https://file.aiquickdraw.com/mathbot/file/1726931450391ijgwl8xt.jpg

In this image, the problem involves calculating \( i^{\frac{1}{3}} \) and \( \left(1 - \sqrt{3}i\right)^{\frac{1}{2}} \), using complex number operations and Euler's formula. Let's break it down step by step:

### 1. Calculation of \( i^{\frac{1}{3}} \):
- The complex number \( i \) is expressed in polar form:
  \[
  i = \cos\frac{\pi}{2} + i\sin\frac{\pi}{2}
  \]
- To compute \( i^{\frac{1}{3}} \), we use the general formula for the roots of a complex number in polar form, which involves:
  \[
  i^{\frac{1}{3}} = \cos\left(\frac{\pi}{2} + \frac{2k\pi}{3}\right) + i \sin\left(\frac{\pi}{2} + \frac{2k\pi}{3}\right)
  \]
  where \( k = 0, 1, 2 \) gives three distinct cube roots.

- Applying \( k = 0, 1, 2 \) results in three cube roots:
  \[
  i^{\frac{1}{3}} = \left\{ \cos\frac{\pi}{6} + i\sin\frac{\pi}{6}, \cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}, \cos\frac{9\pi}{6} + i\sin\frac{9\pi}{6} \right\}
  \]
  Simplifying these gives the cube roots:
  \[
  i^{\frac{1}{3}} = \left\{ \frac{\sqrt{3}}{2} + \frac{i}{2}, -\frac{\sqrt{3}}{2} + \frac{i}{2}, -i \right\}
  \]

### 2. Calculation of \( \left(1 - \sqrt{3}i\right)^{\frac{1}{2}} \):
- We first express \( 1 - \sqrt{3}i \) in polar form. The modulus \( r \) and argument \( \theta \) are calculated as follows:
  \[
  r = \left| 1 - \sqrt{3}i \right| = 2, \quad \theta = \arg(1 - \sqrt{3}i) = -\frac{\pi}{3}
  \]
  So, the polar form is:
  \[
  1 - \sqrt{3}i = 2\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right)
  \]
- To find the square root, we use:
  \[
  \left(1 - \sqrt{3}i\right)^{\frac{1}{2}} = \sqrt{2} \left(\cos\left(\frac{-\pi + 2k\pi}{2}\right) + i\sin\left(\frac{-\pi + 2k\pi}{2}\right)\right)
  \]
  where \( k = 0, 1 \) gives two distinct square roots:
  \[
  \left(1 - \sqrt{3}i\right)^{\frac{1}{2}} = \left\{ \sqrt{2} \left( \cos\left(-\frac{\pi}{6}\right) + i \sin\left(-\frac{\pi}{6}\right) \right), \sqrt{2} \left( \cos\left(\frac{5\pi}{6}\right) + i \sin\left(\frac{5\pi}{6}\right) \right) \right\}
  \]

Would you like me to continue this solution with more steps, or do you have any specific questions about these steps so far?

Here are five related questions you might consider:
1. How do you convert a complex number from rectangular to polar form?
2. What are the general formulas for finding roots of complex numbers?
3. How do you calculate the modulus and argument of a complex number?
4. Why are there multiple roots for a complex number, and how are they distributed?
5. Can the same methods be applied to other complex powers, like \( i^{1/4} \)?

**Tip:** When working with roots of complex numbers, remember that each distinct root corresponds to a different angle, which is why we often use \( k = 0, 1, 2, \dots \) to represent multiple values.

Find the cube root of i, and the square root of (1 - √3i).

This problem demonstrates how to calculate the cube root of i and the square root of (1 - √3i) using Euler's formula. It covers key concepts in complex numbers, including polar forms, and presents a detailed breakdown of the solution.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation