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Nth Root Theorem - How it Works - Video
Nth Root Theorem - How it Works - Video
Why?
Why?
In the picture on the left we have our one angle, θ = π/4, which is the principal or main argument. If add 2π to the argument then our result is 9π/4, and if we do it again, we get 17π/4. This leads us to repeating and how they are same distance apart just like the roots we will find in the example below.
Formula
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 - Find Roots Using Quadratic Formula
Explanation - Find Roots Using Quadratic Formula
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
Added
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.
Nth Root Formula
Nth Root Formula
Since our exponent is three we have 3 solutions. The angle between each solution is the same angle, θ = 2π/3.
Explanation - Find General Argument
Explanation - Find General Argument
Explanation - Find General Argument:
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.
Example 1
Example 1
Example 1:
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
Added
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
Added
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
Added
Calculated
Multiplied
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Live Worksheet
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