Lesson 13

Multiplying Complex Numbers

13.1: $i$ Squared (5 minutes)


The purpose of this activity is to elicit strategies and understandings students have for multiplying imaginary numbers. Later in this lesson, students will multiply complex numbers and write them in the form \(a+bi\), so it will be helpful for students to see various strategies for multiplying imaginary numbers.

Student Facing

Write each expression in the form \(a+bi\), where \(a\) and \(b\) are real numbers.

  1. \(4i \boldcdot 3i\)
  2. \(4i \boldcdot \text-3i\)
  3. \(\text- 2i \boldcdot \text- 5i\)
  4. \(\text- 5i \boldcdot 5i\)
  5. \((\text-5i)^2\)

Student Response

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Activity Synthesis

Ask students to share their strategies for each problem. Record and display their responses for all to see. Highlight ways that students regrouped factors of \(i\) or \(\text-1\) in order to simplify their answers. If no student mentions the fact that squaring an imaginary number always results in a real number, ask students to discuss this idea.

13.2: Multiplying Imaginary Numbers (15 minutes)


In this partner activity, students take turns matching equivalent expressions using the fact that \(i^2 = \text-1\). They build on a previous activity in which they multiplied imaginary numbers and saw how this changed the numbers’ representation on the complex plane. Fluency with strategically using the fact that \(i^2=\text-1\) will be an important skill in the next activity when students multiply numbers that have both real and imaginary parts. As students trade roles explaining their thinking and listening, they have opportunities to explain their reasoning and critique the reasoning of others (MP3).


Arrange students in groups of 2. Tell students that for each expression in column A, one partner finds an equivalent expression in column B and explains why they think it is equivalent. (One item in column B will not be used.) The partner's job is to listen and make sure they agree. If they don't agree, the partners discuss until they come to an agreement. Students then swap roles. If necessary, demonstrate this protocol before students start working.

Conversing: MLR8 Discussion Supports. As students take turns finding a match and explaining their reasoning to their partner, display the following sentence frames for all to see: “ _____ and _____ are equal because . . .” and “I noticed _____, so I matched . . .” Encourage students to challenge each other when they disagree. This will help students clarify their reasoning about equivalent expressions that involve imaginary numbers.
Design Principle(s): Support sense-making; Maximize meta-awareness
Engagement: Develop Effort and Persistence. Encourage and support opportunities for peer collaboration. When students share their work with a partner, display sentence frames to support conversation such as: “First, I _____ because . . .”, “I noticed _____ so I . . .”, “Why did you . . .?”, “I agree/disagree because . . . .”
Supports accessibility for: Language; Social-emotional skills

Student Facing

Take turns with your partner to match an expression in column A with an equivalent expression in column B.

  • For each match that you find, explain to your partner how you know it’s a match.
  • For each match that your partner finds, listen carefully to their explanation. If you disagree, discuss your thinking and work to reach an agreement.
\(5 \boldcdot 7i\) -9
\(5i \boldcdot 7i\) \(35i\)
\(3i^2\) -35
\((3i)^2\) 1
\(8i^3\) 9
\(i^4\) -3
\(\text- i^2\) -1
\((\text- i)^2\) \(\text- 8i\)


Student Response

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Anticipated Misconceptions

If students feel stuck with the expressions \(i^3\) or \(i^4\), suggest replacing exponents with repeated factors (\(i \boldcdot i \boldcdot i\)) and rewriting the expressions a pair at a time.

Activity Synthesis

Once all groups have completed the matching, here are some questions for discussion: 

  • “Which matches were tricky? Explain why.” (\((\text- i)^2\) was tricky because there were several negative signs to keep track of and I had to double-check whether the final answer was negative or positive.)
  • “Did you need to make adjustments in your matches? What might have caused an error? What adjustments were made?” (I thought \(3i^2\) was -9 because I thought the 3 was squared too. I had to go back and just square the \(i\) and then multiply it by 3.)

Ask groups to explain their reasoning for several matches, especially why \(i^4=1\). If not brought up by students, make sure to discuss that it’s possible to use the fact that \(i^2=\text-1\) to make equivalent expressions.

13.3: Multiplying Complex Numbers (15 minutes)


The purpose of this activity is for students to build fluency expressing the product of two complex numbers in the form \(a+bi\), where \(a\) and \(b\) are real numbers. In order to do this, students must use the fact that \(i^2=\text-1\).

Look for students who answer the last question as \(13+0i\) as opposed to 13 to highlight during discussion.

Representation: Internalize Comprehension. Activate or supply background knowledge. Display the table provided in the launch. As students share their explanations, use color-coding and annotations to highlight the connections between the expression, the values in the table, and the final product after combining like terms. Students can refer back to the chart throughout the activity.
Supports accessibility for: Visual-spatial processing


Tell students that they are now going to multiply complex numbers together. Display the expression \((3 + 2i)(\text- 4 - 5i)\) for all to see and give students 1 minute of quiet think time to consider how they would find the product. Then, display this table for all to see:

3 \(\boldsymbol{2i}\)
-4 -12 \(\text- 8i\)
\(\boldsymbol{\text- 5i}\) \(\text- 15i\) \(\text- 10i^2\)

After a brief time to consider the diagram, select students to explain where each of the values came from. Ask students, “Now that we have \(\text- 12 - 8i - 15i - 10i^2\), what do we do in order to write the number in the form \(a+bi\)?” (We know \(i^2 = \text-1\), so the sum of these is \(\text- 12 - 8i - 15i +10 = \text- 2 - 23i\).)

Student Facing

Write each product in the form \(a + bi\), where \(a\) and \(b\) are real numbers.

  1. \((\text- 3 + 9i)(5i)\)
  2. \((8 + i)(\text- 5 + 3i)\)
  3. \((3 + 2i)^2\)
  4. \((3 + 2i)(3 - 2i)\)

Student Response

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Student Facing

Are you ready for more?

On October 16, 1843, while walking across the Broom Bridge in Dublin, Ireland, Sir William Rowan Hamilton came up with an idea for numbers that would work sort of like complex numbers. Instead of just the number \(i\) (and its opposite \(\text- i\)) squaring to give -1, he imagined three numbers \(i\), \(j\), and \(k\) (each with an opposite) that squared to give -1.

The way these numbers multiplied with each other was very interesting. \(i\) times \(j\) would give \(k\), \(j\) times \(k\) would give \(i\), and \(k\) times \(i\) would give \(j\). But the multiplication he imagined did not have a commutative property. When those numbers were multiplied in the opposite order, they’d give the opposite number. So \(j\) times \(i\) would give \(\text- k\), \(k\) times \(j\) would give \(\text- i\), and \(i\) times \(k\) would give \(\text- j\). A quaternion is a number that can be written in the form \(a+bi+cj+dk\) where \(a\), \(b\), \(c\), and \(d\) are real numbers.

Let \(w=2+3i-j\) and \(z=2i+3k\). Write each given expression in the form \(a+bi+cj+dk\).

  1. \(w+z\)
  2. \(wz\)
  3. \(zw\)

Student Response

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Activity Synthesis

The key takeaway is that the product of complex numbers is another complex number, and we can see this by using usual arithmetic along with the fact that \(i^2=\text-1\) to write products in the form \(a+bi\), where \(a\) and \(b\) are real numbers.

Select previously identified students to share their responses to the last question, for which \(13+0i\) is the technically correct response. Discuss the idea that numbers like 13 or \(5i\) don’t need to be written as \(13+0i\) and \(0+5i\) in order to be recognizable as complex numbers. Writing complex numbers as a single term is okay; it’s something they did for a long time before they knew that all real numbers are complex numbers of the form \(a+bi\) where \(b=0\).

Lesson Synthesis

Lesson Synthesis

In this lesson, students learned about multiplication with complex numbers. Display this equation for all to see, leaving room to record student thinking for all to see during the discussion:

\(\displaystyle (2+2i)(2-3i) = (3+2i) + (7+4i) \)

Ask students, “Mentally, without going through all the trouble of writing each side in the form \(a+bi\), what about the structure of the expressions on each side of the equals sign tells you that the equation is true or false?” (Looking at the left hand side, we can make the imaginary part by adding the product \(2i \boldcdot 2\) with the product \(2 \boldcdot \text-3i\), which gives \(4i-6i=\text-2i\). On the right hand side, however, the imaginary part is \(2i+4i=6i\), so the equation is false. The two complex numbers don’t have equal imaginary parts, so they can’t be equal.)

Then, display this equation for all to see:

\(\displaystyle (3-5i)(\text-2+3i) = (\text-10+6i) + (4+13i) \)

Again, ask students to use arguments about the structure of the expressions on each side of the equation to mentally determine whether or not it is true. (The imaginary parts match because on the left hand side, \(3 \boldcdot 3i + \text-5i \boldcdot \text-2 = 9i+10i=19i\), and on the right hand side, \(6i+13i=19i\). The real parts don’t match, however. On the right hand side, the real part is \(\text-10+4=\text-6\). On the left hand side, the real parts of the two factors also multiply to -6, but there are multiples of \(i^2\) that are real that haven’t been accounted for.)

If time allows, ask students to verify the arguments for one or both equations by writing each side in the form \(a+bi\), where \(a\) and \(b\) are real numbers.

13.4: Cool-down - Squares and Imaginary Numbers (5 minutes)


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Student Lesson Summary

Student Facing

To multiply two complex numbers, we use the distributive property:

\(\displaystyle (2 + 3i )(4 + 5i) = 8 + 10i + 12i + 15i^2\)

Remember that \(i^2 = \text- 1\), so:

\(\displaystyle (2 + 3i )(4 + 5i) = 8 + 10i + 12i - 15\)

When we add the real parts together and the imaginary parts together, we get:

\(\displaystyle (2 + 3i )(4 + 5i) = \text- 7 + 22i\)