# Roots of Complex Numbers

The phrase "square roots" is familiar but I'm sure the phrase "complex numbers" is most likely unfamiliar and scary. Scary things can be fun, hereafter you would learn and have fun finding the roots of complex numbers.

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## What is the calculation of roots of complex numbers?

The calculation of roots of complex numbers is the process of finding the roots (square, cube, etc.) of complex numbers in the form:

$a+bi$

where, a is the real component of the complex numbers,

b is the imaginary component of the complex number, and

i is id="2777604" role="math" $\sqrt{-1}$.

For recollection purposes, it is necessary we remind you about what complex numbers are before advancing to finding their roots.

### What are complex numbers?

Complex numbers are numbers that are expressed in the form:

$a+bi$

with a being a real number and bi being an imaginary value used to represent the root of negative values (e.g. $\sqrt{-16},\sqrt{-5}$ etc.). Also, i represents iota and is sometimes represented by j, with both meaning $\sqrt{-1}$.

In Mathematics, there is no solution for finding the roots of negative numbers, this is why the idea of complex numbers came about, to go further into solving the roots of negative numbers. Here, the root of a negative number like $\sqrt{-16}$ can be solved as:

$\sqrt{-16}=\sqrt{-1}×\sqrt{16}$

where $\sqrt{-1}$ is given the value i, then,

$\sqrt{-16}=i×\sqrt{16}\phantom{\rule{0ex}{0ex}}\sqrt{-16}=i×4\phantom{\rule{0ex}{0ex}}\sqrt{-16}=4i\phantom{\rule{0ex}{0ex}}$

Note that:

$i=\sqrt{-1}\phantom{\rule{0ex}{0ex}}{i}^{2}={\sqrt{-1}}^{2}=\sqrt{-1}×\sqrt{-1}=-1\phantom{\rule{0ex}{0ex}}{i}^{3}={\sqrt{-1}}^{3}=\sqrt{-1}×\sqrt{-1}×\sqrt{-1}=-\sqrt{-1}=-i$

### Calculating the square root of complex numbers

Among the calculations for the roots of complex numbers, finding square roots of complex numbers can be achieved with a more direct approach rather than the application of a formula. The example below would see you through this.

Find the square root of the following.

$3+4i$

Solution:

The first step is to equate the complex number to the square of the general equation, therefore,

$3+4i={\left(a+bi\right)}^{2}\phantom{\rule{0ex}{0ex}}3+4i={a}^{2}+2abi+{b}^{2}{i}^{2}\phantom{\rule{0ex}{0ex}}{i}^{2}=-1\phantom{\rule{0ex}{0ex}}3+4i={a}^{2}+2abi-{b}^{2}$

Recall that all expressions with i as an imaginary expression, hence, rearrange the expressions by bringing like terms:

$3+4i={a}^{2}+2abi-{b}^{2}\phantom{\rule{0ex}{0ex}}3+4i={a}^{2}-{b}^{2}+2abi$

On both sides of the equation, we have real and imaginary expressions. Equate the expressions based on similar components (i.e. real to real and imaginary to imaginary). So for the real component we have:

$3={a}^{2}-{b}^{2}$

and for the imaginary component we have:

$4i=2abi\phantom{\rule{0ex}{0ex}}\frac{4i}{2i}=\frac{2abi}{2i}\phantom{\rule{0ex}{0ex}}\frac{{}^{2}\overline{)4i}}{{}^{1}\overline{)2i}}=\frac{{}^{1}\overline{)2}ab\overline{)i}}{{}^{1}\overline{)2}\overline{)i}}\phantom{\rule{0ex}{0ex}}2=ab\phantom{\rule{0ex}{0ex}}a=\frac{2}{b}$

Substitute the value of a as$\frac{2}{b}$into the equation, $3={a}^{2}-{b}^{2}$. Therefore,

$3={a}^{2}-{b}^{2}\phantom{\rule{0ex}{0ex}}a=\frac{2}{b}\phantom{\rule{0ex}{0ex}}3={\left(\frac{2}{b}\right)}^{2}-{b}^{2}\phantom{\rule{0ex}{0ex}}3=\frac{4}{{b}^{2}}-{b}^{2}$

Multiply through by b2,

$3=\frac{4}{{b}^{2}}-{b}^{2}\phantom{\rule{0ex}{0ex}}3×{b}^{2}={b}^{2}×\left(\frac{4}{{b}^{2}}-{b}^{2}\right)\phantom{\rule{0ex}{0ex}}3{b}^{2}=4-{b}^{4}$

Let

$x={b}^{2}$

Then

$3{b}^{2}=4-{b}^{4}\phantom{\rule{0ex}{0ex}}x={b}^{2}\phantom{\rule{0ex}{0ex}}3x=4-{x}^{2}\phantom{\rule{0ex}{0ex}}{x}^{2}+3x-4=0\phantom{\rule{0ex}{0ex}}{x}^{2}+4x-x-4=0\phantom{\rule{0ex}{0ex}}\left({x}^{2}+4x\right)+\left(-x-4\right)=0\phantom{\rule{0ex}{0ex}}x\left(x+4\right)-1\left(x+4\right)=0\phantom{\rule{0ex}{0ex}}\left(x-1\right)\left(x+4\right)=0\phantom{\rule{0ex}{0ex}}x=1\mathrm{or}x=-4$

Therefore, the value of x is 1 or -4. Recall that:

$x={b}^{2}$

Thus,

$x={b}^{2}\phantom{\rule{0ex}{0ex}}\sqrt{x}=\sqrt{{b}^{2}}\phantom{\rule{0ex}{0ex}}b=\sqrt{x}$

