HSN.CN.A.2Common CoreMathNumber and QuantityGrades 9-12
HSN.CN.A.2: Adding, Subtracting and Multiplying Complex Numbers
In plain English: HSN.CN.A.2 is the Common Core number standard that asks students to add, subtract and multiply complex numbers using i² = -1 together with the commutative, associative and distributive properties. Students combine real parts with real parts and imaginary parts with imaginary parts, distribute products such as (a + bi)(c + di), and replace i² with -1 to write each result as a + bi. It is usually taught in Algebra II.
Use the relation i² = -1 and the commutative, associative, and distributive properties to add, subtract, and multiply complex numbers.
Common Core State Standards for Mathematics · Domain: The Complex Number System (CN) · Cluster: Perform arithmetic operations with complex numbers. Also written as HSN-CN.A.2 or N-CN.2 · Official standard
Students add, subtract and multiply complex numbers and justify each step with the commutative, associative and distributive properties, together with the relation i² = -1. The lesson builds on what students already do with polynomials: complex numbers are combined the same way as binomials in x, with one extra step, because i² is the real number -1.
Students first add and subtract by grouping real parts and imaginary parts, then multiply a complex number by a real or imaginary number, and finally multiply two binomials a + bi and c + di. Every result is written in the form a + bi. Division by a complex number, which uses conjugates, belongs to the (+) standard HSN.CN.A.3.
Learning Objectives
By the end of this lesson, students will be able to:
Add complex numbers by grouping real parts and imaginary parts, naming the commutative and associative properties used
Subtract complex numbers by distributing the subtraction over both parts of the second number
Multiply complex numbers with the distributive property, replace i² with -1, and write the product as a + bi
Compare products of complex numbers with products of binomials and explain the role of i² = -1
Prior Knowledge Required
Students should already be comfortable with:
The number i and the form a + bi HSN.CN.A.1
Adding, subtracting and multiplying polynomials HSA.APR.A.1
Using properties of operations to write equivalent expressions 6.EE.A.3
Post two polynomial problems. Students simplify them, then repeat the work with i in place of x:
Warm-Up Prompt
"Simplify (4 + 3x) + (2 - 5x) and (4 + 3x)(2 - 5x). Now replace every x with i and use i² = -1. Which answer changes, and why?"
The sum is 6 - 2x, and with i it is simply 6 - 2i: nothing new happens. The product is 8 - 14x - 15x², but with i the last term becomes -15i² = +15, so the product is 23 - 14i. Use this contrast to set the goal: addition and subtraction work exactly as with like terms, while multiplication needs one extra step.
Direct Instruction20-25 minutes
Part 1: Adding and subtracting. The commutative and associative properties of addition let us reorder and regroup terms, so the real terms can be placed together and the imaginary terms together. The distributive property then combines the imaginary terms: 2i + 5i = (2 + 5)i = 7i. For subtraction, distribute the negative sign over both parts of the second number first. In general, (a + bi) + (c + di) = (a + c) + (b + d)i and (a + bi) - (c + di) = (a - c) + (b - d)i.
Part 2: Multiplying. Follow these steps and name the property at each one:
Distribute: multiply each term of the first number by each term of the second (distributive property), giving four products for two binomials.
Multiply the coefficients: reorder and regroup factors such as (-3i)(4i) = (-3 · 4)(i · i) using the commutative and associative properties of multiplication.
Replace i² with -1: this turns the i² term into a real number.
Group real parts and imaginary parts and write the answer as a + bi.
Adding
Find (6 - 2i) + (-1 + 7i), grouping like parts with the commutative and associative properties.
Find (2 - 3i)(5 + 4i) with the box model in Diagram 1.
Equation: 10 + 8i - 15i - 12i² = 22 - 7i
Squaring a complex number
Find (3 + 2i)², writing it as (3 + 2i)(3 + 2i).
Equation: 9 + 6i + 6i + 4i² = 5 + 12i
After the examples, show Diagram 2: the same product in x and in i. With x the work stops at 3 + 5x + 2x², but with i the term 2i² is the real number -2, so the answer 1 + 5i has only two parts. Stress that the answer to every problem in this standard is a single complex number a + bi.
