9.5 The many merry cube roots of -1

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

You might suspect that $x^3=n$ 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, $x^3=\mathrm{–1}$ has only one solution.

But if we allow for complex answers, $x^3=\mathrm{–1}$ 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 $(a+bi)$ , where $a$ and $b$ are real numbers. So if there is some complex number that solves $x^3=\mathrm{–1}$ , then we can find it by solving the $a$ and $b$ that will make the following equation true:

$(a+bi{)}^3=\mathrm{–1}$

You are going to solve that equation now. When you find $a$ and $b$ , 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.

Now, we are trying to solve the equation $(a+bi{)}^3=\mathrm{–1}$ . So take the formula you just generated in number 4, and set it equal to $\mathrm{–1}$ . 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.

OK. If you did everything right, one of your two equations factors as $b(3a^2–b^2)=0$ . 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 $b=0$ : we’ll chase that down later. The other possibility is that $3a^2–b^2=0$ , which means $3a^2=b^2$ .