A complex number is written $a + bi$: a real part $a$ and an imaginary part $b$. You add and multiply them with the algebra you already know from multiplying binomials, plus one new fact from The Imaginary Unit i: $i^2 = -1$.
Idea clave
Treat $i$ like a variable: combine real parts with real parts and $i$ terms with $i$ terms. When you multiply, every $i^2$ becomes $-1$, which turns that term real. Always finish in the form $a + bi$.
Por qué funciona
Complex numbers follow the same rules as polynomials in $i$ (commutative, associative, distributive). So $(2 + 3i) + (4 - i)$ works just like $(2 + 3x) + (4 - x)$. The only difference shows up in multiplication: an $x^2$ term would stay, but $i^2$ is the number $-1$. In the plane, adding complex numbers adds arrows tip to tail, exactly like adding vectors.
Worked examples
Ejemplo resuelto: Adding and subtracting
Simplify $(5 - 2i) - (3 + 4i)$.
- Distribute the minus sign: $5 - 2i - 3 - 4i$.
- Real parts: $5 - 3 = 2$. Imaginary parts: $-2i - 4i = -6i$.
Respuesta: $2 - 6i$
Ejemplo resuelto: Multiplying
Simplify $(2 + 3i)(4 - i)$.
- FOIL: $8 - 2i + 12i - 3i^2$.
- Replace $i^2$ with $-1$: $-3i^2 = +3$.
- Combine: $8 + 3 + 10i = 11 + 10i$.
Respuesta: $11 + 10i$
Ejemplo resuelto: A number times its conjugate
Simplify $(3 + 4i)(3 - 4i)$.
- FOIL: $9 - 12i + 12i - 16i^2$.
- The $i$ terms cancel, and $-16i^2 = 16$: $9 + 16 = 25$.
Respuesta: $25$, a real number. In general $(a + bi)(a - bi) = a^2 + b^2$.
Error común
Leaving $i^2$ in the answer, or replacing it with $+1$. In $(2 + 3i)(4 - i)$ the last product is $3i \cdot (-i) = -3i^2 = -3(-1) = +3$. If you got $8 - 3 + 10i = 5 + 10i$, you used $i^2 = 1$. A quick check: $|2 + 3i|^2 \cdot |4 - i|^2 = 13 \cdot 17 = 221$, and $11^2 + 10^2 = 221$ too.