Math Problem Statement
Find all the cube roots of -1 + v5i.
Solution
To find all the cube roots of , let's proceed by expressing this complex number in polar form and then using De Moivre's theorem.
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Convert to polar form:
For a complex number , we can express it in polar form as: where: and
For :
So,
Now, calculate : Since lies in the second quadrant, we adjust the angle accordingly: Let’s denote this as for simplicity.
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Apply the formula for cube roots:
The cube roots of are given by: for .
- Here, .
- For , we get different values of the angle.
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Calculate the cube roots:
For each :
Now we can substitute the values for each to obtain the three distinct cube roots of .
Would you like a more detailed calculation for each cube root value?
Here are some questions to explore further:
- How does converting a complex number to polar form help in finding its roots?
- What are the steps to calculate the magnitude and angle for a complex number?
- How does De Moivre's theorem simplify the process of finding roots of complex numbers?
- Can the cube roots of a complex number have both real and imaginary components?
- How does the choice of values affect the distinctiveness of roots in complex numbers?
Tip: When finding roots of complex numbers, always ensure the angle is correctly placed in the appropriate quadrant for accurate polar representation.
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Math Problem Analysis
Mathematical Concepts
Complex Numbers
Polar Form
Cube Roots
De Moivre's Theorem
Formulas
r = √(x² + y²)
θ = arctan(y/x)
Cube Root Formula: z_k = √[3]{r} (cos((θ + 2kπ)/3) + i sin((θ + 2kπ)/3))
Theorems
De Moivre's Theorem
Suitable Grade Level
Grades 11-12