HSN.CN.A.3Common CoreMathNumber and QuantityGrades 9-12
HSN.CN.A.3: Complex Conjugates, Moduli and Quotients
In plain English: HSN.CN.A.3 is an advanced (+) Common Core number and quantity standard, usually taught in Algebra II or Precalculus. Students find the conjugate a - bi of a complex number a + bi, use the real product (a + bi)(a - bi) = a² + b² to find the modulus, and divide complex numbers by multiplying the numerator and denominator by the conjugate of the denominator.
(+) Find the conjugate of a complex number; use conjugates to find moduli and quotients of 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.3 or N-CN.3 · Official standard
Students already add, subtract and multiply complex numbers. This lesson adds one new tool, the conjugate: the conjugate of z = a + bi is z̄ = a - bi. The key fact is that z·z̄ = a² + b² is always a nonnegative real number. That single product does two jobs. Its square root is the modulus, |z| = √(z·z̄), the distance from z to 0 on the complex plane. And multiplying a denominator c + di by its conjugate turns it into the real number c² + d², which is how complex numbers are divided.
Students see the conjugate as a reflection across the real axis, derive z·z̄ = a² + b², compute moduli, and divide complex numbers step by step, checking every quotient by multiplying back.
Learning Objectives
By the end of this lesson, students will be able to:
Find the conjugate of any complex number, including real and pure imaginary numbers
Show that z·z̄ = a² + b² for z = a + bi and explain why the product is a nonnegative real number
Use the product z·z̄ to find the modulus |z| = √(z·z̄)
Divide complex numbers by multiplying the numerator and denominator by the conjugate of the denominator, and write the quotient in a + bi form
Check a quotient by multiplying it by the divisor
Prior Knowledge Required
Students should already be comfortable with:
Writing complex numbers in the form a + bi HSN.CN.A.1
Multiplying complex numbers using i² = -1 HSN.CN.A.2
Rationalizing a denominator such as 1/(2 + √3)
Using the Pythagorean Theorem to find a distance 8.G.B.8
Open with two products that look like they should be complex but are not.
Warm-Up Prompt
"Multiply (4 + 3i)(4 - 3i) and (2 + 5i)(2 - 5i). What kind of number is each answer? Can you predict (a + bi)(a - bi) without multiplying?"
Students should get 16 - 9i² = 25 and 4 - 25i² = 29. The middle terms cancel and i² = -1 turns the last term positive, so each product is a² + b². Record the pattern; it drives the whole lesson.
Direct Instruction20 minutes
Part 1: The conjugate and the modulus. The conjugate of z = a + bi is z̄ = a - bi: keep the real part and change the sign of the imaginary part. On the complex plane (Diagram 1) the conjugate is the reflection of z across the real axis. Then show the general product and what it gives:
Find the conjugate: change only the sign of the imaginary part. A real number is its own conjugate, and the conjugate of bi is -bi.
Multiply z by z̄: (a + bi)(a - bi) = a² - abi + abi - b²i² = a² + b². The result is real and never negative.
Take the square root: |z| = √(z·z̄) = √(a² + b²). This is the distance from 0 to z, the hypotenuse of a right triangle with legs |a| and |b|.
Divide: to find (a + bi)/(c + di), multiply the numerator and the denominator by c - di. The denominator becomes c² + d², a real number (Diagram 2).
Write and check: divide both parts of the new numerator by c² + d² to get x + yi, then multiply x + yi by c + di to confirm you get a + bi.
Part 2: Why the method works. Multiplying by (c - di)/(c - di) is multiplying by 1, so the value does not change; only the form does. This is the same idea students used to rationalize 1/(2 + √3). Work the examples below.
Conjugates
Write the conjugate of 5 - 2i, of -3i and of 7.
Equation: 5 + 2i, 3i and 7
Modulus from the conjugate
Find |20 + 21i| by multiplying 20 + 21i by its conjugate.
Pairs work each row on mini whiteboards. One partner finds the conjugate or sets up the product, and the other finishes and checks.
Guided practice
Task
Work
Result
Conjugate and modulus of 9 + 4i
(9 + 4i)(9 - 4i) = 81 + 16
9 - 4i, √97 ≈ 9.85
Modulus of -15 + 8i
(-15 + 8i)(-15 - 8i) = 225 + 64
√289 = 17
(10 - 5i)/(2 - i)
(10 - 5i)(2 + i)/5 = 25/5
5
(1 + 4i)/(3 + 2i)
(1 + 4i)(3 - 2i)/13 = (11 + 10i)/13
11/13 + (10/13)i
The third row surprises students: the quotient is real because 10 - 5i = 5(2 - i). Watch for students who multiply only the denominator by the conjugate, and for students who write i² = 1 when they expand.
Independent Practice15 minutes
Students work alone: (1) the conjugate of -4 + 7i and the product (-4 + 7i)(-4 - 7i) (-4 - 7i and 65); (2) |9 - 12i| using the conjugate (15); (3) (5 + 5i)/(1 - 2i) (-1 + 3i); (4) (4 - 2i)/(3 + i) (1 - i); (5) 1/(2 + 5i) (2/29 - (5/29)i). As a stretch item, find every complex number z with real part 3 and z·z̄ = 25 (3 + 4i and 3 - 4i).
Closure5 minutes
Exit ticket: (1) Write the conjugate of -2 - 9i. (Answer: -2 + 9i.) (2) Use it to find |-2 - 9i|. (Answer: √85.) (3) Divide (8 + i)/(2 - i). (Answer: (15 + 10i)/5 = 3 + 2i.)
Differentiation Strategies
For Struggling Students
Give a template for division: numerator times conjugate on top, c² + d² on the bottom, then split into two fractions
Have students plot z and z̄ before computing, so the reflection makes the sign change visible
Start with denominators such as 1 + i and 1 - i, where c² + d² = 2 keeps the arithmetic small
For Advanced Students
Ask students to prove that the conjugate of zw is z̄·w̄ and that |zw| = |z|·|w|, using z·z̄ = |z|²
Ask students to show that 1/z = z̄/|z|² and use it to find 1/(3 - 4i) in one step
Ask students to find every complex number that equals its own conjugate, and every one that equals the negative of its conjugate
Assessment Guidance
What to Look For
Check that students change only the sign of the imaginary part when they conjugate, and that they use z·z̄, not z², to find a modulus. In division, look for the conjugate of the denominator applied to both the numerator and the denominator, a real denominator c² + d², and a final answer in a + bi form. Students should check at least one quotient by multiplying back.
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Classroom Activities
3 Activities
1
Conjugate and Modulus Match
15 minPairs
Pairs get 16 cards: 4 number cards, 4 conjugate cards, 4 product cards (z·z̄) and 4 modulus cards. They build 4 rows, one for each number, and justify each match.
Match each number with its conjugate, then compute z·z̄ to find its product card
Take the square root of the product to find the modulus card
Plot each number and its conjugate on grid paper and confirm that the pair is symmetric across the real axis
Discussion Questions
Why is the number 11 its own conjugate?
Why is every product card a positive real number?
2
Division Relay
20 minGroups of 4
Each group gets a relay worksheet with 3 quotients. Every member does one step and passes the sheet on, so every step of the method gets said out loud.
The Steps
Student 1 writes the conjugate of the denominator
Student 2 multiplies the numerator by it
Student 3 multiplies the denominator by it and confirms the result is real
Student 4 writes the quotient in a + bi form and checks it by multiplying by the original denominator
The 3 Quotients
(9 + 7i)/(3 - i) = (20 + 30i)/10 = 2 + 3i
(-10 + 5i)/(4 + 3i) = (-25 + 50i)/25 = -1 + 2i
13/(2 + 3i) = 13(2 - 3i)/13 = 2 - 3i
Modification for Distance Learning
Put the worksheet in a shared document. Each student types one step in a different color, and the last student posts the check.
3
Why Is z·z̄ Always Real?
15 minSmall groups
Groups test the product of a number and its conjugate on examples, prove the pattern in general, and connect it to the length of a segment on the complex plane.
Procedure
Each member picks a complex number with nonzero real and imaginary parts and multiplies it by its conjugate. Compare: is every result real and positive?
Expand (a + bi)(a - bi) as a group and name the step where the imaginary terms cancel and the step where i² = -1 is used
Plot z = 1 + 7i. Draw the right triangle with legs 1 and 7 and use the Pythagorean Theorem to find the distance from 0 to z. Compare with √(z·z̄) = √50 = 5√2
Challenge Variation
Show that z + z̄ = 2a and z - z̄ = 2bi, and use these to explain which complex numbers equal their own conjugate.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: A Complex Number, Its Conjugate and Their Modulus
z = 4 + 3i and its conjugate z̄ = 4 - 3i are reflections of each other across the real axis. Both are 5 units from 0, and z·z̄ = 16 + 9 = 25 = 5², so |z| = √(z·z̄). Drawn to scale, one grid square is one unit.
Diagram 2: Dividing with the Conjugate
To divide (7 + i)/(1 + i), multiply the numerator and the denominator by 1 - i, the conjugate of the denominator. The denominator becomes the real number 2, so the quotient is 4 - 3i. Multiplying 4 - 3i by 1 + i gives back 7 + i.
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Homework Assignment
~30 min
HSN.CN.A.3 Homework: Conjugates, Moduli and Quotients
Directions: Show every product you use. Write each quotient in the form a + bi with exact fractions, and check at least two quotients by multiplying back.
Part 1: Conjugates and Moduli (Problems 1-3)
Write the conjugate of each number, then find the product of the number and its conjugate: (a) 8 + 3i (b) -5 - 6i (c) 10i (d) -4.
Use the product z·z̄ to find the modulus of each number. Give exact answers: (a) 24 - 7i (b) -1 + 3i (c) 6 + 6i.
Let z = a + bi with a and b real. Find every z for which z² = z·z̄, showing your expansion of each side. Then explain why |z| and |z̄| are always equal.
Part 2: Quotients (Problems 4-6)
Write each quotient in the form a + bi: (a) (11 + 3i)/(2 + i) (b) (3 - i)/(1 + 3i) (c) (4 + 7i)/(-3i).
A student wrote (6 + 4i)/(2 + i) = 3 + 4i by dividing the real parts and the imaginary parts separately. Multiply (2 + i)(3 + 4i) to show that the answer is wrong, then find the correct quotient.
Let z = 2 + i and w = 1 - 3i. (a) Find z/w in the form a + bi. (b) Find |z/w| and compare it with |z|/|w|.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Conjugates
Every conjugate correct, including real and pure imaginary numbers
One sign error
Both signs changed or method missing
Moduli
z·z̄ used and the square root simplified exactly
Correct product, root missing or not simplified
z² used or no method
Quotients
Conjugate applied to numerator and denominator, answer in a + bi form
Correct method with one arithmetic error
Parts divided separately
Reasoning
Clear proof in Problem 3 and a correct check in Problem 5
Partial explanation
No explanation
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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
What is the conjugate of -7 + 4i?
Answer: B
Keep the real part and change the sign of the imaginary part: -7 + 4i becomes -7 - 4i. Choice A changes both signs, which gives the opposite -z, not the conjugate. Choice C changes the sign of the real part only.
Question 2 of 20 · Multiple Choice
What is the conjugate of 9i?
Answer: D
9i = 0 + 9i, so its conjugate is 0 - 9i = -9i, its reflection across the real axis. Choice A would be right only for a real number. Choices B and C drop the i, which changes the number to a real one.
Question 3 of 20 · Multiple Choice
What is (5 - 4i)(5 + 4i)?