Guided Practice15 minutes
Pairs solve each problem on a whiteboard and write the name of the property next to the step it justifies. Reveal the answers one at a time:
Guided practice problems and answers
Problem
Answer
Key step
(7 + i) + (2 - 6i)
9 - 5i
Group 7 + 2 and i - 6i
(-4 + 5i) - (-4 - 2i)
7i
Distribute the minus: -4 + 5i + 4 + 2i
(1 - i)(1 + 3i)
4 + 2i
1 + 3i - i - 3i², and -3i² = +3
-2i(6 + i)
2 - 12i
-12i - 2i², and -2i² = +2
(5 - i)²
24 - 10i
25 - 5i - 5i + i², and i² = -1
Listen for three errors: subtracting only the real part of the second number, treating i² as +1, and combining a real term with an imaginary term, such as writing 2 - 12i as -10i.
Independent Practice15 minutes
Students work alone on six problems and check with a partner at the end: (10 - 3i) + (-2 + 3i) = 8; (1/2 + 2i) - (3/2 - i) = -1 + 3i; 3(2 - i) + i(4 + i) = 5 + i; (4 + 3i)(4 - 3i) = 25; (2i)³ = 8i³ = -8i; and (1 + i)(3 + 2i)(1 - i). For the last one, students use the commutative and associative properties to multiply (1 + i)(1 - i) = 2 first, which gives 2(3 + 2i) = 6 + 4i. Ask students to notice which answers are real numbers and why.
Closure5-10 minutes
Exit ticket: (1) Find (9 - 4i) - (3 + 2i). (Answer: 6 - 6i.) (2) Find (1 + 2i)(4 - i). (Answer: 4 - i + 8i - 2i² = 6 + 7i.) (3) In part (2), which step used i² = -1, and which property let you split the product into four parts? (The distributive property.)
Differentiation Strategies
For Struggling Students
Use the box model from Diagram 1 for every product, with the i² cell shaded as a reminder to replace it with -1
Have students color-code real terms and imaginary terms before grouping them
For subtraction, have students rewrite the problem as adding the opposite, (a + bi) + (-c - di), before combining
For Advanced Students
Ask students to derive a general formula for the real part and the imaginary part of (a + bi)(c + di), and use it to check their products
Ask students to find all real numbers k for which (k + i)² is pure imaginary, and explain their reasoning
Ask students to show that multiplying by i twice is the same as multiplying by -1, using any complex number a + bi
Assessment Guidance
What to Look For
Check that students distribute a subtraction over both parts of the second number and that every product with an i² term ends up with that term moved into the real part. Ask students to name the property behind a step: the grouping of real and imaginary terms comes from the commutative and associative properties, and splitting a product into four terms comes from the distributive property. Final answers should be written as a single a + bi, not left as 10 + 8i - 15i + 12.
02
Classroom Activities
3 Activities
1
Property Proof Strips
15 minPairs
Each pair gets an envelope of paper strips. Some strips show a step in a computation and others name a property or the relation i² = -1. Pairs put the steps in order and match each step with its justification.
Strip Set A: A Sum
(3 + 5i) + (6 - 8i)
= 3 + 6 + 5i - 8i (commutative and associative properties of addition)
Which step in Set B would be different if you were multiplying (1 + 4x)(3 - x)?
Could you skip the commutative property in Set A? What would the expression look like?
2
Complex Number Relay
20 minGroups of 4
Each group receives a pair of complex numbers z and w. Student 1 finds z + w, Student 2 finds z - w, Student 3 finds zw, and Student 4 checks the product by computing wz in the other order. Roles rotate each round.
Rounds and Answers
Round 1: z = 2 + i, w = 3 - 2i. Sum 5 - i, difference -1 + 3i, product 8 - i
Round 2: z = -1 + 4i, w = 2 + 3i. Sum 1 + 7i, difference -3 + i, product -14 + 5i
Round 3: groups write their own z and w and trade them with another group
Procedure
Each student writes one step per line and passes the paper on
Student 4 compares zw and wz; if they differ, the group finds the error before moving on
Groups record which property made zw and wz equal
Modification for Distance Learning
Run the relay in a shared document with one column per student. Each student fills in a column in turn, and the group uses a video call to discuss the check in column 4.
