The complex cube roots of 63 +6 i.
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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Question
![**Problem Statement:**
Find all the complex roots. Leave your answer in polar form with the argument in degrees.
The complex cube roots of \( 6\sqrt{3} + 6i \).
**Solution:**
To find the complex cube roots of the given complex number, follow these steps:
1. **Convert to Polar Form:**
- Calculate the magnitude \( r \) of the complex number:
\[
r = \sqrt{(6\sqrt{3})^2 + (6)^2} = \sqrt{108 + 36} = \sqrt{144} = 12
\]
- Calculate the argument \( \theta \):
\[
\theta = \tan^{-1}\left(\frac{6}{6\sqrt{3}}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = 30^\circ
\]
The polar form is \( 12 \text{cis} 30^\circ \), where "cis" is shorthand for \( \cos \theta + i \sin \theta \).
2. **Find the Cube Roots:**
The formula for finding the \( n \)-th roots of a complex number in polar form is given by:
\[
z_k = r^{1/n} \text{cis} \left(\frac{\theta + 360^\circ k}{n}\right)
\]
- For cube roots (\( n = 3 \)), compute:
\[
r^{1/3} = 12^{1/3} = 2\sqrt{3}
\]
- Calculate each root:
\[
z_0 = 2\sqrt{3} \text{cis} \left(\frac{30^\circ}{3}\right) = 2\sqrt{3} \text{cis} 10^\circ
\]
\[
z_1 = 2\sqrt{3} \text{cis} \left(\frac{30^\circ + 360^\circ}{3}\right) = 2\sqrt{3} \text{cis} 130^\circ
\]
\[
z_2 = 2\sqrt{3} \text{cis} \left(\frac{30^\circ + 720^\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb48d2c1d-e6bb-45e6-b6d7-f9e59d1357b7%2F055bf3d6-191d-44a7-9803-1e8c690708ba%2Favqk6sf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find all the complex roots. Leave your answer in polar form with the argument in degrees.
The complex cube roots of \( 6\sqrt{3} + 6i \).
**Solution:**
To find the complex cube roots of the given complex number, follow these steps:
1. **Convert to Polar Form:**
- Calculate the magnitude \( r \) of the complex number:
\[
r = \sqrt{(6\sqrt{3})^2 + (6)^2} = \sqrt{108 + 36} = \sqrt{144} = 12
\]
- Calculate the argument \( \theta \):
\[
\theta = \tan^{-1}\left(\frac{6}{6\sqrt{3}}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = 30^\circ
\]
The polar form is \( 12 \text{cis} 30^\circ \), where "cis" is shorthand for \( \cos \theta + i \sin \theta \).
2. **Find the Cube Roots:**
The formula for finding the \( n \)-th roots of a complex number in polar form is given by:
\[
z_k = r^{1/n} \text{cis} \left(\frac{\theta + 360^\circ k}{n}\right)
\]
- For cube roots (\( n = 3 \)), compute:
\[
r^{1/3} = 12^{1/3} = 2\sqrt{3}
\]
- Calculate each root:
\[
z_0 = 2\sqrt{3} \text{cis} \left(\frac{30^\circ}{3}\right) = 2\sqrt{3} \text{cis} 10^\circ
\]
\[
z_1 = 2\sqrt{3} \text{cis} \left(\frac{30^\circ + 360^\circ}{3}\right) = 2\sqrt{3} \text{cis} 130^\circ
\]
\[
z_2 = 2\sqrt{3} \text{cis} \left(\frac{30^\circ + 720^\
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