Add like arrows, multiply with i² = −1, divide by clearing i from the bottom.
Once a complex number has an address on the plane — a point, an arrow, a length and a direction (that was 31.2) — the natural next question is: can we do arithmetic with these new numbers? The answer is a clean yes, and three of the four operations cost nothing new. Addition and subtraction are exactly the triangle rule for arrows you met with vectors (vector operations): add real to real, imaginary to imaginary. Multiplication looks like ordinary binomial expansion — with a single twist, the defining rule i² = −1, doing all the work. And division, the only one that needs a trick, clears i from the bottom by multiplying through by the conjugate. By the end every sum, product, power of i, and quotient of complex numbers will itself be a complex number — ℂ never spills outside itself.
A complex number carries two independent amounts: a real part and an imaginary part. They never mix, so to add two complex numbers you simply add the matching parts:
(a + b i) + (c + d i) = (a + c) + (b + d) i, and (a + b i) − (c + d i) = (a − c) + (b − d) i.
For example, with z = 3 + i and w = 1 + 2i,
z + w = (3 + 1) + (1 + 2) i = 4 + 3i, z − w = (3 − 1) + (1 − 2) i = 2 − i.
Because a complex number is the arrow from O to its point, this component arithmetic is exactly the triangle rule from Stage 27: lay w's tail at z's tip, and the arrow from O to the far end is z + w. Subtraction z − w is the arrow that runs from the tip of w to the tip of z (it points "from w to z").
Adding or subtracting complex numbers is two real additions at once — one on the real axis, one on the imaginary axis — which is precisely why the picture is the triangle rule. Nothing about i² is needed yet.
Multiplying two complex numbers is the one operation that is genuinely new — and yet it asks nothing you don't already know. Treat a + b i like an ordinary binomial, multiply everything out, and then use the only special fact about i to finish: i² = −1.
Take z = 2 + 3i and w = 1 − i. Expand term by term:
(2 + 3i)(1 − i) = 2·1 + 2·(−i) + 3i·1 + 3i·(−i) = 2 − 2i + 3i − 3i².
Now swap i² = −1, so −3i² = −3·(−1) = +3. Collect the real parts (2 + 3) and the imaginary parts (−2i + 3i):
= (2 + 3) + (−2 + 3) i = 5 + i.
The same expansion in letters gives the general rule, with its tell-tale cross terms:
(a + b i)(c + d i) = (ac − bd) + (ad + bc) i.
Two traps live here. First, i² = −1, not +1 — forgetting the sign flips the whole real part. Second, do not drop the cross terms ad + bc: writing only (ac) + (bd) i ignores the middle of the expansion. Multiply like a careful binomial, every time.
Multiplication has one beautiful by-product: the powers of i never sprawl. Start multiplying i by itself and the four answers chase their tails:
i¹ = i, i² = −1, i³ = i²·i = −i, i⁴ = i²·i² = (−1)(−1) = 1.
Then it repeats: i⁵ = i⁴·i = i, i⁶ = −1, and so on, looping 1, i, −1, −i like the four seasons. So to find any power iⁿ, you only need n mod 4 — the remainder when n is divided by 4. For instance,
i²⁰²³ : 2023 = 4·505 + 3, so 2023 mod 4 = 3, hence i²⁰²³ = i³ = −i.
Geometrically the four values are the four points where the unit circle meets the axes — 1 (east), i (north), −1 (west), −i (south) — and each extra factor of i steps you one quarter-turn around. (Why a quarter-turn? That is the rotation story of 31.5.)
Division is the only operation that needs a small move, and the move is the workhorse of the whole stage: the conjugate. The conjugate of c + d i is c − d i (same real part, imaginary part negated). Its magic is that a number times its conjugate is real:
(c + d i)(c − d i) = c² − (d i)² = c² + d² = |w|², a real number.
So to divide z by w, multiply top and bottom by the conjugate of the denominator w̄ — which is multiplying by w̄/w̄ = 1, changing nothing — and the bottom turns real:
z / w = z · w̄|w|².
Worked out, with z = 2 + 3i and w = 1 − i (so w̄ = 1 + i):
2 + 3i1 − i = (2 + 3i)(1 + i)(1 − i)(1 + i) = −1 + 5i2 = −0.5 + 2.5i.
