = 3 + 6i, then 2x – y = 3 and x + y = 6, so x = 3, y = 3.

Arithmetic: treat i like a letter, then replace every i² by –1.

• Add or subtract part by part: (4 + 3i) + (2 – 5i) = 6 – 2i.
• Multiply like brackets: (2 + 3i)(4 – i) = 8 – 2i + 12i – 3i² = 11 + 10i.
• Squares and cubes use the usual identities: (2 – i)² = 4 – 4i + i² = 3 – 4i, and (1 + 2i)³ = (–3 + 4i)(1 + 2i) = –11 – 2i.

Powers of i go round in a cycle of four: i, –1, –i, 1, then again i, –1, ... To find iⁿ, divide n by 4 and use the remainder:

• i⁵⁰: 50 = 4 × 12 + 2, so i⁵⁰ = i² = –1.
• i⁻³⁵ = 1/i³⁵; 35 leaves remainder 3, so i³⁵ = –i, and 1/(–i) = i/(–i × i) = i.
• Any four consecutive powers add up to 0, because iⁿ(1 + i + i² + i³) = iⁿ(1 + i – 1 – i) = 0. So i²¹ + i²² + i²³ + i²⁴ = 0.

One trap: √a × √b = √(ab) only when at least one of a, b is not negative. For two negatives it fails: √–4 × √–9 = (2i)(3i) = 6i² = –6, not √36 = 6.