This means that b is $\sqrt{1}$ or $\sqrt{-4}$, we know that $\sqrt{-4}$ is invalid but there is a solution for $\sqrt{1}$. Therefore, b is +1 or -1. Recall that:

$a=\frac{2}{b}$

Therefore,

$a=\frac{2}{b}\phantom{\rule{0ex}{0ex}}a=\frac{2}{1}\phantom{\rule{0ex}{0ex}}a=2$

or

$a=\frac{2}{b}\phantom{\rule{0ex}{0ex}}a=\frac{2}{-1}\phantom{\rule{0ex}{0ex}}a=-2$

Recall that:

$3+4i={\left(a+bi\right)}^{2}$

By substituting the values of a and b to get the root of the complex number, we have;

$3+4i={\left(a+bi\right)}^{2}\phantom{\rule{0ex}{0ex}}\sqrt{3+4i}=\sqrt{{\left(a+bi\right)}^{2}}\phantom{\rule{0ex}{0ex}}\sqrt{3+4i}=±\left(a+bi\right)\phantom{\rule{0ex}{0ex}}\sqrt{3+4i}=2+i$

or

$\sqrt{3+4i}=-2-i$

Therefore the square roots of the complex number$3+4i$is$2+i$or$-2-i$.

## What are roots of complex numbers in polar form?

To derive the roots of complex numbers in polar form, you would need to understand how complex numbers are expressed in polar forms. With a good understanding of this, it would become easier to determine the roots of complex numbers in polar forms.

Consider Z to be the complex number a + bi, the distance between the origin and point Z is regarded as the modulus. This is represented as |Z| and is equal to the square root of the sum of squares of a and b.

$\left|Z\right|=\sqrt{{a}^{2}+{b}^{2}}$

An angle θ is subtended between the positive real axis and the line of the modulus Z whose endpoint is located at the origin. θ is referred to as the argument of the modulus z.

$\mathrm{tan}\left(\theta \right)=\frac{b}{a}\phantom{\rule{0ex}{0ex}}\theta ={\mathrm{tan}}^{-1}\frac{b}{a}$

Where,

$-180°\le \theta \le 180°\phantom{\rule{0ex}{0ex}}-\mathrm{\pi }\le \mathrm{\theta }\le \mathrm{\pi }$

An illustration of the modulus and argument of a complex number - StudySmarter Originals

From the above illustration with b being the imaginary axis and a being the real positive axis, we can then define the complex number, Z in terms of its modulus |Z| and argument θ as:

$a=|Z|\mathrm{cos}\left(\theta \right)\phantom{\rule{0ex}{0ex}}b=|Z|\mathrm{sin}\left(\theta \right)\phantom{\rule{0ex}{0ex}}Z=|Z|\mathrm{cos}\left(\theta \right)+|Z|\mathrm{sin}\left(\theta \right)i$

However, the modulus of a complex number |Z| is often referred to as r with the argument θ, this gives rise to the polar form of a complex number represented using (r,θ). Therefore, the complex number Z can be rewritten in terms of its modulus r and argument θ as:

$Z=|Z|\mathrm{cos}\left(\theta \right)+|Z|\mathrm{sin}\left(\theta \right)i\phantom{\rule{0ex}{0ex}}Z=r\mathrm{cos}\left(\theta \right)+r\mathrm{sin}\left(\theta \right)i\phantom{\rule{0ex}{0ex}}Z=r\left(\mathrm{cos}\left(\theta \right)+i\mathrm{sin}\left(\theta \right)\right)$

This equation is the expression of a complex number in its polar form.

Express the complex number, $Z=8+6i$ in polar form.

Solution:

The polar form of a complex number is expressed using (r,θ). To derive both, firstly, we need to find the modulus r, therefore:

$r=\left|Z\right|\phantom{\rule{0ex}{0ex}}\left|Z\right|=\sqrt{{a}^{2}+{b}^{2}}\phantom{\rule{0ex}{0ex}}r=\sqrt{{a}^{2}+{b}^{2}}\phantom{\rule{0ex}{0ex}}a=8\phantom{\rule{0ex}{0ex}}b=6\phantom{\rule{0ex}{0ex}}r=\sqrt{{8}^{2}+{6}^{2}}\phantom{\rule{0ex}{0ex}}r=\sqrt{64+36}\phantom{\rule{0ex}{0ex}}r=\sqrt{100}\phantom{\rule{0ex}{0ex}}r=10$

Next, we need to find the argument, θ. Thus,

$\theta ={\mathrm{tan}}^{-1}\left(\frac{b}{a}\right)\phantom{\rule{0ex}{0ex}}\theta ={\mathrm{tan}}^{-1}\left(\frac{6}{8}\right)\phantom{\rule{0ex}{0ex}}\theta ={\mathrm{tan}}^{-1}\left(0.75\right)\phantom{\rule{0ex}{0ex}}\theta =36.87°$

Now, that both values of the modulus and argument have been derived, we can put it into the polar form equation, Thus,

$Z=r\left(\mathrm{cos}\left(\theta \right)+i\mathrm{sin}\left(\theta \right)\right)\phantom{\rule{0ex}{0ex}}Z=10\left(\mathrm{cos}\left(36.87°\right)+i\mathrm{sin}\left(36.87°\right)\right)$

## Roots of complex numbers formula

In order to derive a formula that can guide you through the successful calculation of the roots of complex numbers, you must understand the multiplication of complex numbers in their polar form. Therefore, for two complex numbers, Z1 and Z2 expressed as:

