math-2017:N.CN.2math-2017:N.CN.1
math-2017:N.CN.2 Use the relation 𝑖𝑖^2= −1 and the commutative, associative, and distributive properties to add, subtract, and multiply complex numbers.
math-2017:N.CN.1 Know there is a complex number 𝑖𝑖 such that 𝑖𝑖^2 = −1, and every complex number has the form 𝑎𝑎 + 𝑏𝑏𝑏𝑏 with 𝑎𝑎 and 𝑏𝑏 real.
Learning goal: Use the relation i² = −1 together with the commutative, associative and distributive properties to add, subtract and multiply complex numbers, and write every result in the form a + bi.
The guiding idea. Complex numbers are not a new kind of algebra; they are ordinary algebra with one extra fact. Treat a + bi the way you would treat a binomial such as a + bx, apply the properties you already know, and then replace every i² with −1. Because the commutative property lets us reorder terms, the associative property lets us regroup them and the distributive property lets us multiply through parentheses, every expression built from complex numbers with addition, subtraction and multiplication can be collapsed back to a single number of the form a + bi. That closing step, "collect real parts and imaginary parts separately," is what makes the answer a complex number in standard form rather than a pile of terms.
Addition. To add (a + bi) + (c + di), use the commutative and associative properties of addition to gather the real parts and the imaginary parts: (a + c) + (bi + di). Then the distributive property factors out i: (a + c) + (b + d)i. So real parts add and imaginary parts add. For example, (3 + 2i) + (−5 + 7i) = (3 + (−5)) + (2 + 7)i = −2 + 9i. There is no carrying between the two parts, because i is not a power of ten; it is a separate number. A misconception to head off: adding 3 + 2i to −5 + 7i does not give 5 + 4i by "adding all the numbers together." The real part −2 and the imaginary part 9 are computed independently.
Subtraction. Subtraction is addition of the opposite: (a + bi) − (c + di) = (a − c) + (b − d)i. The sign must be distributed to both parts of the number being subtracted. For example, (4 − i) − (6 − 3i) = 4 − i − 6 + 3i = (4 − 6) + (−1 + 3)i = −2 + 2i. The most frequent error is subtracting only the real part and adding the imaginary part, which would give 4 − 6 and −1 − 3, an incorrect −2 − 4i. Writing the subtraction out with the sign distributed avoids that.
Multiplication by a real number or by i alone. A real number distributes over both parts: 4(2 − 5i) = 8 − 20i. Multiplying by i also distributes, and then i² = −1 is used: 3i(2 − 5i) = 6i − 15i² = 6i − 15(−1) = 15 + 6i. Notice the reordering: the product is written with the real part first, 15 + 6i, which uses the commutative property of addition.
Multiplying two complex numbers. Use the distributive property twice, exactly as when multiplying two binomials, then simplify i²: (a + bi)(c + di) = ac + adi + bci + bdi² = ac + adi + bci − bd = (ac − bd) + (ad + bc)i. The formula is worth understanding rather than memorizing: the real part is ac − bd because the i² term contributes a real number with its sign flipped, and the imaginary part ad + bc collects the two middle terms.
A worked model inside the lesson. Multiply (2 + 3i)(4 − i). Distribute: 2·4 + 2·(−i) + 3i·4 + 3i·(−i) = 8 − 2i + 12i − 3i². Now use the relation i² = −1: −3i² = −3(−1) = +3. Collect: real parts 8 + 3 = 11, imaginary parts −2i + 12i = 10i. The product is 11 + 10i. Checking against the formula with a = 2, b = 3, c = 4, d = −1: ac − bd = 8 − (3)(−1) = 8 + 3 = 11 and ad + bc = (2)(−1) + (3)(4) = −2 + 12 = 10. Both routes give 11 + 10i.
Squaring. (3 − 2i)² = (3 − 2i)(3 − 2i) = 9 − 6i − 6i + 4i² = 9 − 12i − 4 = 5 − 12i. The i² term is 4i² = −4, which reduces the real part from 9 to 5. Another example: (1 + i)² = 1 + 2i + i² = 1 + 2i − 1 = 2i. So 1 + i is a square root of 2i, a result that would have been impossible to state without i.
A product that comes out real. (5 + 2i)(5 − 2i) = 25 − 10i + 10i − 4i² = 25 + 4 = 29. The middle terms cancel and the i² term adds 4. Whenever a number a + bi is multiplied by a − bi the result is a² + b², a real number; that pair is studied as conjugates in the next lesson. Here it is enough to see it happen and to see why: the imaginary parts are opposites, so they cancel, and −bd = −(b)(−b) = b².
Order of operations still applies. In 2(3 − i) − i(4 + i), do each multiplication first: 2(3 − i) = 6 − 2i and i(4 + i) = 4i + i² = 4i − 1 = −1 + 4i. Then subtract: (6 − 2i) − (−1 + 4i) = 6 − 2i + 1 − 4i = 7 − 6i. The associative and commutative properties are what allow the regrouping in the final line.
Why the properties are worth naming. The standard form a + bi is not just a convention of appearance. When we say (a + bi) + (c + di) = (a + c) + (b + d)i we are using the commutative property to move c next to a, the associative property to group them, and the distributive property to write bi + di as (b + d)i. When we multiply, the distributive property is used twice, then i² = −1, then the collecting step again. Naming the properties shows that complex arithmetic is forced by the rules we already accept; nothing new was invented except i itself. That is also why complex arithmetic never produces a contradiction with real arithmetic: any calculation with imaginary parts equal to 0 is the real calculation it always was.