3
Find the Error Gallery Walk
15 minGroups of 3
Four posters around the room each show a worked problem with one error. Groups rotate, find the error, write the correction on a sticky note and name the property or relation that was misused.
The Four Posters
(5 + 2i) - (1 - 3i) = 4 - i. Error: the subtraction was not distributed to -3i; correct answer 4 + 5i
(3i)(4i) = 12. Error: i · i was treated as 1; correct answer 12i² = -12
(6 - i)(1 + i) = 6 + 6i - i - i² = 5 + 5i. Error: -i² was replaced with -1 instead of +1; correct answer 7 + 5i
Procedure
Groups spend 3 minutes at each poster and add their sticky note below the previous group's note
If a group disagrees with an earlier note, it writes why
Close with a class discussion of which error each group found easiest to miss
Challenge Variation
Each group writes a fifth poster with a new, subtle error for another group to find, such as a sign error in only one of the four products.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: A Box Model for Multiplying Complex Numbers
Each cell holds the product of its row term and its column term, which is the distributive property in picture form. The shaded cell contains i², so it turns into a real number. Adding the four cells and grouping real and imaginary parts gives (2 - 3i)(5 + 4i) = 22 - 7i.
Diagram 2: The Same Product in x and in i
Multiplying complex numbers follows the same distributive steps as multiplying binomials. The difference is the last step: x² stays as a new term, but i² is the real number -1, so the product (3 + 2i)(1 + i) simplifies to the single complex number 1 + 5i.
04
Homework Assignment
~30 min
HSN.CN.A.2 Homework: Adding, Subtracting and Multiplying Complex Numbers
Directions: Show every step and write each final answer in the form a + bi. In Problems 3 and 6, name the property (commutative, associative or distributive) or the relation i² = -1 that justifies each step.
Multiply: (a) (4 - 5i)² (b) (6 + i)(6 - i) (c) i(2 + 3i)(1 - 2i). In part (c), explain which property lets you choose which two factors to multiply first.
Simplify 2(3 - 4i) - (1 + i)(2 - 5i). Show where you used i² = -1 and where you distributed the subtraction.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Adding and Subtracting
Real and imaginary parts grouped correctly, subtraction distributed to both parts
One sign error
Parts mixed or subtraction not distributed
Multiplying
All four products found and i² replaced with -1
Products found, but one i² or sign error
Products missing or i² left in the answer
Justification
Correct property named for each step in Problems 3 and 6
Some properties named or one mislabeled
No properties named
Final Form
Every answer written as a single a + bi
Most answers in a + bi form
Answers left unsimplified
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
Find (4 + 9i) + (3 - 2i).
Answer: A
Group the real parts and the imaginary parts: (4 + 3) + (9 - 2)i = 7 + 7i. Choice B adds 9 and 2 and ignores the minus sign. Choice C subtracts the real parts instead of adding them. Choice D combines real and imaginary terms, which are not like terms.
Question 2 of 20 · Multiple Choice
Find (6 - 5i) - (2 - 8i).
Answer: B
Distribute the subtraction: 6 - 5i - 2 + 8i = (6 - 2) + (-5 + 8)i = 4 + 3i. Choice A subtracts only the real part and adds -8i instead of subtracting it. Choice C adds the real parts. Choice D makes a sign error on the imaginary part.
Question 3 of 20 · Multiple Choice
Find -3i(4 - 2i).
Answer: A
Distribute: -3i · 4 + (-3i)(-2i) = -12i + 6i². Since i² = -1, 6i² = -6, so the product is -6 - 12i. Choice B treats i² as +1. Choice C swaps the real and imaginary parts. Choice D replaces 6i² with -6 but then merges -6 and -12i into -18i, as if a real term and an imaginary term were like terms.
Question 4 of 20 · Multiple Choice
Find (1 + 6i)(2 - i).
Answer: C
Distribute: 2 - i + 12i - 6i² = 2 + 11i + 6 = 8 + 11i, because -6i² = -6(-1) = 6. Choice A drops the i² term instead of replacing it. Choice B treats i² as +1, giving 2 - 6. Choice D uses +i instead of -i in the second product.