(Top: (2 + 3i)(1 + i) = 2 + 2i + 3i + 3i² = 2 + 5i − 3 = −1 + 5i. Bottom: (1 − i)(1 + i) = 1 + 1 = 2.) You can always check a quotient by multiplying back: (−0.5 + 2.5i) · (1 − i) = 2 + 3i. ✓
The conjugate must go on both the top and the bottom — that is what keeps the value unchanged (you multiplied by 1). Multiplying only the bottom would change the number. And the bottom becomes |w|², which is positive and real — never leave an i downstairs.
Add, subtract, multiply, divide (by anything nonzero) — the answer is always another complex number. The number system ℂ is closed under all four operations; we never need to invent anything beyond a + b i again. (And remember: ℂ has no order — you may compare the sizes |z|, but "z < w" between two non-real numbers has no meaning.)
| Operation | Rule | Example |
|---|---|---|
| Add / subtract | (a+bi) ± (c+di) = (a±c) + (b±d) i | (3+i)+(1+2i) = 4+3i |
| Multiply | (a+bi)(c+di) = (ac−bd) + (ad+bc) i | (2+3i)(1−i) = 5+i |
| Powers of i | i, −1, −i, 1 — repeat (use n mod 4) | i²⁰²³ = i³ = −i |
| Divide | z / w = z·w̄ / |w|² | (2+3i)/(1−i) = −0.5+2.5i |
The conjugate w̄ just earned its keep clearing denominators. Next, in 31.4, we study it on its own — the mirror image z̄ = a − b i — and find that z·z̄ = |z|² turns hard problems about z into clean real arithmetic.
Compute (5 + 2i) + (−3 + 4i) and (5 + 2i) − (−3 + 4i).
Add the parts: sum = (5 − 3) + (2 + 4) i = 2 + 6i. Subtract: (5 − (−3)) + (2 − 4) i = 8 − 2i.
Multiply (4 − i)(2 + 3i), showing the cross terms and the i² substitution.
(4 − i)(2 + 3i) = 8 + 12i − 2i − 3i² = 8 + 10i − 3(−1) = (8 + 3) + 10i = 11 + 10i.
Evaluate i¹⁰⁰ and i⁻³.
100 mod 4 = 0, so i¹⁰⁰ = i⁰ = 1. For i⁻³, use −3 mod 4 = 1 (since −3 = 4·(−1) + 1), so i⁻³ = i¹ = i. (Check: i⁻³ = 1/i³ = 1/(−i) = i.)
Show that (1 + i)² = 2i, then use it to find (1 + i)⁴.
(1 + i)² = 1 + 2i + i² = 1 + 2i − 1 = 2i. Then (1 + i)⁴ = (2i)² = 4i² = −4.
Divide (3 + i) / (2 + i), and verify by multiplying back.
Multiply top and bottom by w̄ = 2 − i: top (3 + i)(2 − i) = 6 − 3i + 2i − i² = 6 − i + 1 = 7 − i; bottom (2 + i)(2 − i) = 4 + 1 = 5. So the quotient is (7 − i)/5 = 1.4 − 0.2i. Check: (1.4 − 0.2i)(2 + i) = 2.8 + 1.4i − 0.4i − 0.2i² = 2.8 + i + 0.2 = 3 + i. ✓
Find real numbers x and y with (x + y) + (x − y) i = 4 + 2i.
Match real and imaginary parts: x + y = 4 and x − y = 2. Adding gives 2x = 6, so x = 3; then y = 1. So x = 3, y = 1. (One complex equation is two real equations — the same idea you'll lean on in 31.4.)
Six questions to lock it in. Tap the answer you think is right.
This lesson develops the four arithmetic operations on complex numbers, aligned with CCSS HSN-CN.A.1 (the form a + b i, with i² = −1) and HSN-CN.A.2 (add, subtract, and multiply complex numbers using the commutative, associative, and distributive properties together with i² = −1). The geometric reading of addition and subtraction as the triangle rule connects back to vectors (Stage 27), and clearing the denominator with the conjugate previews HSN-CN.A.3 (conjugates, moduli, quotients), the subject of the next lesson. Encourage students to expand products like ordinary binomials and substitute i² = −1 only at the end, and to always sanity-check a quotient by multiplying back.