Worked addition and subtraction: Compute (7 − 3i) + (−2 + 8i) and (4 + i) − (4 − i). For the sum, real parts 7 + (−2) = 5 and imaginary parts −3 + 8 = 5, so the sum is 5 + 5i. For the difference, distribute the sign: 4 + i − 4 + i = (4 − 4) + (1 + 1)i = 0 + 2i = 2i. The difference is pure imaginary because the real parts were equal.
Worked product: Compute (2 + 5i)(3 − i). Distribute: 2·3 + 2·(−i) + 5i·3 + 5i·(−i) = 6 − 2i + 15i − 5i². Use i² = −1: −5i² = 5. Collect: 6 + 5 = 11 and −2i + 15i = 13i. The product is 11 + 13i. Formula check with a = 2, b = 5, c = 3, d = −1: ac − bd = 6 − (−5) = 11; ad + bc = −2 + 15 = 13.
Worked square: Compute (3 + 4i)². Expand: 9 + 12i + 12i + 16i² = 9 + 24i − 16 = −7 + 24i. The i² term is 16i² = −16, which turns the real part negative. Check by the formula with a = c = 3 and b = d = 4: ac − bd = 9 − 16 = −7 and ad + bc = 12 + 12 = 24.
Worked cube: Compute (1 + 2i)³. First square: (1 + 2i)² = 1 + 4i + 4i² = 1 + 4i − 4 = −3 + 4i. Then multiply by 1 + 2i: (−3 + 4i)(1 + 2i) = −3 − 6i + 4i + 8i² = −3 − 2i − 8 = −11 − 2i. The associative property lets us compute the cube as (square)·(1 + 2i); the result is −11 − 2i.
Worked misconception: A student writes (2 + 3i)(4 − i) = 8 − 3i² = 8 + 3 = 11, multiplying only the first terms and the last terms. The distributive property requires all four products; the two middle products −2i and 12i contribute the imaginary part 10i. The full product is 11 + 10i, and the student's answer is missing the imaginary part entirely.
Property-by-property record: Compute (−2 + 3i)(−2 + 3i) writing one line per step and, beside each line, the property that justifies it (distributive, commutative, associative, or the relation i² = −1). Then do the same for (5 − i) + (−5 + 4i) − (2 − 2i). Finish each computation with a boxed answer in the form a + bi and a formula check using ac − bd and ad + bc where a product occurred.
Independent performance: Choose two complex numbers z = a + bi and w = c + di of your own, with none of a, b, c, d equal to 0 and at least one negative. Compute z + w, z − w, w − z, zw and z² by hand, showing every distributed term and every use of i² = −1. Then verify one product two ways: by distribution and by the formula (ac − bd) + (ad + bc)i. Finally, write a paragraph explaining why z − w and w − z are opposites and why zw and wz are equal, naming the property responsible for each. Success criteria: all five results are in the form a + bi, the two methods for the product agree, and each named property is the one actually used.
An adult may read the prompts aloud. Answer independently with words, objects, drawings or a demonstration before opening the answer guide. Award 1 point for a correct result or claim and 1 point for a matching explanation or demonstration; where a question asks only for a result (a corrected sentence, a spelling, a count), a fully correct result earns both points, and the answer guide shows what a complete response contains. Record prompting separately. The total (two points per question) is local practice feedback, not an Ohio proficiency level.
Compute (6 − 4i) + (−9 + i) and name the properties that let you group the real parts together.
Real parts 6 + (−9) = −3; imaginary parts −4 + 1 = −3; sum −3 − 3i. The commutative and associative properties of addition allow the regrouping, and the distributive property writes −4i + i as (−4 + 1)i.
Compute (3 + 5i) − (7 − 2i) and explain the sign error a student might make.
Distribute the subtraction to both parts: 3 + 5i − 7 + 2i = −4 + 7i. Forgetting to change the sign of −2i gives 3 + 5i − 7 − 2i = −4 + 3i, which is wrong.
Multiply (1 − i)(3 + 2i) and write the answer in the form a + bi.
Distribute: 3 + 2i − 3i − 2i² = 3 − i − 2(−1) = 3 − i + 2 = 5 − i. The i² term contributes +2 to the real part.
Compute (2 − 3i)² and explain why the real part is negative.
(2 − 3i)² = 4 − 12i + 9i² = 4 − 12i − 9 = −5 − 12i. The term 9i² equals −9, which is larger in size than 4, so the real part becomes −5.
Show that (6 − 2i)(6 + 2i) is a real number and explain why the imaginary terms disappear.
36 + 12i − 12i − 4i² = 36 + 4 = 40. The middle terms +12i and −12i are opposites and cancel, and −4i² = +4 is real, so the product is 40.
Compute i(5 − 2i) and write it in standard form.
i·5 − i·2i = 5i − 2i² = 5i + 2 = 2 + 5i. The distributive property gives the two terms; i² = −1 turns −2i² into +2; the commutative property puts the real part first.
Use the formula (a + bi)(c + di) = (ac − bd) + (ad + bc)i to check that (2 + 3i)(4 − i) = 11 + 10i.
With a = 2, b = 3, c = 4, d = −1: ac − bd = 8 − (3)(−1) = 11 and ad + bc = (2)(−1) + (3)(4) = 10. So the product is 11 + 10i, matching the distributed result.
Compute (1 − i)⁴ by squaring twice.
(1 − i)² = 1 − 2i + i² = 1 − 2i − 1 = −2i. Then (−2i)² = 4i² = −4. So (1 − i)⁴ = −4, a real number; the associative property lets the fourth power be computed as the square of the square.