ACT Math study guide · Chapter 3
Real and Complex Numbers
A rational number can be written as a ratio of two integers. This includes every integer, every terminating or repeating decimal, and every fraction, even ones like 22/7 that are commonly used to approximate an irrational number. An irrational number cannot be written as a ratio of integers; its decimal expansion never terminates and never repeats. A number's rationality depends on what it equals, not on how familiar or unfamiliar it looks.
The imaginary unit i is defined by i² = −1. Powers of i cycle in a pattern of four: i¹ = i, i² = −1, i³ = −i, i⁴ = 1, and then the pattern repeats. A high power of i is reduced by dividing the exponent by 4 and using only the remainder, since i to any multiple of 4 equals 1.
A complex number has the form a + bi, where a and b are real numbers. Adding or subtracting complex numbers combines the real parts together and the imaginary parts together, separately. Multiplying complex numbers uses ordinary distribution, followed by replacing every i² with −1. Dividing by a complex number requires multiplying the numerator and denominator by the conjugate of the denominator, which clears the imaginary part from the denominator entirely.
Where this appears as a model: a quadratic model that predicts no real outcome. If a launched object's height is modeled by −16t²+20t+5, solving for when it reaches 50 feet gives a negative discriminant, −2480; the complex result confirms the object never actually reaches that height.
▸ Powers of i
Worked Example Simplify i²⁹.
Divide the exponent by 4 and use the remainder.
29 = 4(7) + 1
i²⁹ = (i⁴)⁷·i¹ = 1·i = i
▸ Complex arithmetic and the discriminant
A quadratic equation has complex solutions exactly when its discriminant is negative. This connects complex numbers directly to a skill from the next Part: a question can ask for the range of a parameter that forces complex solutions, which reduces to solving an inequality built from the discriminant.
Worked Example For what values of c does x² + 6x + c = 0 have complex solutions?
Complex solutions require a negative discriminant.
36 − 4c < 0
c > 9
▸ Practice
1. Simplify (2 + 3i) + (4 − 5i).
A) −2 + 8i B) 6 + 8i C) 6 − 2i D) −2 − 2i
2. Which of the following is irrational?
F) √7 G) 0.454545... H) √16 J) 22/7
3. Simplify (1 + i)(1 − i).
A) 0 B) 2 C) 2i D) −2i
4. Which expression equals i²³?
F) i G) 1 H) −i J) −1
5. For which values of k does x² − 4x + k = 0 have complex solutions?
A) k < 4 B) k > 4 C) k = 4 D) k ≥ 4
6. Divide: (6 + 2i) ÷ (2i).
F) 1 + 3i G) 1 − 3i H) 3 + i J) −1 − 3i
7. Simplify (3 + 4i) − (1 − 2i).
A) 6 + 2i B) 2 + 6i C) −2 + 6i D) 2 − 6i
8. Which of the following is rational?
F) √0.25 G) √2 H) π J) e
9. Simplify i⁴².
A) 1 B) i C) −1 D) −i
10. Simplify (2 − i)(2 + i).
F) 5i G) −5 H) 0 J) 5
11. Is 22/7 rational or irrational?
A) Irrational B) Undefined C) Imaginary D) Rational
12. Divide: (8 − 4i) ÷ (2i).
F) 2 + 4i G) −2 + 4i H) −2 − 4i J) 2 − 4i
13. For what values of b does x² + bx + 25 = 0 have complex solutions?
A) b > 10 B) b < −10 C) b = 10 D) −10 < b < 10
14. Simplify i⁷ · i¹³.
F) −1 G) 1 H) i J) −i
15. Simplify (7 − 2i) − (3 + 4i).
A) 4 − 6i B) 10 − 6i C) 4 + 6i D) 10 + 2i
16. Is √16 rational or irrational?
F) Irrational G) Rational H) Undefined J) Imaginary
17. Simplify i⁵⁵.
A) −i B) i C) 1 D) −1
18. Simplify (5 + 3i)(2 − 4i).
F) 10 − 12i G) 22 + 14i H) −22 − 14i J) 22 − 14i
19. For what values of k does 2x² + kx + 18 = 0 have complex solutions?
A) −12 < k < 12 B) k > 12 C) k < −12 D) k = ±12
20. Divide: (12 − 6i) ÷ (3 + i).
F) 3 − 3i G) 3 + 3i H) −3 − 3i J) 6 − 3i
▸ Answers and Explanations
1: C. Combine real parts (2 + 4 = 6) and imaginary parts (3 − 5 = −2) separately: 6 − 2i. Choice A subtracts the real parts and adds the imaginary parts, the operations reversed.
