Math Problem Statement

root(-1+i, 3)

Solution

To find the cube roots of a complex number such as z=1+iz = -1 + i, we can follow these steps:

  1. Convert the complex number into polar form:
    A complex number z=a+biz = a + bi can be expressed as z=r(cosθ+isinθ)z = r (\cos \theta + i \sin \theta), where rr is the modulus and θ\theta is the argument (angle).

    • Modulus rr is given by:
      r=a2+b2=(1)2+12=2r = \sqrt{a^2 + b^2} = \sqrt{(-1)^2 + 1^2} = \sqrt{2}

    • Argument θ\theta is the angle formed with the positive real axis:
      θ=tan1(ba)=tan1(11)=tan1(1)=3π4\theta = \tan^{-1} \left( \frac{b}{a} \right) = \tan^{-1} \left( \frac{1}{-1} \right) = \tan^{-1} (-1) = \frac{3\pi}{4} (We use 3π4\frac{3\pi}{4} because 1+i-1 + i is in the second quadrant.)

    Therefore, the polar form of 1+i-1 + i is:
    z=2(cos3π4+isin3π4)z = \sqrt{2} \left( \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \right)

  2. Find the cube roots:
    The formula for the nn-th roots of a complex number in polar form is:
    z1/n=r1/n(cosθ+2kπn+isinθ+2kπn)fork=0,1,,n1z^{1/n} = r^{1/n} \left( \cos \frac{\theta + 2k\pi}{n} + i \sin \frac{\theta + 2k\pi}{n} \right) \quad \text{for} \, k = 0, 1, \dots, n-1 Here, n=3n = 3, r=2r = \sqrt{2}, and θ=3π4\theta = \frac{3\pi}{4}.

    Thus, the cube roots are:
    zk=23(cos3π/4+2kπ3+isin3π/4+2kπ3)fork=0,1,2z_k = \sqrt[3]{\sqrt{2}} \left( \cos \frac{3\pi/4 + 2k\pi}{3} + i \sin \frac{3\pi/4 + 2k\pi}{3} \right) \quad \text{for} \, k = 0, 1, 2

  3. Calculate each root:

    • For k=0k = 0:
      [ 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) ] z0=26(22+i22)z_0 = \sqrt[6]{2} \left( \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right)

    • For k=1k = 1:
      z1=26(cos11π12+isin11π12)z_1 = \sqrt[6]{2} \left( \cos \frac{11\pi}{12} + i \sin \frac{11\pi}{12} \right)

    • For k=2k = 2:
      z2=26(cos19π12+isin19π12)z_2 = \sqrt[6]{2} \left( \cos \frac{19\pi}{12} + i \sin \frac{19\pi}{12} \right)

Thus, the three cube roots of 1+i-1 + i are given by the above expressions for z0z_0, z1z_1, and z2z_2.

Would you like a more detailed explanation of any of these steps?


Related questions:

  1. What is the process to find the nth roots of a complex number in general?
  2. How can we convert a complex number from rectangular form to polar form?
  3. What is the geometric interpretation of complex roots on the complex plane?
  4. How do cube roots of unity relate to finding roots of complex numbers?
  5. 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.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

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