Question 5 of 20 · Multiple Choice
Find (5 + 3i)².
Answer: C
(5 + 3i)(5 + 3i) = 25 + 15i + 15i + 9i² = 25 + 30i - 9 = 16 + 30i. Choice A treats i² as +1. Choice B squares each term separately and misses the middle terms 15i + 15i. Choice D squares each term separately and also leaves out i².
Question 6 of 20 · Multiple Choice
Find (7 - 2i)(7 + 2i).
Answer: B
49 + 14i - 14i - 4i² = 49 + 4 = 53. The middle terms cancel and -4i² = +4. Choice A treats i² as +1, giving 49 - 4. Choice C multiplies only the first terms and the last terms and keeps i. Choice D makes a sign error, so the middle terms do not cancel.
Question 7 of 20 · Multiple Choice
Which properties justify rewriting (3 + 2i) + (5 + 4i) as (3 + 5) + (2i + 4i)?
Answer: D
Moving 5 next to 3 changes the order of the terms (commutative property), and regrouping them into two sums uses the associative property. Choice A is the property used in a later step, when the imaginary terms are combined into one term. Choice B is not used, because no i² appears in a sum.
Question 8 of 20 · Multiple Choice
Which property justifies the step 9i - 4i = (9 - 4)i?
Answer: B
Factoring the common factor i out of 9i - 4i is the distributive property read from right to left: (9 - 4)i = 9i - 4i. Choice C only regroups terms and cannot combine them. Choice D is not used, since no i² appears.
Question 9 of 20 · Multiple Choice
Simplify i(3 + i) - (2 - 5i).
Answer: A
i(3 + i) = 3i + i² = -1 + 3i. Then -1 + 3i - 2 + 5i = -3 + 8i. Choice B treats i² as +1. Choice C does not distribute the subtraction to -5i. Choice D makes both errors.
Question 10 of 20 · Multiple Choice
A student simplified (2i)(5i) and wrote 10. What is the correct value?
Answer: B
(2i)(5i) = (2 · 5)(i · i) = 10i² = -10. The student treated i · i as 1 (Choice C). Choices A and D keep a single i, but the product of two factors of i is i².
Question 11 of 20 · Multiple Choice
Find (-8 + i) - (-8 - i).
Answer: D
Distribute the subtraction: -8 + i + 8 + i = 0 + 2i = 2i. Choice C subtracts only the real part and then combines i - i. Choices A and B add -8 and -8 instead of subtracting.
Question 12 of 20 · Multiple Choice
Find (1 - i)².
Answer: A
(1 - i)(1 - i) = 1 - i - i + i² = 1 - 2i - 1 = -2i. Choice B treats i² as +1. Choice C squares each term separately, 1² + i² = 0, and misses the middle terms. Choice D makes a sign error on the middle terms.
Question 13 of 20 · Multiple Choice
Let z = 3 - i and w = -2 + 4i. Find zw.
Answer: C
(3 - i)(-2 + 4i) = -6 + 12i + 2i - 4i² = -6 + 14i + 4 = -2 + 14i. Choice A treats i² as +1. Choice B multiplies only the real parts and only the imaginary parts. Choice D uses -2i instead of +2i for (-i)(-2).
Question 14 of 20 · Multiple Choice
Which expression is equal to a real number?
Answer: D
Adding gives (2 + 3) + (1 - 1)i = 5 + 0i = 5, a real number. Choice A equals 6 - 2i + 3i - i² = 7 + i. Choice B equals -1 + 2i, because the subtraction doubles the imaginary part instead of canceling it. Choice C equals 2i + i² = -1 + 2i.
Question 15 of 20 · Short Answer
Find (12 - 7i) - (5 - 10i). Write the answer in the form a + bi.
Distribute the subtraction: 12 - 7i - 5 + 10i. Group the parts: (12 - 5) + (-7 + 10)i = 7 + 3i.
Multiply (a + bi)(c + di), where a, b, c and d are real numbers, and write the product in the form (real part) + (imaginary part)i.
