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Daily Math Minute

Quadratic Functions & Complex Numbers

Operations with Complex Numbers

Adding, subtracting, multiplying, and dividing complex numbers.

Advanced20 min lesson4 min readUpdated August 12, 2026Author not yet attributed

Prerequisites

  • The Imaginary Unit

Arithmetic That Already Knows the Rules

Before reading on, predict how you'd add (3 + 4i) + (2 − 5i), and how you'd multiply (3 + 4i)(2 − 5i). Try treating i as if it were an ordinary variable, like x, for both — then remember one extra fact about it that a normal variable doesn't have.

Complex numbers add and multiply exactly like binomials, since a + bi really is a two-term expression — combine like terms for addition, and use FOIL for multiplication, with one extra simplification step: any i² that appears gets replaced with −1, since that's the one fact ordinary variables don't satisfy.

Worked Example — Adding and Subtracting Complex Numbers

Add (3 + 4i) + (2 − 5i) and subtract (3 + 4i) − (2 − 5i). Adding: combine real parts and imaginary parts separately: (3 + 2) + (4 − 5)i = 5 − i. Subtracting: distribute the negative sign first: (3 − 2) + (4 + 5)i = 1 + 9i.

Worked Example — Multiplying Complex Numbers

Multiply (3 + 4i)(2 − 5i). Use FOIL: 6 − 15i + 8i − 20i². Combine the imaginary terms: 6 − 7i − 20i². Replace i² with −1: 6 − 7i − 20(−1) = 6 − 7i + 20 = 26 − 7i.

Definition — Complex Conjugate

The complex conjugate of a + bi is a − bi — the same real part, with the imaginary part's sign flipped. Multiplying a complex number by its conjugate always produces a real number: (a + bi)(a − bi) = a² + b².

That last fact isn't a coincidence — it's the difference-of-squares pattern from Algebra I, (a + b)(a − b) = a² − b², applied with b replaced by bi: (a + bi)(a − bi) = a² − (bi)² = a² − b²i² = a² − b²(−1) = a² + b². The conjugate is exactly the partner that makes the imaginary part vanish, which is why it's the standard tool for dividing complex numbers — multiplying a fraction's numerator and denominator by the denominator's conjugate clears i from the bottom entirely.

Worked Example — Dividing Complex Numbers Using the Conjugate

Divide (5 + 2i) ÷ (1 − 3i). Multiply numerator and denominator by the denominator's conjugate, 1 + 3i: [(5 + 2i)(1 + 3i)] / [(1 − 3i)(1 + 3i)]. Numerator: 5 + 15i + 2i + 6i² = 5 + 17i − 6 = −1 + 17i. Denominator: 1² + 3² = 10. Result: (−1 + 17i)/10, or −1/10 + (17/10)i.

Complex Plane

3 + 4i = 5∠53.1301°

1 - 2i = 2.2361∠-63.4349°

Operations

z₁ + z₂ =

4+2i4 + 2i
Step-by-step
  1. z₁ + z₂ = (3 + 1) + (4 + -2)i
  2. = 4 + 2i
Forms — z₁
Rectangular
z1=3+4iz₁ = 3 + 4i
Polar
z₁ = 5∠53.1301°
Trigonometric
z₁ = 5(cos 53.1301° + i sin 53.1301°)
Exponential (Euler)
z1=5e0.9273iz₁ = 5e^{0.9273i}

Tip

Multiplying by the conjugate always turns the denominator into a² + b², a real number, exactly because the conjugate pair matches the difference-of-squares pattern — no complex denominator ever survives this step.

Common Mistakes

  • Multiplying two complex numbers by multiplying only the real parts and only the imaginary parts together, instead of using full FOIL.

    Complex multiplication needs all four FOIL products, not just a term-by-term pairing — (3 + 4i)(2 − 5i) requires all of First, Outer, Inner, and Last, the same as any binomial multiplication.

  • Forgetting to multiply both the numerator and denominator by the conjugate, changing the value of the expression.

    Multiplying only the denominator by its conjugate changes the fraction's value — multiply both the numerator and denominator by the same conjugate, which is really just multiplying by a disguised form of 1.

Key Takeaways

  • Complex numbers add, subtract, and multiply like binomials, with i² replaced by −1 whenever it appears.
  • A complex number's conjugate flips the sign of its imaginary part, and their product is always real, via the difference-of-squares pattern.
  • Dividing complex numbers uses the conjugate to clear i from the denominator.

Summary

Complex arithmetic reuses familiar algebra, with one new simplification rule for i². With complex numbers fully defined, the next lesson returns to the discriminant to complete the classification of a quadratic's solutions.

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