Dividing Complex Numbers

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SWBAT use the properties of complex conjugates to divide two complex numbers. They will also understand and explain the corresponding graphical representation.

Big Idea

Division of complex numbers is best understood in its relation to multiplication and transformations of the complex plane.

Lost in the Shuffle

10 minutes

N.B.  This lesson addresses a content standard typically reserved for Year 4 courses. For more advanced classes, I would use this as an extension lesson.

Set the stage for division of complex numbers.

Problem #1:  What do we call two numbers like (4 + 9i) & (4 - 9i)?  What is their product? A real number! (That is important.)

Problem #2:  Is it possible to factor the difference of two squares (4 - 9)?  Is it possible to factor the sum of two squares (4 + 9)?

Problem #3:  Let's multiply (7-8i)(2+11i).  Notice what happens to the -88i^2 term; it gets combined with the other real term!  Looking at our final number, 102 + 61i, we've lost some infomation.  The part of that 102 which was originally real is mixed up with the part that was orignially imaginary!

Division of Complex Numbers

20 minutes

By now we've established the basic problem: information is lost in the multiplication of complex numbers.  At this point, I'd put the following problem on the board:

(3 + 2i)(z) = (22 -7i)

This is a division problem, but I've expressed it in such a way that students are led to think of it in terms of the missing factor.  And so I'll say, "Your job is to find the missing factor."

Students will be given 5 minutes of individual time (MP 1) followed by about 15 minutes of group time.  Hopefully that will be enough for a handful of students to find both a graphical and algebraic solution to the problem.

If students are stuck, I might try the following suggestions:

1) Make use of the graphical representation.  You know how multiplication works graphically, so you can think about it in those terms.

2) Express z as a ratio or fraction and then simplify.  How?  Find a way to make the denominator a real number.  How?  Recall that the product of two complex conjugates is always a real number.

Please see the video for some more details on this section.

Student Solutions

15 minutes

At this point, it is time for students to present their solutions to the division problem to their peers.  During the work period, I would have pre-selected the students I intend to call on because their work exhibited certain insights, ways of thinking, or even instructive errors.  These students would be asked to use the document camera to show their classmates their work and to carefully explain their thought process.   The rest of the class would be encouraged to ask questions until they were confident that they understood the solution being presented.  (MP 3)  For more details on this, please see my strategy video on Student Presentations.

I am looking for two key insights here.  The first is that the graphical representation of multiplication of complex numbers can fairly easily be "undone" to identify the missing factor.  First, the argument of the missing factor is the difference of the arguments of the product and the given factor.  Second, the magnitude of the missing factor is the quotient of the magnitudes of the product and the given factor.  With both the argument and the magnitude known, the missing factor is known.

The second key insight is that the missing factor can be found algebraically by simplifying the ratio of the product and the given factor.  When multiplied by its complex conjugate, the denominator can always be converted into a real number.  Once the denominator is a real number, the quotient is fairly easy to obtain.

This algebraic method will ultimately be the most useful for my students, but it's vitally important that they understand the graphical method, as well.

All of this being done, tonight's homework will be four problems of complex division.  Since there are so few, I'll just write them on the board for the class to copy down.  They are the following:

(2 + i)(z) = (5 + 5i)

(3 - i)(z) = (17 + i)

(29 + 11i) / (5 + 7i) = z

(-6 + 18i) / (2 + 2i) = z

Alternatively, you might just assign one of these problems with the added instructions to solve it both graphically and algebraically.