The number that had to be invented
Every number system in your course was invented because the previous one could not solve some equation. x + 5 = 3 forced negative numbers into existence; 2x = 3 forced fractions; x² = 2 forced the irrationals. And x² = −1 forces one more step.
No real number squares to a negative, so we define a new one. The imaginary unit i is defined by i² = −1, and every square root of a negative number can then be written in terms of it: √(−9) = √9 × √(−1) = 3i.
A complex number is anything of the form z = a + bi, where a and b are real. a is the real part Re(z), and b — note, not bi — is the imaginary part Im(z). Every real number is complex with b = 0, so ℝ sits inside ℂ exactly as ℚ sits inside ℝ.
The rule √(ab) = √a·√b breaks here
That rule requires both numbers to be non-negative. Applied carelessly, √(−4) × √(−9) would give √36 = 6, but the correct answer is 2i × 3i = 6i² = −6. Always convert to i first, then multiply. This is the single most common error in the chapter.
Arithmetic: three operations that are ordinary, one that is not
Addition, subtraction and multiplication work exactly as they do for any algebraic expression in one letter. Treat i as if it were a variable, and at the very end replace i² with −1.
Division is the only operation needing a new idea, and the idea is one you have already met when rationalising a surd denominator.
| Operation | Rule | Example |
|---|---|---|
| Add | add real to real, imaginary to imaginary | (3+2i) + (1+5i) = 4 + 7i |
| Subtract | the same, minding the signs | (3+2i) − (1+5i) = 2 − 3i |
| Multiply | expand, then replace i² with −1 | (3+2i)(1+5i) = 3 + 17i + 10i² = −7 + 17i |
| Divide | multiply top and bottom by the conjugate | see the worked example below |
Express (4 + 3i) / (2 − i) in the form a + bi.
- The conjugate of
2 − iis2 + i. Multiply numerator and denominator by it.The conjugate flips the sign of the imaginary part, and a number times its conjugate is always real. - Denominator:
(2 − i)(2 + i) = 4 − i² = 4 + 1 = 5.This is a² + b², not a² − b², because the i² turns the subtraction into an addition. - Numerator:
(4 + 3i)(2 + i) = 8 + 4i + 6i + 3i² = 8 + 10i − 3 = 5 + 10i.Expand fully before simplifying, and deal with i² last. (5 + 10i)/5 = 1 + 2i.Divide both parts by the real denominator. The answer must be written in a + bi form to get the final mark.
1 + 2i
The conjugate and the modulus
The conjugate of z = a + bi is z̄ = a − bi. It has three properties that get used constantly: z + z̄ is real (equal to 2a), z − z̄ is purely imaginary, and z · z̄ = a² + b², which is real and never negative.
That last product is where the modulus comes from. |z| = √(a² + b²) is the distance of the point from the origin on the Argand diagram, so |z|² = z z̄.
Switch to z + w. Adding complex numbers is the parallelogram rule — exactly the same picture as adding two vectors. Then try z × w: the lengths multiply and the angles add, which addition never does.
Solving equations that have no real solutions
The whole point of ℂ is that every polynomial equation now has solutions. A quadratic with a negative discriminant, which you previously described as having "no real roots", has two complex roots instead — and they always come as a conjugate pair.
Solve x² − 4x + 13 = 0.
- Discriminant:
b² − 4ac = 16 − 52 = −36.Negative, so there are no real roots — but two complex ones. - Quadratic formula:
x = [4 ± √(−36)] / 2.The formula still works; only the square root needs new treatment. √(−36) = 6i, sox = (4 ± 6i)/2.Convert to i before doing anything else with the root.x = 2 ± 3i.The two roots 2 + 3i and 2 − 3i are conjugates of each other, which is always true when the coefficients are real.
x = 2 + 3i or x = 2 − 3i
A free check on any quadratic with complex roots
For ax² + bx + c = 0, the roots still satisfy sum = −b/a and product = c/a. Here the sum is (2+3i) + (2−3i) = 4 = −(−4)/1 ✓ and the product is (2+3i)(2−3i) = 4 + 9 = 13 = c/a ✓. Both come out real, as they must.
Before you leave this chapter
- i² = −1, and the powers of i cycle with period 4: i, −1, −i, 1.
- Convert every negative square root into i form before multiplying anything.
- The imaginary part of a + bi is b, not bi.
- Divide by multiplying top and bottom by the conjugate of the denominator.
- |z| = √(a² + b²) and z z̄ = |z|². Complex roots of a real quadratic always come in conjugate pairs.