Simplifying (3 + 2i)(4 - 5i)

Algebra 2 · Axiom Academy

EXAMPLE Simplifying (3 + 2i)(4 - 5i) Multiply two complex numbers with FOIL, and see exactly where i^2 = -1 comes into play. Simplify the product of two complex numbers, (3 + 2i)(4 - 5i) , and write the result in standard form a + bi . Nice work — you multiplied two complex numbers with FOIL and used the one identity that makes it different from ordinary algebra. The pieces worth keeping: FOIL still applies: a complex number a+bi has the same two-term shape as any binomial, so First, Outer, Inner, Last works exactly as it does in real-number algebra. The critical step is i^2 = -1 : substituting it turns the Last term from imaginary into real — miss this step and the answer stays wrong. Standard form is a+bi : combine the real terms into a and the imaginary terms into the coefficient b , keeping them separate until the final step. Result: (3+2i)(4-5i) = 22 - 7i . This same FOIL-then-substitute routine works for any product of two complex numbers — practice it on a few more to make it automatic.

This is the written version of the interactive lesson above. See the full Algebra 2 course.