Distribute: ac + adi + bci + bdi². Since i² = -1, bdi² = -bd. Group with the commutative and associative properties: (ac - bd) + (ad + bc)i. The real part is ac - bd and the imaginary part is ad + bc.
Question 18 of 20 · Short Answer
Simplify i(4 - i) + (6 - 2i), and name each property or relation you use.
A student wrote (3 - 2i) - (1 + 4i) = 2 + 2i. Find the error and give the correct answer.
The student did not distribute the subtraction to 4i and added it instead. Correct work: 3 - 2i - 1 - 4i = (3 - 1) + (-2 - 4)i = 2 - 6i.
Question 20 of 20 · Short Answer
Find the real number x for which the product (x + 2i)(3 - i) is a real number, and find that product.
Multiply: 3x - xi + 6i - 2i² = (3x + 2) + (6 - x)i. The product is real when the imaginary part is 0, so 6 - x = 0 and x = 6. The product is then 3(6) + 2 = 20. Check: (6 + 2i)(3 - i) = 18 - 6i + 6i - 2i² = 20.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSN.CN.A.2 mean?
HSN.CN.A.2 means students can add, subtract and multiply complex numbers and can explain the work with i² = -1 and the commutative, associative and distributive properties. Every answer is written as a single complex number a + bi. Dividing complex numbers is not part of this standard; it is in the (+) standard HSN.CN.A.3.
Is HSN.CN.A.2 Algebra 1 or Algebra 2?
It is usually taught in Algebra II, right after students meet i in HSN.CN.A.1. In integrated courses it typically appears in Math III. Students need polynomial arithmetic from Algebra I before starting it.
How do you add complex numbers?
Add the real parts, then add the imaginary parts. For example, (2 + 5i) + (1 - 3i) = (2 + 1) + (5 - 3)i = 3 + 2i. The commutative and associative properties allow the regrouping, and the distributive property combines 5i - 3i into 2i.
Why do you have to distribute the minus sign when subtracting complex numbers?
Because you subtract the whole second number, both its real part and its imaginary part. For example, (5 + i) - (2 - 4i) = 5 + i - 2 + 4i = 3 + 5i. A frequent error is subtracting only the real part, which here would give 3 - 3i.
How do you multiply two complex numbers?
Use the distributive property to multiply each term of the first number by each term of the second, replace i² with -1, and group the real and imaginary parts. For example, (1 + 2i)(2 + i) = 2 + i + 4i + 2i² = 2 + 5i - 2 = 5i. A box model like Diagram 1 keeps the four products organized.
Why does i² turn into -1 in the middle of a multiplication?
Because i is defined as a number whose square is -1. Any term with i² is therefore a real number: 7i² = -7 and -3i² = 3. Replacing i² this way is what makes the product end up in the form a + bi instead of having three kinds of terms.
Is multiplying complex numbers the same as multiplying binomials?
Almost. The distributive steps are identical to multiplying binomials in HSA.APR.A.1, and the same patterns, such as the square of a sum, still apply. The only new step is replacing i² with -1 at the end. Diagram 2 shows the two computations side by side.
Why does the standard mention the commutative, associative and distributive properties?
Because those properties are the reason the procedures work. Complex numbers obey the same properties of operations as real numbers, so students can regroup, reorder and distribute exactly as they do with real expressions. Asking students to name the properties keeps the work from becoming a list of memorized rules.
Can the product of two nonreal complex numbers be a real number?
Yes. For example, (3i)(-2i) = -6i² = 6, and (1 + i)(1 - i) = 1 - i² = 2. In general a product is real whenever its imaginary part equals 0. Pairs of the form a + bi and a - bi, called conjugates, always multiply to a real number, which is the key idea of HSN.CN.A.3.
What mistakes do students often make with complex number operations?
Common ones are: treating i² as +1 or leaving it in the answer; subtracting only the real part of the second number; squaring a + bi as a² + (bi)² and losing the middle terms; and combining a real term with an imaginary term, such as writing 3 + 4i as 7i. Asking students to label each step with a property catches many of these.
07
Related Standards
5 standards
These standards connect to HSN.CN.A.2: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
HSN.CN.A.1Prerequisite
Know that i² = -1 and every complex number has the form a + bi with a and b real