${Z}_{1}={r}_{1}\left(\mathrm{cos}\left({\theta }_{1}\right)+i\mathrm{sin}\left({\theta }_{1}\right)\right)\phantom{\rule{0ex}{0ex}}{Z}_{2}={r}_{2}\left(\mathrm{cos}\left({\theta }_{2}\right)+i\mathrm{sin}\left({\theta }_{2}\right)\right)$

Then, the product of Z1 and Z2 becomes:

${Z}_{1}×{Z}_{2}={r}_{1}\left(\mathrm{cos}\left({\theta }_{1}\right)+i\mathrm{sin}\left({\theta }_{1}\right)\right)×{r}_{2}\left(\mathrm{cos}\left({\theta }_{2}\right)+i\mathrm{sin}\left({\theta }_{2}\right)\right)\phantom{\rule{0ex}{0ex}}{Z}_{1}×{Z}_{2}={r}_{1}×{r}_{2}\left(\mathrm{cos}\left({\theta }_{1}\right)+i\mathrm{sin}\left({\theta }_{1}\right)\right)\left(\mathrm{cos}\left({\theta }_{2}\right)+i\mathrm{sin}\left({\theta }_{2}\right)\right)\phantom{\rule{0ex}{0ex}}{Z}_{1}×{Z}_{2}={r}_{1}×{r}_{2}\left(\mathrm{cos}{\theta }_{1}\mathrm{cos}{\theta }_{2}+i\mathrm{cos}{\theta }_{2}\mathrm{sin}{\theta }_{1}+i\mathrm{cos}{\theta }_{1}\mathrm{sin}{\theta }_{2}+{i}^{2}\mathrm{sin}{\theta }_{1}\mathrm{sin}{\theta }_{2}\right)\phantom{\rule{0ex}{0ex}}{i}^{2}=-1\phantom{\rule{0ex}{0ex}}{Z}_{1}×{Z}_{2}={r}_{1}×{r}_{2}\left(\mathrm{cos}{\theta }_{1}\mathrm{cos}{\theta }_{2}+i\mathrm{cos}{\theta }_{2}\mathrm{sin}{\theta }_{1}+i\mathrm{cos}{\theta }_{1}\mathrm{sin}{\theta }_{2}-\mathrm{sin}{\theta }_{1}\mathrm{sin}{\theta }_{2}\right)\phantom{\rule{0ex}{0ex}}{Z}_{1}×{Z}_{2}={r}_{1}×{r}_{2}\left(\mathrm{cos}{\theta }_{1}\mathrm{cos}{\theta }_{2}-\mathrm{sin}{\theta }_{1}\mathrm{sin}\theta +i\mathrm{cos}{\theta }_{2}\mathrm{sin}{\theta }_{1}+i\mathrm{cos}{\theta }_{1}\mathrm{sin}{\theta }_{2}\right)\phantom{\rule{0ex}{0ex}}i\mathrm{cos}{\theta }_{2}\mathrm{sin}{\theta }_{1}+i\mathrm{cos}{\theta }_{1}\mathrm{sin}{\theta }_{2}=i\left(\mathrm{sin}{\theta }_{1}\mathrm{cos}{\theta }_{2}+\mathrm{sin}{\theta }_{2}\mathrm{cos}{\theta }_{1}\right)\phantom{\rule{0ex}{0ex}}{Z}_{1}×{Z}_{2}={r}_{1}×{r}_{2}\left(\mathrm{cos}{\theta }_{1}\mathrm{cos}{\theta }_{2}-\mathrm{sin}{\theta }_{1}\mathrm{sin}\theta +i\left(\mathrm{sin}{\theta }_{1}\mathrm{cos}{\theta }_{2}+\mathrm{sin}{\theta }_{2}\mathrm{cos}{\theta }_{1}\right)\right)\phantom{\rule{0ex}{0ex}}\mathrm{cos}{\theta }_{1}\mathrm{cos}{\theta }_{2}-\mathrm{sin}{\theta }_{1}\mathrm{sin}\theta =\mathrm{cos}\left({\theta }_{1}+{\theta }_{2}\right)\phantom{\rule{0ex}{0ex}}\mathrm{sin}{\theta }_{1}\mathrm{cos}{\theta }_{2}+\mathrm{sin}{\theta }_{2}\mathrm{cos}{\theta }_{1}=\mathrm{sin}\left({\theta }_{1}+{\theta }_{2}\right)\phantom{\rule{0ex}{0ex}}{Z}_{1}×{Z}_{2}={r}_{1}×{r}_{2}\left(\mathrm{cos}\left({\theta }_{1}+{\theta }_{2}\right)+i\mathrm{sin}\left({\theta }_{1}+{\theta }_{2}\right)\right)$

We shall dwell on only the multiplication of complex numbers in their polar forms as it remains more relevant to us than division operations. Now you have seen what happens with the product operation, can you now imagine how roots are gotten? Before advancing to roots, let us see what happens to exponents of polar forms.

