# Nth Root Theorem

## Nth Root Theorem - How it Works - Video

### Formula

Formula:

Here is the formula in radians and degrees, where n is the number roots that you want to find. And k is the number of root that you want to find starting with 0. So k = 0 is the first root, k = 1 is the second root, and ...

### Explanation

Explanation:

Our equation is z3 = 1. If we move all the variables and constants to one side we get z3 - 1 = 0. We can rewrite that as (z - 1)(z2 + 2z +1) = 0, because it is a difference of cubes.

So so we have our first answer z = 1. For the other the two we have to use the quadratic equation. Our results will be two complex numbers, z = [-1 + sqrt(3)i ] / 2 and z = [-1 - sqrt(3)i ] / 2.

Let's graph and label.

Let's use z1 = 1.

Let's use z2 = [-1 + sqrt(3)i ] / 2.

Let's use z3 = [-1 + sqrt(3)i ] / 2.

Now that we can graph and labeled. Let's find the modulus of z, r, and the argument, θ. Let's use z2 = [-1 + sqrt(3)i ] / 2 because we want to find an angle that is not 0 nor negative and less than π.

(-1/2)2 + [sqrt(3)/2]2 = c2

sqrt({(-1/2)2 + [sqrt(3)/2]}2 ) = c

sqrt(1/4 + 3/4) = c

sqrt(1) = c

c = 1

Pythagorean Theorem

Substituted for a and b

Squared both numbers

Found square root

tanθ = b/a

tanθ = [sqrt(3)/2] / (-1/2)

tanθ = -sqrt(3) / 2

θ = tan-1(-sqrt(3) / 3)

θ = 2π/3

Substituted for a and b

Divided

Used the inverse tangent on both sides

Calculated

Since our exponent is three we have 3 solutions. The angle between each solution is the same angle, θ = 2π/3.

### Explanation

Since our exponent is three we have 3 solutions. The angle between each solution is the same angle, θ = 2π/3.

In the picture on the left we have our one angle, θ = 9π/4. If we multiply by the same complex number, we add our argument, 2π, and our result is 17π/4. If we divide by the same complex number, we subtract our argument, 2π, and our result is π/4.

### Example 1

Example 1:

Here we take a look on why the θ is added 2π to find the next root.

We have cos(3θ) = cos(0) and sin(3θ) = sin(0). Cos(0) = 1 and sin(0) = 0.

If we look at the cosine graph, the first time when cosine is 1 is (0, 1). The next time is (2π, 1), and the next one is (4π, 1).

If we look at the sine graph, the first time when sine is 0 is (0, 0). The next time is (π, 0), and the next one is (2π, 0).

We have to be true for both so that is why we add 2πk to our θ.

To find the 1st root, we start when is k = 0.

z1 = r1/n * {cos[(θ + 2 * k * π)/n] + i * sin[(θ + 2 * k * π)/n]}

z1 = 11/3 * {cos[(0 + 2 * 0 * π)/3] + i * sin[(0 + 2 * 0 * π)/3]}

z1 = 1 * {cos[(0 + 0)/3] + i * sin[(0 + 0)/3]}

z1 = 1 * {cos(0) + i * sin(0)}

z1 = 1 * {1 + i *0}

z1 = 1

Nth Root Theorem

Substituted argument and k

Multiplied

Calculated

Distributed

To find the 2nd root, we start when is k = 1.

z2 = r1/n * {cos[(θ + 2 * k * π)/n] + i * sin[(θ + 2 * k * π)/n]}

z2 = 11/3 * {cos[(0 + 2 * 1 * π)/3] + i * sin[(0 + 2 * 1 * π)/3]}

z2 = 1 * {cos[(0 + 2π))/3] + i * sin[(0 +2π)/3]}

z2 = 1 * {cos(2π/3) + i * sin(2π/3)}

z2 = 1 * {(-1/2) + i * [sqrt(3)/2]}

z2 = (-1/2) + [sqrt(3)/2]i

Nth Root Theorem

Substituted argument and k

Multiplied

Calculated

Multiplied

To find the 3rd root, we start when is k = 2.

z3 = r1/n * {cos[(θ + 2 * k * π)/n] + i * sin[(θ + 2 * k * π)/n]}

z3 = 11/3 * {cos[(0 + 2 * 2 * π)/3] + i * sin[(0 + 2 * 2 * π)/3]}

z3 = 1 * {cos[(0 + 4π))/3] + i * sin[(0 +4π)/3]}

z3 = 1 * {cos(4π/3) + i * sin(4π/3)}

z3 = 1 * {(-1/2) + i * [-sqrt(3)/2]}

z3 = (-1/2) - [sqrt(3)/2]i

Nth Root Theorem

Substituted argument and k

Multiplied