>Calculations>What Are the Roots of a Complex Number?

What Are the Roots of a Complex Number?

The number $- 1$ has no real square roots because there are no real numbers that when multiplied by themselves give a negative product. The solution to this barrier is to define the imaginary number $i$ as the square root of $- 1$ . By using $i$ , it can be shown that you can find general $n$ th roots of all numbers, including the complex numbers:

Theory

Definition of $n$ th Root

The complex number $w$ is an $n$ th root of $z$ if $w$ satisfies

w^{n} = z .

An $n$ th root of the number $z$ is a number that when multiplied by itself $n$ times yields $z$ . To find the $n$ th roots of a complex number $z$ , it is smart to write $z$ in polar form. If you write $z = r e^{i 𝜃}$ , you can easily find a number $w_{0}$ which fulfills the definition of $n$ th root by using normal power rules:

w_{0} = {(r e^{i 𝜃})}^{\frac{1}{n}} = r^{\frac{1}{n}} e^{i \frac{𝜃}{n}} .

To find an $n$ th root of $z$ , you take the $n$ th root of the norm of $z$ and divide the argument of $z$ by $n$ . It is also possible to find other $n$ th roots of $z$ , because the polar form of $z$ is not unique.

A rotation of $2 π$ radians does not change the value of $z$ , so you can write

z = r e^{i (𝜃 + 2 π \cdot k)}

for an arbitrary integer $k \in ℤ$ . By using the power rules, you can get a general expression for the $n$ th roots of $z$ :

w_{k} = r^{\frac{1}{n}} e^{i (\frac{𝜃}{n} + \frac{2 π}{n} \cdot k)} .

Every integer $k$ gives you a new $n$ th root $w_{k}$ , as long as $k = 0, 1, 2, \dots, n - 1$ . When $k$ is $n$ or greater, you will no longer get new $n$ th roots. This is because you now have rotated $2 π$ radians and are back at the starting point $w_{0}$ .

Formula

$n$ th Roots

For every complex number $z = r e^{i 𝜃} \neq 0$ , there exist $n$ different $n$ th roots $w_{0}, w_{1}, w_{2}, \dots, w_{n - 1}$ given by

w_{k} = r^{\frac{1}{n}} e^{i (\frac{𝜃}{n} + \frac{2 π}{n} \cdot k)},

for integer $k = 0, 1, 2, \dots, n - 1 .$

Every $n$ th root of $z$ has the same norm, and the difference in argument between $w_{k}$ and $w_{k - 1}$ is $\frac{2 π}{n}$ . If you know $w_{k}$ , you can find $w_{k + 1}$ by

w_{k + 1} = w_{+} \cdot w_{k},

where $w_{+} = e^{i \frac{2 π}{n}}$ . This is because a multiplication with $w_{+}$ can be thought of as a rotation of $\frac{2 π}{n}$ radians in the complex plane.

The strategy to find the $n$ th roots of $z$ is to first find $w_{0} = z^{\frac{1}{n}}$ and then the other $n - 1$ roots by multiplying by $w_{+}$ .

Example 1

Find the fourth roots of $z = - 81$

To find the fourth roots it is smart to write $z$ in polar form. Here is the norm $r = 81$ , and the argument $𝜃 = π$ . In polar form $z$ is therefore

z = 81 e^{i π} .

Next, you can find $w_{0}$ :

w_{0} = z^{\frac{1}{4}} = 3 e^{i \frac{π}{4}},

and $w_{+}$ :

w_{+} = e^{i \frac{2 π}{4}} = e^{i \frac{π}{2}} .

Now you can find the other roots by using the connection $w_{k + 1} = w_{+} \cdot w_{k}$ :

\begin{array}{l} w_{1} & = w_{+} \cdot w_{0} \\ = e^{i \frac{π}{2}} \cdot 3 e^{i \frac{π}{4}} \\ = 3 e^{i \frac{3 π}{4}} \\ w_{2} & = w_{+} \cdot w_{1} \\ = e^{i \frac{π}{2}} \cdot 3 e^{i \frac{3 π}{4}} \\ = 3 e^{i \frac{5 π}{4}} \\ w_{3} & = w_{+} \cdot w_{2} \\ = e^{i \frac{π}{2}} \cdot 3 e^{i \frac{5 π}{4}} \\ = 3 e^{i \frac{7 π}{4}} . \end{array}

\begin{array}{l} w_{1} & = w_{+} \cdot w_{0} = e^{i \frac{π}{2}} \cdot 3 e^{i \frac{π}{4}} = 3 e^{i \frac{3 π}{4}} \\ w_{2} & = w_{+} \cdot w_{1} = e^{i \frac{π}{2}} \cdot 3 e^{i \frac{3 π}{4}} = 3 e^{i \frac{5 π}{4}} \\ w_{3} & = w_{+} \cdot w_{2} = e^{i \frac{π}{2}} \cdot 3 e^{i \frac{5 π}{4}} = 3 e^{i \frac{7 π}{4}} . \end{array}

Since the exercise gave you the number

z

in Cartesian form, you should also give the answer in Cartesian form. In Cartesian form the fourth roots of

z

are:

\begin{array}{l} w_{0} & = \frac{3 \sqrt{2}}{2} + \frac{3 \sqrt{2}}{2} i, \\ w_{1} & = - \frac{3 \sqrt{2}}{2} + \frac{3 \sqrt{2}}{2} i, \\ w_{2} & = - \frac{3 \sqrt{2}}{2} - \frac{3 \sqrt{2}}{2} i, \\ w_{3} & = \frac{3 \sqrt{2}}{2} - \frac{3 \sqrt{2}}{2} i . \end{array}

Since all $n$ th roots of a number have the same norm, the $n$ th roots $w_{0}, w_{1}, w_{2}, \dots, w_{n - 1}$ are uniformly distributed around a circle in the complex plane. The fourth roots of $81$ from Example 1 lie on a circle in the complex plane with radius $3$ , and the angle between the roots is $\frac{π}{2}$ radians:

The fourth roots of 81 visualized in the complex plane.

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