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In our previous blog post, we were answering the question "Why is it useful to write complex numbers in trigonometric form?"

We saw how squaring a complex number can be simplified using trig, and we saw the same thing is true for multiplying any two complex numbers. In this blog post we'll explore how trigonometric form for complex numbers can help us take roots (cube roots, fourth roots, etc).

Before we get into taking roots of numbers, I'd like to point out something that you might not have realized: there are an infinite number of ways to write a complex number in cis form, because cos and sin are periodic functions:

rcis(xº) = rcis(xº + 360º) = rcis(xº + 720º) = etc.

Okay, tuck that information away in your memory banks; we'll get back to it shortly. Now let's talk about taking a cube root.

Let's suppose we wanted to take the cube root of the number 8...

"Well, that's silly," you might argue, "I already know the cube root of 8. It's TWO!"

Ah...but you only *think* you know. Bear with me, and I'll show you something interesting.

How would you write 8 as a complex number? It's 8 + 0i. Which is the same as 8cis0º. Now we know from the previous post, that if rcisx cubed equals 8cis0º, then r^{3} = 8, and 3x = 0º. Or, since it's the same value, 3x = 360º, or 3x = 720º.

Thus, x = 0º, x = 120º, or x = 240º.

Therefore, the cube roots of 8cis0º are 2cis0º, 2cis120º, and 2cis120º.

But what are those numbers?

Well, 2cis0º is easy; it's just 2 + 0i = 2. "Ah ha!" you might say, "I told you it was two!"

Very good, but let's keep going:

2cis120º = 2cos120º + 2isin120º = -1 + 1.732i

2cis240º = 2cos240º + 2isin240º = -1 - 1.732i

So it turns out that 8 has three cube roots, and you found them using trigonometry!

It turns out that you could have found those roots algebraically, by setting up the following equation:

x^{3} = 8

x^{3} - 8 = 0

(x - 2)(x^{2} + 2x + 4) = 0

So either x - 2 = 0 or x^{2} + 2x + 4 = 0

The first equation gives us the root we expected: x = 2.

The second equation can be solved using the quadratic formula:

x = (-2 +/- 3.4641i)/2, which simplifies to:

x = -1 + 1.732i or x = -1 - 1.732i

We obtain the same result algebraically as we did using trigonometry. You might argue that it was easier to use the algebraic method, but that's only because you know how to factor x^{3} - 8, and the factors are a linear factor and a quadratic. But suppose I'd asked you to find the fifth root of 32...what would you have done then? Or what if I'd asked you to find the cube root of 1 - 2i?

In these cases, algebraic solutions would be much more challenging, and the the trigonometry solution simplifies the process significantly.