8.8Complex & Imaginary Numbers
When the discriminant is negative, the square root of a negative number introduces i = √(−1). Complex numbers a + bi extend the real number system and give every quadratic equation exactly two solutions.
Why This Matters
Imaginary numbers complete the number system — they arise naturally when the discriminant is negative and are essential in electrical engineering, signal processing, and advanced mathematics.
Workbook
Lesson, vocabulary, worked examples, and practice problems.
Essential Question
What is the imaginary unit i, how do complex numbers extend the real number system, and how do you find complex solutions of a quadratic equation?
Lesson Overview
The imaginary unit i is defined as i = √(−1), so i² = −1. A complex number has the form a + bi where a is the real part and b is the imaginary part. When b = 0 the number is real; when a = 0 it is pure imaginary. The powers of i cycle with period 4: i¹ = i, i² = −1, i³ = −i, i⁴ = 1. Complex numbers are added by combining like parts and multiplied using FOIL (replacing i² with −1). When a quadratic has D < 0, its two solutions are complex conjugates of the form a ± bi.
The Number System — Hierarchy
Complex Numbers: a + bi
All numbers of the form a + bi where a, b are real
Real Numbers (b=0)
Pure Imaginary (a=0)
Powers of i — The Cycle of 4
i¹
i
i²
−1
i³
−i
i⁴
1
The pattern repeats every 4 powers. To find iⁿ, divide n by 4 and use the remainder: remainder 1 → i, remainder 2 → −1, remainder 3 → −i, remainder 0 → 1.
Example: i¹⁷ → 17 ÷ 4 = 4 remainder 1 → i¹⁷ = i
Worked Examples
Simplify √(−16)
√(−16) = √(16 · −1)
= √16 · √(−1)
= 4 · i
Simplify √(−12)
√(−12) = √(4 · 3 · −1)
= √4 · √3 · √(−1)
= 2√3 · i
Add: (3 + 2i) + (1 − 5i)
Combine real parts: 3 + 1 = 4
Combine imaginary parts: 2i + (−5i) = −3i
Multiply: (2 + 3i)(1 − i)
FOIL: 2(1) + 2(−i) + 3i(1) + 3i(−i)
= 2 − 2i + 3i − 3i²
Replace i² = −1: = 2 + i − 3(−1)
= 2 + i + 3 = 5 + i
Solve x² + 4x + 13 = 0
a=1, b=4, c=13
D = 16 − 52 = −36
x = (−4 ± √(−36)) / 2
√(−36) = 6i
x = (−4 ± 6i) / 2 = −2 ± 3i
Guided Practice
Guided Practice Video: Complex and Imaginary Numbers
Review imaginary unit i, simplifying square roots of negatives, and arithmetic with complex numbers before completing the guided problems below.
Video by Sang Real Math
Watch on YouTube ↗Simplify √(−49).
Hint: √(−49) = √49 · √(−1). What is √49?
Add: (5 − 3i) + (−2 + 7i).
Hint: Combine real parts: 5+(−2). Combine imaginary parts: −3i+7i.
Multiply: (1 + 2i)(3 + i).
Hint: Use FOIL. Remember i² = −1. Combine real and imaginary parts.
Solve x² + 6x + 13 = 0.
Hint: Compute D = 36 − 52 = −16. Then x = (−6 ± √(−16))/2 = (−6 ± 4i)/2.
Simplify i²³.
Hint: Divide 23 by 4: 23 = 4·5 + 3. Remainder 3 → i³ = −i.
Key Vocabulary
Imaginary Unit (i)
i = √(−1), so i² = −1. The basis of all imaginary and complex numbers.
Example: √(−9) = 3i
Complex Number
A number of the form a + bi where a and b are real numbers. a is the real part; b is the imaginary part.
Example: 3 + 4i, −1 − 2i
Real Part
The a in a + bi. The part of a complex number that lies on the real number line.
Imaginary Part
The b in a + bi (the coefficient of i). Note: the imaginary part is a real number.
Complex Conjugates
Two complex numbers of the form a + bi and a − bi. Their product is always real: (a+bi)(a−bi) = a²+b².
Example: 3+2i and 3−2i
Pure Imaginary Number
A complex number with a = 0: 0 + bi = bi. Has no real part.
Example: 5i, −3i, 2i√3
Practice Questions
Interactive Practice — 5 Questions
What is the value of i²?
Simplify √(−25).
Add: (4 + 3i) + (2 − 7i)
Multiply: (2 + i)(3 − 2i)
Solve x² + 9 = 0.
Independent Practice
Independent Practice
Simplify: √(−36)
Add: (2 + 5i) + (4 − 3i)
Multiply: (3 − i)(2 + 4i)
Solve: x² + 2x + 5 = 0. Express solutions in a + bi form.
Challenge: Simplify (1 + i)⁴ using repeated multiplication. What do you notice?
ChallengeCommon Mistakes
Writing √(−16) = −4 instead of 4i.
√(−16) = 4i. The square root of a negative number is imaginary, not negative. √(−16) ≠ −4.
Forgetting to replace i² with −1 when multiplying complex numbers.
After FOIL, every i² term must be replaced with −1. This changes the sign of that term.
Adding complex numbers incorrectly — combining a real part with an imaginary part.
Only combine like parts: real with real, imaginary with imaginary. (3+2i)+(1+4i) = 4+6i, not 10i.
Writing complex solutions as x = ±bi instead of x = a ± bi.
Complex solutions from the quadratic formula are a ± bi where a = −b/(2a) from the formula. Both the real and imaginary parts must be included.
Math Tips
Remember the cycle: i¹=i, i²=−1, i³=−i, i⁴=1. For any power, divide the exponent by 4 and use the remainder.
Complex solutions always come in conjugate pairs: if a+bi is a solution, then a−bi is also a solution.
To multiply complex numbers, use FOIL exactly as with binomials — then substitute i²=−1 at the end.
The product of complex conjugates is always real: (a+bi)(a−bi) = a²+b². This is used to simplify complex fractions.