We know generally exponents would be in the form, Zn. Apparently, the simplest exponent to deal with in this case is Z2. Surely, you agree that:

${Z}^{2}=Z×Z$

Where,

${Z}_{1}={Z}_{2}\phantom{\rule{0ex}{0ex}}{Z}_{1}×{Z}_{2}={Z}_{1}×{Z}_{1}\phantom{\rule{0ex}{0ex}}{Z}_{1}×{Z}_{1}={{Z}_{1}}^{2}\phantom{\rule{0ex}{0ex}}$

Also, this works for the argument, θ1 and θ2

${\theta }_{1}={\theta }_{2}$

This implies that:

${Z}_{1}×{Z}_{2}={r}_{1}×{r}_{2}\left(\mathrm{cos}\left({\theta }_{1}+{\theta }_{2}\right)+i\mathrm{sin}\left({\theta }_{1}+{\theta }_{2}\right)\right)\phantom{\rule{0ex}{0ex}}{Z}_{1}={Z}_{2}\phantom{\rule{0ex}{0ex}}{\theta }_{1}={\theta }_{2}\phantom{\rule{0ex}{0ex}}{r}_{1}={r}_{2}\phantom{\rule{0ex}{0ex}}{{Z}_{1}}^{2}={Z}_{1}×{Z}_{1}={r}_{1}×{r}_{1}\left(\mathrm{cos}\left({\theta }_{1}+{\theta }_{1}\right)+i\mathrm{sin}\left({\theta }_{1}+{\theta }_{1}\right)\right)\phantom{\rule{0ex}{0ex}}{{Z}_{1}}^{2}={{r}_{1}}^{2}\left(\mathrm{cos}\left(2{\theta }_{1}\right)+i\mathrm{sin}\left(2{\theta }_{1}\right)\right)$

Therefore, for the exponent of a complex number, Zn, we can say,

${{Z}_{1}}^{n}={{r}_{1}}^{n}\left(\mathrm{cos}\left(n{\theta }_{1}\right)+i\mathrm{sin}\left(n{\theta }_{1}\right)\right)$

The above formula is what we regard as de Moivre's Theorem. We would discuss this later on.

Meanwhile, our quest is to determine the roots of complex numbers in their polar form, not their exponents. Nonetheless, this theorem is useful as it helps us determine roots. How?

Interestingly, you know that the exponent of the complex number Z is in the form Zn, the roots of Z would be in the form Z1/n. Thus, by simply applying the theorem, we can get very close to completing our task of finding the roots of complex numbers in their polar form. Thus:

${Z}^{n}={r}^{n}\left(\mathrm{cos}\left(n\theta \right)+i\mathrm{sin}\left(n\theta \right)\right)\phantom{\rule{0ex}{0ex}}{Z}^{\frac{1}{n}}={r}^{\frac{1}{n}}\left(\mathrm{cos}\left(\frac{\theta }{n}\right)+i\mathrm{sin}\left(\frac{\theta }{n}\right)\right)\phantom{\rule{0ex}{0ex}}$

Moreover, this formula cannot be effective because dealing with roots means several other numbers which have to be accounted for. In a circle below,

An illustration on the roots of complex numbers in a coordinate plane, StudySmarter Originals

You would infer that the distance between successive roots is given by$\frac{2\mathrm{\pi }}{n}$ radians. Therefore, in other to account for other roots, the roots of complex numbers were effectively adjusted to become,

${Z}^{\frac{1}{n}}={r}^{\frac{1}{n}}\left(\mathrm{cos}\left(\frac{\theta +2k\mathrm{\pi }}{n}\right)+i\mathrm{sin}\left(\frac{\theta +2k\mathrm{\pi }}{n}\right)\right)$

where k is 0, 1, 2, 3..., n-2, and n-1.

This means that the roots of complex numbers in their polar form is calculated using the formula,

${Z}^{\frac{1}{n}}={r}^{\frac{1}{n}}\left(\mathrm{cos}\left(\frac{\theta +2k\mathrm{\pi }}{n}\right)+i\mathrm{sin}\left(\frac{\theta +2k\mathrm{\pi }}{n}\right)\right)$

Note that the values for k ranges begin with 0 and end with n-1 because when k is 0 you have the first root while when k is n-1, you have the last root.

## How can we use de Moivre's Theorem to solve equations?

Earlier on, we mentioned the de Moivre's Theorem,

${Z}^{n}={r}^{n}\left(\mathrm{cos}\left(n\theta \right)+i\mathrm{sin}\left(n\theta \right)\right)$

We shall now apply de Moivre's Theorem in determining the roots of complex numbers in their polar form. Meanwhile, de Moivre's Theorem is applicable to all rational numbers and fractions.

Find the fourth roots of the equation.

${Z}^{4}=256$.

Solution:

Firstly, we need to express the complex number in its polar form. But to achieve this, the argument, θ has to be gotten. In line with the general format:

$Z=a+bi$

Then,

$256=256+0i\phantom{\rule{0ex}{0ex}}r=\sqrt{{256}^{2}+{0}^{2}}=256\phantom{\rule{0ex}{0ex}}\theta ={\mathrm{tan}}^{-1}\left(\frac{0}{256}\right)=\pi$

Since the argument θ is π, then the polar form of the complex number Z4 can be expressed as,

${Z}^{4}=256\left(\mathrm{cos}\left(\pi \right)+i\mathrm{sin}\left(\pi \right)\right)$

In finding the roots of the complex number in its polar form we apply the formula:

${Z}^{\frac{1}{n}}={r}^{\frac{1}{n}}\left(\mathrm{cos}\left(\frac{\theta +2k\mathrm{\pi }}{n}\right)+i\mathrm{sin}\left(\frac{\theta +2k\mathrm{\pi }}{n}\right)\right)$

Because we need to find values for Z and we have Z4, it means we have four roots. This means that you would have to solve for situations where k is 0, 1, 2 and 3.

