# How do I find the cube root of a complex number?

Nov 4, 2015

See expanation.

#### Explanation:

To find a cubic root (or generally root of degree $n$) you have to use de'Moivre's formula:

${z}^{\frac{1}{n}} = | z {|}^{\frac{1}{n}} \cdot \left(\cos \left(\frac{\phi + 2 k \pi}{n}\right) + i \sin \left(\frac{\phi + 2 k \pi}{n}\right)\right)$ for $k \in \left\{0 , 1 , 2 , \ldots , n - 1\right\}$

From tis formula you can see, that every complex number has $n$ roots of degree $n$

So to calculate root of a complex number you first have to write the number in a trigonometric form

May 12, 2018

There are three cube roots:

${\left(r {e}^{i \theta}\right)}^{\frac{1}{3}} = \sqrt[3]{r} \setminus {e}^{i \left(\frac{\theta}{3} + \frac{2 \pi k}{3}\right)} \quad$ for $k = - 1 , 0 , \mathmr{and} 1.$

#### Explanation:

This is an important question, or is at least hinting at one. Unfortunately I can only get to the question in the usual space allotted. The answer will have to wait.

I'm never sure what precalculus is. What's left after Algebra I, Algebra II, Geometry and Trigonometry? But let's assume it's a time to get deep about what the math we've been doing really means.

Taking the cube root is easy if we have our complex number in polar coordinates. I'll write the polar form as

$z = r {e}^{i \theta}$

(Hopefully they do it this way in precalc; it makes everything easy).

Every non-zero complex number has three cube roots. In general, any non-integer exponent, like $\frac{1}{3}$ here, gives rise to multiple values. The way we find them is by multiplying $z$ by 1 before exponentiating. We write $1$ using Euler's Identity to the $2 k$ power for integer $k$. That's

${e}^{2 \pi k i} = 1$

Now

${z}^{\frac{1}{3}} = {\left(r {e}^{i \theta} {e}^{2 \pi k i}\right)}^{\frac{1}{3}} = \sqrt[3]{r} \setminus {e}^{i \left(\frac{\theta}{3} + \frac{2 \pi k}{3}\right)}$

Those are some symbols that's say if you want to take the cube root of a complex number, take the (real) cube root of its magnitude, and divide the angle by three. That's one cube root. Then the same with the angle $\setminus \pm {120}^{\circ}$ are the other two cube roots.

In theory this procedure works if we're given $z$ in rectangular coordinates. We convert to polar, take the cube root, convert back. It's a magical walk through a transcendental tunnel.

In practice this is a bit problematic. It's a big shortcut to the answer, but it generally extracts a cost: we end up either with an approximation or a result expressed with trig functions. Neither is ideal.

The deep question we want to address is how to express $\quad \sqrt[3]{a + b i} \quad$ as $\quad x + y i .$

This is related to the denesting of radicals, because $i = \sqrt{- 1}$.

We're especially interested in the case where $a$ and $b$ are integers or rationals or quadratic surds ( $g + h \sqrt{d}$ for rational $g , h , d$ where $d$ isn't a perfect square ) and we want to know if there are rationals or surds for $x$ and $y$.

I'm being warned this is getting long. We've barely gotten to the issue. The most pressing place this comes up is in the cubic formula, which is like the quadratic formula for cubic equations. The simplest form is called the depressed cubic:

${x}^{3} + 3 p x = 2 q$ has solutions

$x = \sqrt[3]{q + \setminus \sqrt{{q}^{2} + {p}^{3}}} + \sqrt[3]{q - \setminus \sqrt{{q}^{2} + {p}^{3}}}$

We see the cube root of a square root of what is often a negative number, so the cube root of a complex number in rectangular form.

Let's make up a cubic of this form with one root we know, say $x = 2.$

$8 + 6 p = 2 q$

We'll skip $p = - 1 , q = 1$ because ${q}^{2} + {p}^{3}$ isn't negative. Let's go with

$p = - 2 , q = - 2 , {q}^{2} + {p}^{3} = - 4$

${x}^{3} - 6 x = - 4$

Applying the formula,

$x = \sqrt[3]{- 2 + \setminus \sqrt{- 4}} + \sqrt[3]{- 2 - \setminus \sqrt{- 4}}$

$x = \sqrt[3]{- 2 + 2 i} + \sqrt[3]{- 2 - 2 i}$

Here we're supposed to get two but we get the sum of two cube roots of complex numbers. I want to show you how to denest these, which is kinda similar to factoring.

But Socratic says I'm intimidating you all, so I'll stop here and continue this in some future answer. In the meantime, get a pencil and see if you can make this expression equal to two. Other questions to ponder: What are the other two solutions? Each cube root has three values, what are they? Isn't this nine combinations for the sum $x$, how can this be when a cubic has three solutions?