Multiplying (2+3i)(4-i)

Complex Analysis · Axiom Academy

EXAMPLE Multiplying Complex Numbers Use FOIL to multiply two complex numbers, then simplify with i^2 = -1 to land in standard form a + bi . Multiply the two complex numbers (2 + 3i)(4 - i) and write the result in standard form a + bi . Nice work — you multiplied two complex numbers exactly the way you multiply binomials, with one extra rule. FOIL still applies: a complex number is just a binomial, so (2+3i)(4-i) expands to four products: 8 , -2i , 12i , and -3i^2 . The one new rule: i^2 = -1 , which turns -3i^2 into +3 — an imaginary-squared term becomes a real number. Result: collect like terms — real 8 + 3 = 11 , imaginary -2i + 12i = 10i — so (2+3i)(4-i) = 11 + 10i . Because i^2 always collapses to a real number, the product of two complex numbers is always another complex number a + bi .

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