8. Find (-1+√3i) by converting to trigonometric form and using the formula for raising powers (DeMoivre's Theroem). Leave your answer in the form a +bi. No decimals. No calculators for this problem.
8. Find (-1+√3i) by converting to trigonometric form and using the formula for raising powers (DeMoivre's Theroem). Leave your answer in the form a +bi. No decimals. No calculators for this problem.
Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE:
1. Give the measures of the complement and the supplement of an angle measuring 35°.
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Question
![### Complex Numbers and DeMoivre's Theorem
#### Problem Statement:
8. Find \( \left( -1 + \sqrt{3}i \right)^{10} \) by converting to trigonometric form and using the formula for raising powers (DeMoivre’s Theorem). Leave your answer in the form \( a + bi \). No decimals. No calculators for this problem.
#### Step-by-Step Solution:
1. **Convert to Trigonometric (Polar) Form**:
- Identify the magnitude (r) and argument (θ) of the complex number \( -1 + \sqrt{3}i \).
\[
r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2
\]
\[
\theta = \tan^{-1}\left(\frac{\sqrt{3}}{-1}\right)
\]
Since the complex number \( -1 + \sqrt{3}i \) lies in the second quadrant, where the tangent is positive but the x-component is negative, the angle is:
\[
\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}
\]
Thus, the trigonometric form is:
\[
-1 + \sqrt{3}i = 2 \left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right)
\]
2. **Apply DeMoivre’s Theorem**:
- DeMoivre's Theorem states that for a complex number in polar form \( r (\cos \theta + i \sin \theta) \),
\[
\left[r (\cos \theta + i \sin \theta)\right]^n = r^n \left(\cos (n\theta) + i \sin (n\theta)\right)
\]
Here, \( r = 2 \), \( \theta = \frac{2\pi}{3} \), and \( n = 10 \):
\[
[2 (\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}) ]^{10} = 2^{10} (\cos (\frac](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd88595c7-c499-4476-ae94-4573188e2c46%2Fd8c2870a-325c-4c56-a8a1-18e54eeb1722%2Fpc5t58x_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Complex Numbers and DeMoivre's Theorem
#### Problem Statement:
8. