standard_description: Understand ordering and absolute value of rational numbers.
problem_activity: Activity
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
Launch
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             |   |                     |
| \(\boldsymbol{\text- 5i}\) |   |                     |
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.
\((\text- 3 + 9i)(5i)\)
\((8 + i)(\text- 5 + 3i)\)
\((3 + 2i)^2\)
\((3 + 2i)(3 - 2i)\)
Student Response
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for free access to Student Response.
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
\(a+bi+cj+dk\)
.
\(w+z\)
\(wz\)
\(zw\)
Student Response
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click here to register or sign in
for free access to Extension Student Response.
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
\(b=0\)
.


aligns: no