https://r2.erweima.ai/mathbot/image/1727030947169zauhsaxp.jpg

The expression in the image shows \((-1 + i)^{1/3}\), which represents the cube root of the complex number \(-1 + i\). To solve this, we need to express the complex number in polar form and then apply De Moivre's Theorem to find the cube roots.

### Step-by-step solution:

1. **Convert the complex number to polar form**:
   - The complex number is \(-1 + i\), where the real part \(a = -1\) and the imaginary part \(b = 1\).
   - The modulus \(r\) of the complex number is:
     \[
     r = \sqrt{a^2 + b^2} = \sqrt{(-1)^2 + 1^2} = \sqrt{2}
     \]
   - The argument \(\theta\) (angle) is:
     \[
     \theta = \tan^{-1} \left( \frac{b}{a} \right) = \tan^{-1} \left( \frac{1}{-1} \right) = \tan^{-1}(-1)
     \]
     Since \(-1 + i\) is in the second quadrant, \(\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}\).

2. **Polar form**:
   In polar form, the complex number is expressed as:
   \[
   -1 + i = \sqrt{2} \left( \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \right)
   \]

3. **Cube roots using De Moivre's Theorem**:
   To find the cube roots of a complex number, we use the formula:
   \[
   z^{1/n} = r^{1/n} \left( \cos \frac{\theta + 2k\pi}{n} + i \sin \frac{\theta + 2k\pi}{n} \right), \quad k = 0, 1, 2, \dots, (n-1)
   \]
   Here \(r = \sqrt{2}\), \(\theta = \frac{3\pi}{4}\), and \(n = 3\).

   The three cube roots will be:

   - For \(k = 0\):
     \[
     z_0 = \sqrt[3]{\sqrt{2}} \left( \cos \frac{3\pi}{12} + i \sin \frac{3\pi}{12} \right)
     = \sqrt[6]{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right)
     = \sqrt[6]{2} \left( \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right)
     \]
     \[
     z_0 = \frac{\sqrt{2}}{2} \left(1 + i \right)
     \]

   - For \(k = 1\):
     \[
     z_1 = \sqrt[3]{\sqrt{2}} \left( \cos \frac{3\pi + 2\pi}{12} + i \sin \frac{3\pi + 2\pi}{12} \right)
     = \sqrt[6]{2} \left( \cos \frac{5\pi}{12} + i \sin \frac{5\pi}{12} \right)
     \]

   - For \(k = 2\):
     \[
     z_2 = \sqrt[3]{\sqrt{2}} \left( \cos \frac{3\pi + 4\pi}{12} + i \sin \frac{3\pi + 4\pi}{12} \right)
     = \sqrt[6]{2} \left( \cos \frac{7\pi}{12} + i \sin \frac{7\pi}{12} \right)
     \]

These are the three cube roots of the complex number \((-1 + i)^{1/3}\).

Would you like further details on any specific step? Here are some related questions for further exploration:

1. How do you convert complex numbers to polar form in general?
2. Can you explain De Moivre's Theorem in more detail?
3. What are cube roots of unity, and how are they related to this problem?
4. How do arguments of complex numbers change when taking roots?
5. Can you show how to visualize these cube roots on the complex plane?

**Tip:** When working with roots of complex numbers, always remember to express them in polar form first. This makes the calculation of roots much more straightforward!

Find the cube roots of the complex number (-1 + i).

Learn how to find the cube roots of the complex number (-1 + i) using De Moivre's Theorem. This problem involves converting complex numbers to polar form and calculating the cube roots step-by-step. Suitable for students in Grades 11-12 and college level.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation