Math Problem Statement
root(-1+i, 3)
Solution
To find the cube roots of a complex number such as , we can follow these steps:
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Convert the complex number into polar form:
A complex number can be expressed as , where is the modulus and is the argument (angle).-
Modulus is given by:
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Argument is the angle formed with the positive real axis:
(We use because is in the second quadrant.)
Therefore, the polar form of is:
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Find the cube roots:
The formula for the -th roots of a complex number in polar form is:
Here, , , and .Thus, the cube roots are:
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Calculate each root:
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For :
[ z_0 = \sqrt[6]{2} \left( \cos \frac{3\pi}{12} + i \sin \frac{3\pi}{12} \right) = \sqrt[6]{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) ] -
For :
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For :
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Thus, the three cube roots of are given by the above expressions for , , and .
Would you like a more detailed explanation of any of these steps?
Related questions:
- What is the process to find the nth roots of a complex number in general?
- How can we convert a complex number from rectangular form to polar form?
- What is the geometric interpretation of complex roots on the complex plane?
- How do cube roots of unity relate to finding roots of complex numbers?
- What are some real-world applications of finding roots of complex numbers?
Tip: Always express complex numbers in polar form when dealing with powers and roots—it simplifies the process.
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Math Problem Analysis
Mathematical Concepts
Complex Numbers
Polar Coordinates
Roots of Complex Numbers
Formulas
Modulus of a complex number: r = sqrt(a^2 + b^2)
Argument of a complex number: theta = tan^(-1)(b/a)
nth roots of a complex number: z^(1/n) = r^(1/n) * (cos((theta + 2kπ)/n) + i * sin((theta + 2kπ)/n))
Theorems
De Moivre's Theorem
Suitable Grade Level
Grades 11-12