When k is 0, therefore,

${Z}^{\frac{1}{n}}={r}^{\frac{1}{n}}\left(\mathrm{cos}\left(\frac{\theta +2k\mathrm{\pi }}{n}\right)+i\mathrm{sin}\left(\frac{\theta +2k\mathrm{\pi }}{n}\right)\right)\phantom{\rule{0ex}{0ex}}k=0\phantom{\rule{0ex}{0ex}}\theta =\pi \phantom{\rule{0ex}{0ex}}{Z}^{4}=256\left(\mathrm{cos}\left(\pi \right)+i\mathrm{sin}\left(\pi \right)\right)\phantom{\rule{0ex}{0ex}}{\left({Z}^{4}\right)}^{\frac{1}{4}}={256}^{\frac{1}{4}}\left(\mathrm{cos}\left(\frac{\pi +2\left(0\right)\left(\pi \right)}{4}\right)+i\mathrm{sin}\left(\frac{\pi +2\left(0\right)\left(\pi \right)}{4}\right)\right)\phantom{\rule{0ex}{0ex}}Z=4\left(\mathrm{cos}\frac{\pi }{4}+i\mathrm{sin}\frac{\pi }{4}\right)$

This is the first root of the complex number.

To solve for the second root, we solve for k as 1, therefore,

${Z}^{\frac{1}{n}}={r}^{\frac{1}{n}}\left(\mathrm{cos}\left(\frac{\theta +2k\mathrm{\pi }}{n}\right)+i\mathrm{sin}\left(\frac{\theta +2k\mathrm{\pi }}{n}\right)\right)\phantom{\rule{0ex}{0ex}}k=1\phantom{\rule{0ex}{0ex}}\theta =\pi \phantom{\rule{0ex}{0ex}}{Z}^{4}=256\left(\mathrm{cos}\left(\pi \right)+i\mathrm{sin}\left(\pi \right)\right)\phantom{\rule{0ex}{0ex}}{\left({Z}^{4}\right)}^{\frac{1}{4}}={256}^{\frac{1}{4}}\left(\mathrm{cos}\left(\frac{\pi +2\left(1\right)\left(\pi \right)}{4}\right)+i\mathrm{sin}\left(\frac{\pi +2\left(1\right)\left(\pi \right)}{4}\right)\right)\phantom{\rule{0ex}{0ex}}Z=4\left(\mathrm{cos}\frac{3\pi }{4}+i\mathrm{sin}\frac{3\pi }{4}\right)$

This is the second root of the complex number.

To solve for the third root, we solve for k as 2, therefore,

${Z}^{\frac{1}{n}}={r}^{\frac{1}{n}}\left(\mathrm{cos}\left(\frac{\theta +2k\mathrm{\pi }}{n}\right)+i\mathrm{sin}\left(\frac{\theta +2k\mathrm{\pi }}{n}\right)\right)\phantom{\rule{0ex}{0ex}}k=2\phantom{\rule{0ex}{0ex}}\theta =\pi \phantom{\rule{0ex}{0ex}}{Z}^{4}=256\left(\mathrm{cos}\left(\pi \right)+i\mathrm{sin}\left(\pi \right)\right)\phantom{\rule{0ex}{0ex}}{\left({Z}^{4}\right)}^{\frac{1}{4}}={256}^{\frac{1}{4}}\left(\mathrm{cos}\left(\frac{\pi +2\left(2\right)\left(\pi \right)}{4}\right)+i\mathrm{sin}\left(\frac{\pi +2\left(2\right)\left(\pi \right)}{4}\right)\right)\phantom{\rule{0ex}{0ex}}Z=4\left(\mathrm{cos}\frac{5\pi }{4}+i\mathrm{sin}\frac{5\pi }{4}\right)$

This is the third root of the complex number.

To solve for the fourth or last root, we solve for k as 3, therefore,

${Z}^{\frac{1}{n}}={r}^{\frac{1}{n}}\left(\mathrm{cos}\left(\frac{\theta +2k\mathrm{\pi }}{n}\right)+i\mathrm{sin}\left(\frac{\theta +2k\mathrm{\pi }}{n}\right)\right)\phantom{\rule{0ex}{0ex}}k=3\phantom{\rule{0ex}{0ex}}\theta =\pi \phantom{\rule{0ex}{0ex}}{Z}^{4}=256\left(\mathrm{cos}\left(\pi \right)+i\mathrm{sin}\left(\pi \right)\right)\phantom{\rule{0ex}{0ex}}{\left({Z}^{4}\right)}^{\frac{1}{4}}={256}^{\frac{1}{4}}\left(\mathrm{cos}\left(\frac{\pi +2\left(3\right)\left(\pi \right)}{4}\right)+i\mathrm{sin}\left(\frac{\pi +2\left(3\right)\left(\pi \right)}{4}\right)\right)\phantom{\rule{0ex}{0ex}}Z=4\left(\mathrm{cos}\frac{7\pi }{4}+i\mathrm{sin}\frac{7\pi }{4}\right)$

This is the last root of the complex number.

So our roots are:

$z=4\left(\mathrm{cos}\frac{\pi }{4}+i\mathrm{sin}\frac{\pi }{4}\right),z=4\left(\mathrm{cos}\frac{3\pi }{4}+i\mathrm{sin}\frac{3\pi }{4}\right)\phantom{\rule{0ex}{0ex}}z=4\left(\mathrm{cos}\frac{5\pi }{4}+i\mathrm{sin}\frac{5\pi }{4}\right),z=4\left(\mathrm{cos}\frac{7\pi }{4}+i\mathrm{sin}\frac{7\pi }{4}\right)$

Note that, roots of equations in the form of ${z}^{n}=1$, are referred to as roots of unity.

## Further examples on roots of complex numbers

To understand the roots of complex numbers, a few more examples would be handy.

Find the third roots of 8 in rectangular form.

Solution:

The first thing to do is to express 8 in the general form of complex numbers.

