ACT Math: Complex Numbers
Complex numbers appear in just a question or two per ACT, but they follow such predictable rules that they're practically free points once you know the playbook. This free, complete lesson covers the imaginary unit and powers of i, adding, subtracting, and multiplying complex numbers, complex conjugates and division, and complex numbers in quadratic equations. Each idea is followed by a fully worked example, and then a set of original practice problems with instant feedback, in the same four-answer-choice, calculator-allowed format as the current ACT. Everything here is free, and you can keep practicing afterward with the full ACT Math Question Bank linked below.
1. The Imaginary Unit and Powers of i
The imaginary unit i is defined as √(−1), so i² = −1. A complex number is written in standard form a + bi, where a is the real part and b is the imaginary part. Powers of i repeat in a cycle of 4:
i¹ = i i² = −1 i³ = −i i⁴ = 1
For a large exponent, divide by 4 and use the remainder to find the equivalent smaller power.
Worked Example: Simplify i⁷.
7 ÷ 4 leaves a remainder of 3, so i⁷ behaves the same as i³.
i⁷ = i³ = −i
1. Simplify i³.
Explanation: i³ = i² × i = (−1)(i) = −i.
2. Simplify i¹⁰.
Explanation: 10 ÷ 4 leaves a remainder of 2, so i¹⁰ behaves like i² = −1.
3. Simplify i¹³.
Explanation: 13 ÷ 4 leaves a remainder of 1, so i¹³ behaves like i¹ = i.
4. Simplify i⁹⁹.
Explanation: 99 ÷ 4 leaves a remainder of 3 (96 is the nearest multiple of 4), so i⁹⁹ behaves like i³ = −i.
5. Simplify √(−16).
Explanation: √(−16) = √16 × √(−1) = 4i.
2. Adding, Subtracting, and Multiplying Complex Numbers
Add or subtract complex numbers by combining the real parts together and the imaginary parts together, just like combining like terms. Multiply complex numbers the same way you'd multiply two binomials (FOIL), then simplify using i² = −1.
Worked Example: Add (5 − 2i) + (7 + 3i).
Real parts: 5 + 7 = 12. Imaginary parts: −2 + 3 = 1.
Result: 12 + i
1. Add: (3 + 4i) + (6 − 2i)
Explanation: Real: 3 + 6 = 9. Imaginary: 4 − 2 = 2. Result: 9 + 2i.
2. Subtract: (8 + 5i) − (3 − 2i)
Explanation: Real: 8 − 3 = 5. Imaginary: 5 − (−2) = 7. Result: 5 + 7i.
3. Multiply: (2 + 3i)(4 − i)
Explanation: FOIL: 8 − 2i + 12i − 3i² = 8 + 10i − 3(−1) = 8 + 10i + 3 = 11 + 10i.
4. Multiply: (1 + i)(1 − i)
Explanation: FOIL: 1 − i + i − i² = 1 − i² = 1 − (−1) = 2.
5. Simplify: (3 + 2i) + (3 − 2i)
Explanation: Real: 3 + 3 = 6. Imaginary: 2 − 2 = 0. Result: 6, a purely real number.
3. Complex Conjugates and Division
The complex conjugate of a + bi is a − bi, simply flip the sign of the imaginary part. The product of a complex number and its conjugate is always a real number, which is why conjugates are the key to dividing complex numbers: multiply the numerator and denominator by the conjugate of the denominator.
Worked Example: Find the complex conjugate of 3 − 4i, and their product.
Conjugate: 3 + 4i.
Product: (3 − 4i)(3 + 4i) = 9 − 16i² = 9 + 16 = 25
1. Find the complex conjugate of 5 + 7i.
Explanation: Flip the sign of the imaginary part only: 5 − 7i.
2. Find the product of (2 − 3i) and its complex conjugate.
Explanation: (2 − 3i)(2 + 3i) = 4 − 9i² = 4 + 9 = 13.
3. Divide: 4 + 10i2 − 3i, using the complex conjugate.
Explanation: Multiply top and bottom by (2 + 3i). Numerator: (4 + 10i)(2 + 3i) = 8 + 12i + 20i + 30i² = 8 + 32i − 30 = −22 + 32i. Denominator: (2 − 3i)(2 + 3i) = 4 + 9 = 13. Result: −22 + 32i13.
4. What is true about the product of any complex number and its conjugate?
Explanation: (a + bi)(a − bi) = a² − b²i² = a² + b², always a real number, since the imaginary terms cancel out.
5. Simplify: 6i2i
Explanation: The i in the numerator and denominator cancel directly: 6i2i = 3.
4. Complex Numbers in Quadratic Equations
When a quadratic equation's discriminant (b² − 4ac) is negative, its solutions are complex numbers. For a quadratic with real coefficients, complex roots always occur in conjugate pairs.
Worked Example: Solve x² + 9 = 0.
x² = −9, so x = ±3i
1. Solve: x² + 25 = 0
Explanation: x² = −25, so x = ±√(−25) = ±5i.
2. Solve x² + 4 = 0, and express the solutions in a + bi form.
Explanation: x² = −4, so x = ±2i, which has a real part of 0: x = 0 ± 2i.
3. A quadratic equation with real coefficients has one complex root of 3 − 2i. What must its other root be?
Explanation: Complex roots of a quadratic with real coefficients always come in conjugate pairs, so the other root must be 3 + 2i.
4. Use the quadratic formula to solve x² − 2x + 5 = 0.
Explanation: a = 1, b = −2, c = 5. Discriminant: (−2)² − 4(1)(5) = 4 − 20 = −16. x = 2 ± √(−16)2 = 2 ± 4i2 = 1 ± 2i.
Common Mistakes to Avoid
- Forgetting to simplify i² to −1. This is the single most common source of sign errors when multiplying complex numbers, always substitute i² = −1 as your last step.
- Treating i like a normal variable when it comes to its powers. Unlike x, i cycles through only four distinct values (i, −1, −i, 1) no matter how large the exponent gets.
- Forgetting to distribute properly when squaring a binomial with i. (a + bi)² needs full FOIL treatment, a² + 2abi + b²i², not just squaring each term separately.
- Flipping the wrong sign when finding a conjugate. Only the sign of the imaginary part changes; the real part stays exactly the same.
- Forgetting that dividing by i (or any pure imaginary number) still requires a conjugate. Even abi should be cleared of i in the denominator, treat bi's own conjugate as −bi.
Frequently Asked Questions
Why does the pattern for powers of i repeat every 4 exponents?
Since i⁴ = i² × i² = (−1)(−1) = 1, multiplying by i⁴ always returns you to where you started. That means every fourth power resets the cycle, and any exponent's remainder after dividing by 4 tells you exactly where in the cycle you land.
Why does dividing complex numbers require a conjugate?
Multiplying the denominator by its own conjugate always produces a real number, this clears the imaginary part out of the denominator entirely, leaving a standard a + bi form for the final answer.
How do complex numbers connect to the quadratic formula?
Whenever the discriminant (b² − 4ac) is negative, the quadratic formula requires taking the square root of a negative number, which produces two complex conjugate solutions instead of real ones.
Where can I practice more problems like these?
The Complex Numbers quizzes in the ACT Math Question Bank include additional original problems on this topic, along with quizzes covering every other topic tested on the ACT Math section.