To find the nth roots of a complex number, we use a specific formula derived from De Moivre's theorem. This involves converting the complex number into polar form and then applying a formula that accounts for multiple roots.
Steps to Finding the nth Root
Here's how to find the nth roots of a complex number:
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Convert to Polar Form: If your complex number is in the form a + bi, convert it to polar form r(cos θ + i sin θ).
- r = √(a² + b²) (magnitude or modulus)
- θ = arctan(b/a) (angle or argument) - be mindful of the quadrant when calculating the arctangent.
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Apply the nth Root Formula: The nth roots of z = r(cos θ + i sin θ) are given by:
- √ⁿz = √ⁿr [cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)]
Where:
- n is the root you're looking for (e.g., 2 for a square root, 3 for a cube root, etc.).
- k ranges from 0 to n - 1. Each value of k gives you a different root. The video reference highlights that k values run from 0 to n - 1.
- √ⁿr is the real nth root of the modulus r.
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Calculate the Roots: Plug in the values of n, r, and θ, and then calculate the root for each value of k. For instance, to find square roots (n=2), you will calculate roots for k=0 and k=1.
Example
Let's say we want to find the square roots (n=2) of z = 4(cos(60°) + i sin(60°)).
- Step 1: We are already in polar form: r = 4, θ = 60° = π/3.
- Step 2: Applying the formula:
- √²z = √²4 [cos((π/3 + 2kπ)/2) + i sin((π/3 + 2kπ)/2)]
- √²z = 2 [cos((π/3 + 2kπ)/2) + i sin((π/3 + 2kπ)/2)]
- Step 3: Calculate for k = 0 and k = 1:
- For k = 0:
- z₀ = 2 [cos(π/6) + i sin(π/6)]
- z₀ = 2 (√3/2 + i(1/2)) = √3 + i
- For k = 1:
- z₁ = 2 [cos(7π/6) + i sin(7π/6)]
- z₁ = 2 (-√3/2 + i(-1/2)) = -√3 - i
- For k = 0:
Therefore, the square roots of z are √3 + i and -√3 - i.
Key Points
- The nth root of a complex number has n distinct roots.
- The roots are evenly spaced around a circle in the complex plane.
- The formula uses the fact that adding multiples of 2π to an angle doesn't change the value of the sine or cosine.