$z=a+bi$

Thus,

$z=8+0i$

To derive its roots, we need to find values for the r and θ. So,

$r=\sqrt{{a}^{2}+{b}^{2}}\phantom{\rule{0ex}{0ex}}r=\sqrt{{8}^{2}+{0}^{2}}\phantom{\rule{0ex}{0ex}}r=\sqrt{{8}^{2}}\phantom{\rule{0ex}{0ex}}r=8\phantom{\rule{0ex}{0ex}}\theta ={\mathrm{tan}}^{-1}\left(\frac{0}{8}\right)\phantom{\rule{0ex}{0ex}}\theta ={\mathrm{tan}}^{-1}0\phantom{\rule{0ex}{0ex}}\theta =0or\pi$

Now we have values for r and θ, we can find the root in the polar form. Therefore, we have to apply our formula for k values 0, 1 and 2.

${Z}^{\frac{1}{n}}={r}^{\frac{1}{n}}\left(\mathrm{cos}\left(\frac{\theta +2k\mathrm{\pi }}{n}\right)+i\mathrm{sin}\left(\frac{\theta +2k\mathrm{\pi }}{n}\right)\right)\phantom{\rule{0ex}{0ex}}k=0\phantom{\rule{0ex}{0ex}}r=8\phantom{\rule{0ex}{0ex}}n=3\phantom{\rule{0ex}{0ex}}\theta =\pi \phantom{\rule{0ex}{0ex}}{Z}^{\frac{1}{3}}={8}^{\frac{1}{3}}\left(\mathrm{cos}\left(\frac{\pi +2\left(0\right)\left(\mathrm{\pi }\right)}{3}\right)+i\mathrm{sin}\left(\frac{\pi +2\left(0\right)\left(\mathrm{\pi }\right)}{3}\right)\right)\phantom{\rule{0ex}{0ex}}{\mathbit{Z}}^{\frac{\mathbf{1}}{\mathbf{3}}}\mathbf{=}\mathbf{2}\mathbf{\left(}\mathbit{c}\mathbit{o}\mathbit{s}\mathbf{\left(}\frac{\mathbf{\pi }}{\mathbf{3}}\mathbf{\right)}\mathbf{+}\mathbit{i}\mathbit{s}\mathbit{i}\mathbit{n}\mathbf{\left(}\frac{\mathbf{\pi }}{\mathbf{3}}\mathbf{\right)}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}k=1\phantom{\rule{0ex}{0ex}}{Z}^{\frac{1}{3}}={8}^{\frac{1}{3}}\left(\mathrm{cos}\left(\frac{\pi +2\left(1\right)\left(\mathrm{\pi }\right)}{3}\right)+i\mathrm{sin}\left(\frac{\pi +2\left(1\right)\left(\mathrm{\pi }\right)}{3}\phantom{\rule{0ex}{0ex}}{Z}^{\frac{1}{3}}=2\left(cos\left(\frac{3\mathrm{\pi }}{3}\right)+isin\left(\frac{3\mathrm{\pi }}{3}\right)\right)\phantom{\rule{0ex}{0ex}}{\mathbit{Z}}^{\frac{\mathbf{1}}{\mathbf{3}}}\mathbf{=}\mathbf{2}\mathbf{\left(}\mathbit{c}\mathbit{o}\mathbit{s}\mathbit{\pi }\mathbf{+}\mathbit{i}\mathbit{s}\mathbit{i}\mathbit{n}\mathbit{\pi }\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}k=2\phantom{\rule{0ex}{0ex}}{Z}^{\frac{1}{3}}={8}^{\frac{1}{3}}\left(\mathrm{cos}\left(\frac{\pi +2\left(2\right)\left(\mathrm{\pi }\right)}{3}\right)+i\mathrm{sin}\left(\frac{\pi +2\left(2\right)\left(\mathrm{\pi }\right)}{3}\phantom{\rule{0ex}{0ex}}{\mathbit{Z}}^{\frac{\mathbf{1}}{\mathbf{3}}}\mathbf{=}\mathbf{2}\mathbf{\left(}\mathbit{c}\mathbit{o}\mathbit{s}\mathbf{\left(}\frac{\mathbf{5}\mathbf{\pi }}{\mathbf{3}}\mathbf{\right)}\mathbf{+}\mathbit{i}\mathbit{s}\mathbit{i}\mathbit{n}\mathbf{\left(}\frac{\mathbf{5}\mathbf{\pi }}{\mathbf{3}}\mathbf{\right)}\mathbf{\right)}$

The roots we have derived (the answers in bold) are in the polar forms; to convert to the rectangular forms, we need to find the cosine and sine values and input them. Therefore,

${\mathbit{Z}}^{\frac{\mathbf{1}}{\mathbf{3}}}\mathbf{=}\mathbf{2}\mathbf{\left(}\mathbit{c}\mathbit{o}\mathbit{s}\mathbf{\left(}\frac{\mathbf{\pi }}{\mathbf{3}}\mathbf{\right)}\mathbf{+}\mathbit{i}\mathbit{s}\mathbit{i}\mathbit{n}\mathbf{\left(}\frac{\mathbf{\pi }}{\mathbf{3}}\mathbf{\right)}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\mathrm{If}\mathrm{we}\mathrm{consider}\mathrm{\theta }={0}^{°},\phantom{\rule{0ex}{0ex}}{\mathbit{Z}}^{\frac{\mathbf{1}}{\mathbf{3}}}\mathbf{=}\mathbf{2}\mathbf{\left(}\mathbit{c}\mathbit{o}\mathbit{s}\mathbf{\left(}\mathbf{0}\mathbf{°}\mathbf{\right)}\mathbf{+}\mathbit{i}\mathbit{s}\mathbit{i}\mathbit{n}\mathbf{\left(}\mathbf{0}\mathbf{°}\mathbf{\right)}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}{Z}^{\frac{1}{3}}=2\left(1+0i\right)\phantom{\rule{0ex}{0ex}}{\mathbit{Z}}^{\frac{\mathbf{1}}{\mathbf{3}}}\mathbf{=}\mathbf{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

