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The Many Merry Cube Roots of -1

Module by: Kenny M. Felder. E-mail the author

Summary: This module provides a sample problem related to development of concepts related to cubed roots of -1.

When you work with real numbers, x 2 = 1 x 2 =1 has two different solutions (1 and –1). But x 2 = -1 x 2 =-1 has no solutions at all. When you allow for complex numbers, things are much more consistent: x 2 = -1 x 2 =-1 has two solutions, just like x 2 = 1 x 2 =1. In fact, x 2 = n x 2 =n will have two solutions for any number n—positive or negative, real or imaginary or complex. There is only one exception to this rule.

Exercise 1

What is the one exception?

You might suspect that x 3 = n x 3 =n should have three solutions in general—and you would be right! Let’s take an example. We know that when we are working with real numbers, x 3 = –1 x 3 =–1 has only one solution.

Exercise 2

What is the one solution?

But if we allow for complex answers, x 3 = –1 x 3 =–1 has three possible solutions. We are going to find the other two.

How do we do that? Well, we know that every complex number can be written as ( a + b i ) (a+bi), where aa and bb are real numbers. So if there is some complex number that solves x 3 = –1 x 3 =–1, then we can find it by solving the a a and bb that will make the following equation true:

( a + b i ) 3 = –1 (a+bi ) 3 =–1

You are going to solve that equation now. When you find aa and bb, you will have found the answers.

Stop now and make sure you understand how I have set up this problem, before you go on to solve it.

Exercise 3

What is ( a + b i ) 3 (a+bi ) 3 ? Multiply it out.

Exercise 4

Now, rearrange your answer so that you have collected all the real terms together and all the imaginary terms together.

Now, we are trying to solve the equation ( a + b i ) 3 = –1 (a+bi ) 3 =–1. So take the formula you just generated in number 4, and set it equal to –1–1. This will give you two equations: one where you set the real part on the left equal to the real part on the right, and one where you set the imaginary part on the left equal to the imaginary part on the right.

Exercise 5

Write down both equations.

OK. If you did everything right, one of your two equations factors as b ( 3 a 2 b 2 ) = 0 b(3 a 2 b 2 )=0. If one of your two equations doesn’t factor that way, go back—something went wrong!

If it did, then let’s move on from there. As you know, we now have two things being multiplied to give 0, which means one of them must be 0. One possibility is that b = 0 b=0: we’ll chase that down later. The other possibility is that 3 a 2 b 2 = 0 3 a 2 b 2 =0, which means 3 a 2 = b 2 3 a 2 = b 2 .

Exercise 6

Solve the two equations for a a and bb by substituting 3 a 2 = b 2 3 a 2 = b 2 into the other equation

Exercise 7

So…now that you know a a and b b, write down two complex answers to the problem x 3 = –1 x 3 =–1. If you don’t have two answers, look again!

Exercise 8

But wait…shouldn’t there be a third answer? Oh, yeah…what about that b = 0 b=0 business? Time to pick that one up. If b = 0 b=0, what is a a? Based on this aa and b b, what is the third and final solution to x 3 = –1 x 3 =–1?

Exercise 9

Did all that work? Well, let’s find out. Take either of your answers in #7 and test it: that is, cube it, and see if you get –1. If you don’t, something went wrong!

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