Compute the solutions of the following polynomial equation in C 2³ +1 = i for z E C. Sketch the solutions in the complex plane.
Compute the solutions of the following polynomial equation in C 2³ +1 = i for z E C. Sketch the solutions in the complex plane.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Solving a Polynomial Equation in the Complex Plane
**Problem Statement:**
**Compute the solutions of the following polynomial equation in \(\mathbb{C}\)**:
\[ z^3 + 1 = i \quad \text{for} \quad z \in \mathbb{C}. \]
**Task:**
Sketch the solutions in the complex plane.
**Step-by-Step Solution:**
To solve the polynomial equation \( z^3 + 1 = i \), follow these steps:
1. **Rewrite the Equation**:
\[ z^3 = i - 1 \]
2. **Convert \( i - 1 \) to Polar Form**:
- First, express \( -1 + i \) in polar form. The magnitude \( r \) and angle \( \theta \) are given by:
\[
r = \sqrt{(-1)^2 + 1^2} = \sqrt{2}
\]
\[
\tan(\theta) = \frac{1}{-1} = -1 \quad \Rightarrow \theta = \frac{3\pi}{4} \text{ or } -\frac{5\pi}{4}
\]
- Typically, the principal argument between \(-\pi\) and \(\pi\) is preferred, so we take \( -\frac{3\pi}{4} \) (since \(\theta = \frac{3\pi}{4}\) was adjusted by \(\pi\) to \(-\frac{3\pi}{4}\)).
3. **Expressing \( z^3 \) in Polar Form**:
\[ z^3 = \sqrt{2} \text{cis} \left( -\frac{3\pi}{4} \right) \]
4. **Taking the Cube Root**:
- The solutions for \( z \) will be obtained by taking the cube root in polar form.
- To find the roots:
\[
r = \sqrt[3]{\sqrt{2}} = 2^{1/6}
\]
\[
\theta = \frac{-3\pi/4 + 2k\pi}{3}, \quad k = 0, 1, 2
\]
5. **Find \( z \) for](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdf4abcde-0790-4b6d-b8ff-52c0ea5b10aa%2F1409a252-7aaa-4e38-8f1e-a6269f3484a7%2Flba4p6_processed.png&w=3840&q=75)
Transcribed Image Text:### Solving a Polynomial Equation in the Complex Plane
**Problem Statement:**
**Compute the solutions of the following polynomial equation in \(\mathbb{C}\)**:
\[ z^3 + 1 = i \quad \text{for} \quad z \in \mathbb{C}. \]
**Task:**
Sketch the solutions in the complex plane.
**Step-by-Step Solution:**
To solve the polynomial equation \( z^3 + 1 = i \), follow these steps:
1. **Rewrite the Equation**:
\[ z^3 = i - 1 \]
2. **Convert \( i - 1 \) to Polar Form**:
- First, express \( -1 + i \) in polar form. The magnitude \( r \) and angle \( \theta \) are given by:
\[
r = \sqrt{(-1)^2 + 1^2} = \sqrt{2}
\]
\[
\tan(\theta) = \frac{1}{-1} = -1 \quad \Rightarrow \theta = \frac{3\pi}{4} \text{ or } -\frac{5\pi}{4}
\]
- Typically, the principal argument between \(-\pi\) and \(\pi\) is preferred, so we take \( -\frac{3\pi}{4} \) (since \(\theta = \frac{3\pi}{4}\) was adjusted by \(\pi\) to \(-\frac{3\pi}{4}\)).
3. **Expressing \( z^3 \) in Polar Form**:
\[ z^3 = \sqrt{2} \text{cis} \left( -\frac{3\pi}{4} \right) \]
4. **Taking the Cube Root**:
- The solutions for \( z \) will be obtained by taking the cube root in polar form.
- To find the roots:
\[
r = \sqrt[3]{\sqrt{2}} = 2^{1/6}
\]
\[
\theta = \frac{-3\pi/4 + 2k\pi}{3}, \quad k = 0, 1, 2
\]
5. **Find \( z \) for
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