We would repeat the same values of k as 1 and 2, but in the following cases, the first angle is 0° because the first root is located on the positive x-axis and as such the angle formed (θ) is 0° which is the argument, but the next roots have angles greater than 0° and would be derived with$\frac{2k\pi }{n}$, where π is 180°. Thus,

$k=1\phantom{\rule{0ex}{0ex}}{Z}^{\frac{1}{3}}={8}^{\frac{1}{3}}\left(\mathrm{cos}\left(\frac{\theta +2\left(1\right)\left(\mathrm{\pi }\right)}{3}\right)+i\mathrm{sin}\left(\frac{\theta +2\left(1\right)\left(\mathrm{\pi }\right)}{3}\right)\phantom{\rule{0ex}{0ex}}{Z}^{\frac{1}{3}}=2\left(\mathrm{cos}\left(\frac{0°+2\left(1\right)\left(180°\right)}{3}\right)+i\mathrm{sin}\left(\frac{0°+2\left(1\right)\left(180°\right)}{3}\right)\phantom{\rule{0ex}{0ex}}{Z}^{\frac{1}{3}}=2\left(\mathrm{cos}\left(120°\right)+i\mathrm{sin}\left(120°\right)\phantom{\rule{0ex}{0ex}}\mathrm{cos}\left(120°\right)=-\frac{1}{2}\phantom{\rule{0ex}{0ex}}\mathrm{sin}\left(120°\right)=\frac{\sqrt{3}}{2}\phantom{\rule{0ex}{0ex}}{Z}^{\frac{1}{3}}=2\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)\phantom{\rule{0ex}{0ex}}{\mathbit{Z}}^{\frac{\mathbf{1}}{\mathbf{3}}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{+}\sqrt{\mathbf{3}}\mathbit{i}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}k=2\phantom{\rule{0ex}{0ex}}{Z}^{\frac{1}{3}}={8}^{\frac{1}{3}}\left(\mathrm{cos}\left(\frac{\theta +2\left(2\right)\left(\mathrm{\pi }\right)}{3}\right)+i\mathrm{sin}\left(\frac{\theta +2\left(2\right)\left(\mathrm{\pi }\right)}{3}\right)\phantom{\rule{0ex}{0ex}}{Z}^{\frac{1}{3}}=2\left(\mathrm{cos}\left(\frac{0°+2\left(2\right)\left(180°\right)}{3}\right)+i\mathrm{sin}\left(\frac{0°+2\left(2\right)\left(180°\right)}{3}\right)\phantom{\rule{0ex}{0ex}}{Z}^{\frac{1}{3}}=2\left(\mathrm{cos}\left(240°\right)+i\mathrm{sin}\left(240°\right)\phantom{\rule{0ex}{0ex}}\mathrm{cos}\left(240°\right)=-\frac{1}{2}\phantom{\rule{0ex}{0ex}}\mathrm{sin}\left(240°\right)=-\frac{\sqrt{3}}{2}\phantom{\rule{0ex}{0ex}}{Z}^{\frac{1}{3}}=2\left(-\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)\phantom{\rule{0ex}{0ex}}{\mathbit{Z}}^{\frac{\mathbf{1}}{\mathbf{3}}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{-}\sqrt{\mathbf{3}}\mathbit{i}$

Therefore, the complex roots of 8 are: 2, $-1+\sqrt{3}i$ and $-1-\sqrt{3}i$.

Find the fourth roots of the complex number in its polar form:

$-8+8\sqrt{3}i$

Solution:

To derive its roots, we need to find values for the r and θ. So,

$r=\sqrt{{a}^{2}+{b}^{2}}\phantom{\rule{0ex}{0ex}}r=\sqrt{{\left(-8\right)}^{2}+{\left(8\sqrt{3}\right)}^{2}}\phantom{\rule{0ex}{0ex}}r=\sqrt{64+192}\phantom{\rule{0ex}{0ex}}r=\sqrt{256}\phantom{\rule{0ex}{0ex}}\mathbit{r}\mathbf{=}\mathbf{16}\phantom{\rule{0ex}{0ex}}\theta ={\mathrm{tan}}^{-1}\left(\frac{8\sqrt{3}}{-8}\right)\phantom{\rule{0ex}{0ex}}\theta ={\mathrm{tan}}^{-1}\left(-\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}\mathbit{\theta }\mathbf{=}\mathbf{120}\mathbf{°}$

Now we have values for r and θ, we can find the root in the polar form. Therefore, we are to apply our formula for k values 0, 1, 2, and 3. Thus,

