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What are the fourth roots of 2√3 −2i? Enter your answer by filling in the boxes. Enter the roots in order of increasing angle measure.

Sagot :

By using the De Moivre's formula, the quartic roots of the complex numbers (in polar form) are z₁ = (√2, π/6), z₂ = (√2, 2π/3), z₃ = (√2, 7π/6), z₄ = (√2, 5π/3).

How to find the roots of a complex number

Complex numbers are numbers of the form a + i b, where a and b are the real and imaginary component, respectively. In other words, complex numbers are an expansion from real numbers. The n-th root of a complex number is found by using the De Moivre's formula:

[tex]\sqrt[n]{z} = \sqrt[n]{r}\cdot \left[\cos \left(\frac{\theta + 2\pi\cdot k}{n} \right) + i\,\sin \left(\frac{\theta + 2\pi\cdot k}{n}\right)\right][/tex], for i = {0, 1, 2, ...,  n - 1}.

Where:

  • r - Norm of the complex number.
  • θ - Direction of the complex number, in radians.

The norm of the complex number is found by Pythagorean theorem:

[tex]r = \sqrt{(2\sqrt{3})^{2}+(-2)^{2}}[/tex]

r = 4

And the direction is determined below:

[tex]\theta = \tan^{-1} \left(\frac{2\sqrt{3}}{-2} \right)[/tex]

θ = 2π/3 rad

Then, the quartic roots of the complex numbers (in polar form) are z₁ = (√2, π/6), z₂ = (√2, 2π/3), z₃ = (√2, 7π/6), z₄ = (√2, 5π/3).

To learn more on complex numbers: https://brainly.com/question/10251853

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