2: F. √7 does not simplify to an integer, and its decimal never terminates or repeats, so it is irrational. Choice H is a perfect square (√16 = 4), which is rational, not irrational.
3: B. (1 + i)(1 − i) = 1 − i² = 1 − (−1) = 2, the classic difference-of-squares pattern that always produces a real number. Choice C leaves the i² term unresolved instead of replacing it with −1.
4: H. i²³ = i²⁰·i³; since i⁴ = 1, i²⁰ = 1, so this reduces to i³ = −i. Choice F drops the negative sign on i³, the final reduced power.
5: B. Complex roots require a negative discriminant: 16 − 4k < 0, so k > 4. Choice A reverses the inequality direction from the discriminant condition.
6: G. Multiply by −i/−i to clear the imaginary denominator: (6+2i)(−i) ÷ (2i)(−i) = (2−6i) ÷ 2 = 1 − 3i. Choice F drops the negative sign on the imaginary part after dividing by 2.
7: B. Combine real parts (3 − 1 = 2) and imaginary parts (4 − (−2) = 6): 2 + 6i. Choice A swaps the real and imaginary parts.
8: F. √0.25 = 0.5, a ratio of integers (1/2), so it is rational. Choice G is an irrational square root, since 2 is not a perfect square.
9: C. 42 = 4(10) + 2, so i⁴² = (i⁴)¹⁰·i² = 1·(−1) = −1. Choice D reports i¹ instead of completing the reduction to i² = −1.
10: J. (2−i)(2+i) = 4 + 2i − 2i − i² = 4 − (−1) = 5. Choice F leaves an i attached to what is actually a real number.
11: D. 22/7 is a ratio of two integers, which makes it rational by definition, even though it is commonly used to approximate the irrational number π. Choice A confuses 22/7 with the irrational number π that it is commonly used to approximate.
12: H. Multiply by −i/−i: (8−4i)(−i) ÷ (2i)(−i) = (−8i+4i²) ÷ (−2i²) = (−8i−4) ÷ 2 = −2 − 4i. Choice G drops the negative sign on the imaginary part.
13: D. Complex solutions require a negative discriminant: b² − 100 < 0, so −10 < b < 10. Choice A reports only one boundary of the interval instead of the full range.
14: G. i⁷·i¹³ = i²⁰ = (i⁴)⁵ = 1. Choice F reports i² instead of completing the reduction to i⁰ = 1.
15: A. Subtract real parts (7 − 3 = 4) and imaginary parts (−2 − 4 = −6): 4 − 6i. Choice B adds the real parts (7+3=10) instead of subtracting them.
16: G. √16 = 4, a ratio of integers, so it is rational. Choice F assumes any square root is irrational without checking whether the radicand is a perfect square.
17: A. 55 = 4(13) + 3, so i⁵⁵ = (i⁴)¹³·i³ = 1·(−i) = −i. Choice B drops the negative sign on the final reduced power, i³.
18: J. (5+3i)(2−4i) = 10 − 20i + 6i − 12i² = 10 − 14i + 12 = 22 − 14i. Choice G drops the negative sign on the imaginary term after distributing.
19: A. Complex solutions require a negative discriminant: k² − 144 < 0, so −12 < k < 12. Choice B reports only one boundary of the interval instead of the full range.
20: F. Multiply by the conjugate (3−i)/(3−i): (12−6i)(3−i) = 30 − 30i; the denominator (3+i)(3−i) = 10; (30−30i) ÷ 10 = 3 − 3i. Choice G drops the negative sign on the imaginary part after multiplying by the conjugate.