Answer: A
(5 - 4i)(5 + 4i) = 25 + 20i - 20i - 16i² = 25 + 16 = 41, which is a² + b². Choice B treats i² as 1, giving 25 - 16. Choice C multiplies only the first parts and the last parts and forgets that i² = -1.
Question 4 of 20 · Multiple Choice
Use the conjugate to find |16 - 30i|.
Answer: C
(16 - 30i)(16 + 30i) = 256 + 900 = 1156, and √1156 = 34. Choice A adds the parts 16 and 30. Choice B subtracts the squares, 900 - 256. Choice D subtracts the parts.
Question 5 of 20 · Multiple Choice
For z = a + bi, which expression always equals |z|²?
Answer: B
z·z̄ = (a + bi)(a - bi) = a² + b² = |z|². Choice A is wrong: z² = a² - b² + 2abi, which is usually not even real. Choice C equals 2a, twice the real part, and choice D equals 2bi.
Question 6 of 20 · Multiple Choice
What is (-1 - 13i)/(3 - i) in the form a + bi?
Answer: C
Multiply by (3 + i)/(3 + i): (-1 - 13i)(3 + i) = -3 - i - 39i - 13i² = 10 - 40i, and (3 - i)(3 + i) = 9 + 1 = 10. So the quotient is 1 - 4i. Check: (3 - i)(1 - 4i) = 3 - 12i - i - 4 = -1 - 13i. Choice A divides the parts separately. Choice B forgets the denominator. Choice D uses 9 - 1 = 8 for the denominator.
Question 7 of 20 · Multiple Choice
To divide (a + bi)/(c + di), why do you multiply the numerator and the denominator by c - di?
Answer: A
The goal is a real denominator, and c² + d² is real. Since (c - di)/(c - di) = 1, the quotient keeps its value. Choice C is false: (c + di)(c - di) = c² + d², which equals 1 only when |c + di| = 1. Choices B and D are false because the numerator usually stays complex.
Question 8 of 20 · Multiple Choice
What is 1/i?
Answer: B
Multiply by -i/(-i): 1/i = -i/(-i²) = -i/1 = -i. Check: i·(-i) = -i² = 1. Choice A fails the check, because i·i = -1. Choices C and D are real numbers, and no real number times i equals 1.
Question 9 of 20 · Multiple Choice
Use the conjugate to find |-3 - 3i|.
Answer: D
(-3 - 3i)(-3 + 3i) = 9 + 9 = 18, so |-3 - 3i| = √18 = 3√2 ≈ 4.24. Choice A leaves off the square root. Choice B adds the sizes of the parts, 3 + 3. Choice C uses only one of the parts.
Question 10 of 20 · Multiple Choice
A student tries to find |2 + 6i| by squaring: (2 + 6i)² = -32 + 24i. What is |2 + 6i|?
Answer: A
The modulus comes from the conjugate, not the square: (2 + 6i)(2 - 6i) = 4 + 36 = 40, so |2 + 6i| = √40 = 2√10. Choice B follows the student's z², which is not real. Choice C forgets the square root, and choice D adds the parts.
Question 11 of 20 · Multiple Choice
Which expression equals (4 - i)/(2 + 3i) and has a real denominator?
Answer: C
Multiply the numerator and the denominator by 2 - 3i, the conjugate of the denominator: (2 + 3i)(2 - 3i) = 4 + 9 = 13. Choice A multiplies the numerator by the denominator itself. Choice B computes the denominator as 4 - 9, treating i² as 1. Choice D also conjugates the numerator, which changes the value.
Question 12 of 20 · Multiple Choice
What is (-4 + 2i)/(1 - i)?
Answer: D
(-4 + 2i)(1 + i) = -4 - 4i + 2i + 2i² = -6 - 2i, and (1 - i)(1 + i) = 2, so the quotient is -3 - i. Choice A forgets to divide by 2. Choice B divides the parts separately. Choice C has a sign error in the imaginary part.
Question 13 of 20 · Multiple Choice
A pure imaginary number z has a positive imaginary part and z·z̄ = 49. What is z?
