Math Problem Statement

Find all the cube roots of -1 + v5i.

Solution

To find all the cube roots of 1+5i-1 + \sqrt{5}i, let's proceed by expressing this complex number in polar form and then using De Moivre's theorem.

  1. Convert 1+5i-1 + \sqrt{5}i to polar form:

    For a complex number z=x+yiz = x + yi, we can express it in polar form as: z=r(cosθ+isinθ)z = r \left( \cos \theta + i \sin \theta \right) where: r=x2+y2r = \sqrt{x^2 + y^2} and θ=arctan(yx).\theta = \arctan \left( \frac{y}{x} \right).

    For z=1+5iz = -1 + \sqrt{5}i:

    • x=1x = -1
    • y=5y = \sqrt{5}

    So, r=(1)2+(5)2=1+5=6.r = \sqrt{(-1)^2 + (\sqrt{5})^2} = \sqrt{1 + 5} = \sqrt{6}.

    Now, calculate θ\theta: θ=arctan(51).\theta = \arctan \left( \frac{\sqrt{5}}{-1} \right). Since 1+5i-1 + \sqrt{5}i lies in the second quadrant, we adjust the angle accordingly: θ=π+arctan(51).\theta = \pi + \arctan \left( -\frac{\sqrt{5}}{1} \right). Let’s denote this as θ0\theta_0 for simplicity.

  2. Apply the formula for cube roots:

    The cube roots of z=r(cosθ+isinθ)z = r \left( \cos \theta + i \sin \theta \right) are given by: r3(cosθ+2kπ3+isinθ+2kπ3)\sqrt[3]{r} \left( \cos \frac{\theta + 2k\pi}{3} + i \sin \frac{\theta + 2k\pi}{3} \right) for k=0,1,2k = 0, 1, 2.

    • Here, r3=63=66\sqrt[3]{r} = \sqrt[3]{\sqrt{6}} = \sqrt[6]{6}.
    • For k=0,1,2k = 0, 1, 2, we get different values of the angle.
  3. Calculate the cube roots:

    For each kk: zk=66(cosθ0+2kπ3+isinθ0+2kπ3).z_k = \sqrt[6]{6} \left( \cos \frac{\theta_0 + 2k\pi}{3} + i \sin \frac{\theta_0 + 2k\pi}{3} \right).

    Now we can substitute the values for each kk to obtain the three distinct cube roots of 1+5i-1 + \sqrt{5}i.

Would you like a more detailed calculation for each cube root value?


Here are some questions to explore further:

  1. How does converting a complex number to polar form help in finding its roots?
  2. What are the steps to calculate the magnitude and angle for a complex number?
  3. How does De Moivre's theorem simplify the process of finding roots of complex numbers?
  4. Can the cube roots of a complex number have both real and imaginary components?
  5. How does the choice of kk 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