By De Moivre's formula, the cubic roots of the complex number are 3 + i 4, - 4.96 + i 0.60 and 1.96 - i 4.60.
How to find the cube root of a complex number
Herein we have a complex number in rectangular form, from which we need its magnitude (r) and direction (θ) and the De Moivre's formula as well. The root formula is introduced below:
[tex]z^{\frac{1}{n} } = r^{\frac{1}{n} }\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 k ∈ {0, 1, ..., n - 1} (1)
Where n is the grade of the complex root.
The magnitude and direction of the complex number are 125 and 0.886π radians, respectively. Thus, by the De Moivre's formula we obtain the following three solutions:
k = 0
z₁ = 2.997 + i 4.002
k = 1
z₂ = - 4.964 + i 0.595
k = 2
z₃ = 1.967 - i 4.597
By De Moivre's formula, the cubic roots of the complex number are 3 + i 4, - 4.96 + i 0.60 and 1.96 - i 4.60.
To learn more on complex numbers: https://brainly.com/question/10251853
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