${Z}^{\frac{1}{n}}={r}^{\frac{1}{n}}\left(\mathrm{cos}\left(\frac{\theta +2k\mathrm{\pi }}{n}\right)+i\mathrm{sin}\left(\frac{\theta +2k\mathrm{\pi }}{n}\right)\right)\phantom{\rule{0ex}{0ex}}k=0\phantom{\rule{0ex}{0ex}}r=16\phantom{\rule{0ex}{0ex}}n=4\phantom{\rule{0ex}{0ex}}\theta =120°\phantom{\rule{0ex}{0ex}}\mathrm{\pi }=180°\phantom{\rule{0ex}{0ex}}{Z}^{\frac{1}{4}}={16}^{\frac{1}{4}}\left(\mathrm{cos}\left(\frac{120°+2\left(0\right)\left(\mathrm{\pi }\right)}{4}\right)+i\mathrm{sin}\left(\frac{120°+2\left(0\right)\left(\mathrm{\pi }\right)}{4}\right)\right)\phantom{\rule{0ex}{0ex}}{Z}^{\frac{1}{4}}=2\left(cos\left(\frac{120°}{4}\right)+isin\left(\frac{120°}{4}\right)\right)\phantom{\rule{0ex}{0ex}}{\mathbit{Z}}^{\frac{\mathbf{1}}{\mathbf{4}}}\mathbf{=}\mathbf{2}\mathbf{\left(}\mathbit{c}\mathbit{o}\mathbit{s}\mathbf{30}\mathbf{°}\mathbf{+}\mathbit{i}\mathbit{s}\mathbit{i}\mathbit{n}\mathbf{30}\mathbf{°}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}k=1\phantom{\rule{0ex}{0ex}}{Z}^{\frac{1}{4}}={16}^{\frac{1}{4}}\left(\mathrm{cos}\left(\frac{120°+2\left(1\right)\left(180°\right)}{4}\right)+i\mathrm{sin}\left(\frac{120°+2\left(1\right)\left(180°\right)}{4}\right)\phantom{\rule{0ex}{0ex}}{Z}^{\frac{1}{4}}=2\left(cos\left(\frac{480°}{4}\right)+isin\left(\frac{480°}{4}\right)\right)\phantom{\rule{0ex}{0ex}}{\mathbit{Z}}^{\frac{\mathbf{1}}{\mathbf{4}}}\mathbf{=}\mathbf{2}\mathbf{\left(}\mathbit{c}\mathbit{o}\mathbit{s}\mathbf{120}\mathbf{°}\mathbf{+}\mathbit{i}\mathbit{s}\mathbit{i}\mathbit{n}\mathbf{120}\mathbf{°}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}k=2\phantom{\rule{0ex}{0ex}}{Z}^{\frac{1}{4}}={16}^{\frac{1}{4}}\left(\mathrm{cos}\left(\frac{120°+2\left(2\right)\left(180°\right)}{4}\right)+i\mathrm{sin}\left(\frac{120°+2\left(2\right)\left(180°\right)}{4}\right)\phantom{\rule{0ex}{0ex}}{\mathrm{Z}}^{\frac{1}{4}}=2\left(\mathrm{cos}\left(\frac{840°}{4}\right)+\mathrm{isin}\left(\frac{840°}{4}\right)\right)\phantom{\rule{0ex}{0ex}}{\mathbf{Z}}^{\frac{\mathbf{1}}{\mathbf{4}}}\mathbf{=}\mathbf{2}\mathbf{\left(}\mathbf{cos}\mathbf{210}\mathbf{°}\mathbf{+}\mathbf{isin}\mathbf{210}\mathbf{°}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{k}=3\phantom{\rule{0ex}{0ex}}{\mathrm{Z}}^{\frac{1}{4}}={16}^{\frac{1}{4}}\left(\mathrm{cos}\left(\frac{120°+2\left(3\right)\left(180°\right)}{4}\right)+\mathrm{isin}\left(\frac{120°+2\left(3\right)\left(180°\right)}{4}\right)\phantom{\rule{0ex}{0ex}}{\mathrm{Z}}^{\frac{1}{4}}=2\left(\mathrm{cos}\left(\frac{1200°}{4}\right)+\mathrm{isin}\left(\frac{1200°}{4}\right)\right)\phantom{\rule{0ex}{0ex}}{\mathbf{Z}}^{\frac{\mathbf{1}}{\mathbf{4}}}\mathbf{=}\mathbf{2}\mathbf{\left(}\mathbf{cos}\mathbf{300}\mathbf{°}\mathbf{+}\mathbf{isin}\mathbf{300}\mathbf{°}\mathbf{\right)}$

The roots in bold are the four roots of the complex number $-8+8\sqrt{3}i$ in their polar forms.

## Roots of Complex Numbers - Key takeaways

• The calculation of roots of complex numbers is the process of finding the roots (square, cube, etc.) of complex numbers in the form:

$a+bi$

• Finding square roots of complex numbers can be achieved with a more direct approach rather than the application of a formula.
• Complex numbers Z can be rewritten in terms of its modulus r and argument θ as,

$Z=r\left(\mathrm{cos}\left(\theta \right)+i\mathrm{sin}\left(\theta \right)\right)$

• In order to derive a formula that can guide you through the successful calculation of the roots of complex numbers, you must understand the multiplication of complex numbers in their polar form.
• The de Moivre's Theorem is given as,${{Z}_{1}}^{n}={{r}_{1}}^{n}\left(\mathrm{cos}\left(n{\theta }_{1}\right)+i\mathrm{sin}\left(n{\theta }_{1}\right)\right)$
• The root of complex numbers is calculated with:${Z}^{\frac{1}{n}}={r}^{\frac{1}{n}}\left(\mathrm{cos}\left(\frac{\theta +2k\mathrm{\pi }}{n}\right)+i\mathrm{sin}\left(\frac{\theta +2k\mathrm{\pi }}{n}\right)\right)$

#### Flashcards in Roots of Complex Numbers 3

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How do you find the roots of a complex number?

You find the root of complex numbers by applying de Moivre's Theorem. Rather than solve for Zn, you solve for Z1/n.

What is the meaning of roots with complex numbers?

Roots with complex numbers mean roots of numbers that are complex numbers.

What is the formula for solving roots of complex numbers?

The formula for solving roots complex number is;

What are the roots of the complex number rule?

The rule of roots of a complex number is to ensure that you express the number in the complex form Z = a+bi before applying the formula.

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