Imaginary & Complex Numbers
Inventing i so that √(−1) has an answer.
The explanation
No real number squares to a negative, so mathematicians defined a new one: i, with i² = −1.
That makes √(−25) = 5i, since (5i)² = 25i² = −25.
A complex number combines a real and an imaginary part: a + bi. In 3 − 4i, the real part is 3 and the imaginary part is −4.
Powers of i cycle every four: i¹ = i, i² = −1, i³ = −i, i⁴ = 1, then it repeats. To find a high power, divide the exponent by 4 and use the remainder.
Add and subtract by combining like parts. Multiply with FOIL, then replace i² with −1.
The imaginary unit is defined by i² = −1, and ℂ is the set {a + bi : a, b ∈ ℝ}. This extension makes ℂ algebraically closed: every non-constant polynomial has a root, which is the Fundamental Theorem of Algebra.
Powers of i are periodic with period 4, so i^n = i^{n mod 4}.
Arithmetic follows the field axioms with the single extra rule i² = −1. Multiplication uses FOIL and then that substitution.
Division uses the complex conjugate a − bi. Since (a + bi)(a − bi) = a² + b² is real and non-negative, multiplying numerator and denominator by the conjugate rationalises the denominator — structurally identical to rationalising √ denominators with conjugates.
One caution: the identity √a·√b = √(ab) requires non-negative arguments. Writing √(−4)·√(−9) = √36 = 6 is wrong; converting to 2i·3i = 6i² = −6 first is the correct route. Always extract i before multiplying radicals of negatives.
Worked example
Simplify (3 + 2i)(4 − 5i) and find i²³.
- FOIL: 12 − 15i + 8i − 10i².
- Replace i² with −1: 12 − 7i + 10.
- Combine: 22 − 7i.
- 23 ÷ 4 leaves remainder 3, so i²³ = i³ = −i.
Answer: 22 − 7i, and i²³ = −i
Common mistakes
- Leaving i² in an answer instead of replacing it with −1.
- Multiplying √(−4)·√(−9) as √36.