AP PRECALCULUS — PREREQUISITE REVIEW
Complex Numbers
Notes — Prerequisite Topic 8
💡 Learning Objectives
By the end of this lesson you will be able to:
- Define the imaginary unit i and compute powers of i
- Identify the real and imaginary parts of a complex number written as a + bi
- Add, subtract, multiply, and divide complex numbers
- Compute the complex conjugate and use it to rationalize denominators
- Plot complex numbers on the complex (Argand) plane and compute the modulus
- Use complex numbers to express roots of quadratics with negative discriminants
1. Why We Need Complex Numbers
Over the real numbers, equations like x² = −1 have no solution. Extending the number system by defining a new element whose square is −1 repairs this gap and — remarkably — gives us a system in which every polynomial equation has a solution.
💡 The imaginary unit
Define the imaginary unit i by the rule:
i² = −1, equivalently i = √(−1).
Using this, any negative square root can be written with i:
√(−25) = √25 · √(−1) = 5i, √(−18) = 3i√2.
2. Form of a Complex Number
A complex number is an expression:
z = a + bi,
where a and b are real numbers. a is the real part (Re z = a) and b is the imaginary part (Im z = b). If b = 0, z is a real number. If a = 0 and b ≠ 0, z is a pure imaginary number.
Every complex number a + bi corresponds to the point (a, b) in the complex plane, also called the Argand plane. The horizontal axis is the real axis; the vertical axis is the imaginary axis.
3. Powers of i
💡 A repeating pattern
i¹ = i, i² = −1, i³ = −i, i⁴ = 1.
Powers of i cycle through these four values. To evaluate iⁿ for any whole n, divide n by 4 and use the remainder.
📘 Example — Computing high powers
i²⁷: 27 ÷ 4 leaves remainder 3, so i²⁷ = i³ = −i.
i¹⁰⁰: 100 ÷ 4 leaves remainder 0, so i¹⁰⁰ = i⁰ = 1.
i⁻¹: 1/i = i/(i · i) = i/(−1) = −i.
4. Arithmetic with Complex Numbers
4.1 Addition and subtraction
Add and subtract the real and imaginary parts separately, as if i were a variable.
📘 Example — Adding and subtracting
(3 + 4i) + (5 − 2i) = (3 + 5) + (4 − 2)i = 8 + 2i.
(7 − i) − (2 + 6i) = (7 − 2) + (−1 − 6)i = 5 − 7i.
4.2 Multiplication
Multiply as with binomials, remembering to replace i² with −1 at the end.
📘 Example — Multiplying
(2 + 3i)(4 − i):
= 8 − 2i + 12i − 3i²
= 8 + 10i − 3(−1)
= 11 + 10i.
4.3 The complex conjugate
A complex number and its conjugate are reflections of each other across the real axis.
💡 Conjugate properties
The conjugate of z = a + bi is z̄ = a − bi.
• z + z̄ = 2a (a real number)
• z · z̄ = a² + b² (a non-negative real number)
• z̄ reflects z across the real axis
4.4 Division by a conjugate
To divide two complex numbers, multiply the numerator and denominator by the conjugate of the denominator. This forces the denominator to become real.
📘 Example — Dividing complex numbers
Simplify (3 + 2i) / (1 − i).
Multiply by the conjugate (1 + i):
= (3 + 2i)(1 + i) / ((1 − i)(1 + i))
= (3 + 3i + 2i + 2i²) / (1 − i²)
= (3 + 5i − 2) / (1 + 1)
= (1 + 5i) / 2 = 1/2 + (5/2)i.
5. Modulus and Geometric Interpretation
The modulus (or absolute value) of z = a + bi is the distance from the origin to the point (a, b) in the complex plane:
|z| = √(a² + b²).
📘 Example — Computing a modulus
|3 + 4i| = √(9 + 16) = √25 = 5.
|−2 + 5i| = √(4 + 25) = √29.
💡 Useful identities
• |z|² = z · z̄
• |z · w| = |z| · |w|
• |z + w| ≤ |z| + |w| (the triangle inequality)
6. Complex Numbers and Quadratic Equations
If the discriminant Δ = b² − 4ac of a quadratic ax² + bx + c = 0 is negative, the quadratic has no real solutions — but it always has two complex solutions, which come as a conjugate pair.
📘 Example — Solving a quadratic over the complex numbers
Solve x² − 4x + 13 = 0.
Δ = 16 − 52 = −36, so x = (4 ± √(−36)) / 2 = (4 ± 6i) / 2 = 2 ± 3i.
The two solutions, 2 + 3i and 2 − 3i, are complex conjugates.
💡 The conjugate root theorem
If a polynomial has real coefficients and (a + bi) is a root with b ≠ 0, then (a − bi) is also a root. Complex roots of real polynomials come in conjugate pairs.
7. Putting It Together
📘 Example — A longer simplification
Simplify (2 + i)² − 3i(1 − 2i).
(2 + i)² = 4 + 4i + i² = 4 + 4i − 1 = 3 + 4i.
3i(1 − 2i) = 3i − 6i² = 3i + 6 = 6 + 3i.
(3 + 4i) − (6 + 3i) = −3 + i.
8. Summary
- i is defined by i² = −1; powers of i cycle through {i, −1, −i, 1}
- Every complex number has the form a + bi with a and b real
- Add, subtract, and multiply as polynomials; replace i² with −1
- The conjugate of a + bi is a − bi; use conjugates to divide
- The modulus |a + bi| = √(a² + b²) is the distance from the origin
- Quadratics with Δ < 0 have two complex conjugate roots