Answer: B
Write z = bi with b > 0. Then z·z̄ = (bi)(-bi) = b² = 49, so b = 7 and z = 7i. Choice A is real, not pure imaginary. Choice C has a negative imaginary part. Choice D forgets that the product is b², not b.
Question 14 of 20 · Multiple Choice
What is 1/(1 + 2i) in the form a + bi?
Answer: A
1/(1 + 2i) = (1 - 2i)/((1 + 2i)(1 - 2i)) = (1 - 2i)/5 = 1/5 - (2/5)i. Check: (1 + 2i)(1 - 2i)/5 = 5/5 = 1. Choice B takes the reciprocal of each part. Choice C loses the sign of the conjugate. Choice D uses 1 - 4 = -3 as the denominator.
Question 15 of 20 · Short Answer
Let z = 8 - 5i. Find z̄, the product z·z̄, and |z|.
It means students can find the conjugate of a complex number and use it for two jobs: finding the modulus and dividing. The conjugate of a + bi is a - bi, the product (a + bi)(a - bi) = a² + b² gives the modulus √(a² + b²), and multiplying by the conjugate of a denominator makes division possible.
Is HSN.CN.A.3 taught in Algebra 2 or Precalculus?
It is usually taught in Algebra II or Precalculus. The (+) marks it as additional mathematics that Common Core describes for students taking advanced courses, so some Algebra II courses introduce division by conjugates and Precalculus returns to it with the complex plane.
What is the conjugate of a complex number?
The conjugate of a + bi is a - bi: the real part stays and the imaginary part changes sign. For example, the conjugate of 1 - 6i is 1 + 6i. On the complex plane, a number and its conjugate are mirror images across the real axis.
Why is a complex number times its conjugate always real?
Because the imaginary terms cancel and i² = -1 makes the last term positive. (a + bi)(a - bi) = a² - abi + abi - b²i² = a² + b². Since a and b are real, a² + b² is a real number that is never negative, and it is 0 only when z = 0.
How do you find the modulus of a complex number using the conjugate?
Multiply the number by its conjugate and take the square root: |z| = √(z·z̄). For 2 + 3i, the product is 4 + 9 = 13, so the modulus is √13. The result matches the Pythagorean Theorem, because the modulus is the distance from 0 to z.
How do you divide complex numbers?
Multiply the numerator and the denominator by the conjugate of the denominator, then simplify. For example, (1 + i)/(1 - i) = (1 + i)(1 + i)/((1 - i)(1 + i)) = 2i/2 = i. Always finish by writing the answer as a + bi, and check it by multiplying it by the original denominator.
Is dividing complex numbers like rationalizing the denominator?
Yes, it is the same idea. To simplify 1/(2 + √3), you multiply by (2 - √3)/(2 - √3) so that the denominator becomes 4 - 3 = 1. With complex numbers, you multiply by the conjugate so that the denominator becomes the real number c² + d².
What mistakes do students make with conjugates and division?
A common one is dividing the real parts and the imaginary parts separately, which gives a wrong quotient. Others include changing the sign of both parts when conjugating, using z² instead of z·z̄ for the modulus, writing i² = 1, and multiplying only the denominator by the conjugate. Checking by multiplication catches most of these.
Is the modulus the same as absolute value?
Yes, for real numbers they agree. The modulus extends absolute value to complex numbers: both measure distance from 0. For a real number a, the conjugate is a itself, so √(a·a) = |a|.
How does HSN.CN.A.3 connect to later topics?
Conjugates appear again when quadratic equations with real coefficients have complex solutions, which come in conjugate pairs (HSN.CN.C.7). The modulus becomes the distance from 0 in polar form (HSN.CN.B.4), and the modulus of a difference gives the distance between two complex numbers (HSN.CN.B.6).
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Related Standards
6 standards
These standards connect to HSN.CN.A.3: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
HSN.CN.A.2Prerequisite
Use i² = -1 and the properties of operations to add, subtract and multiply complex numbers
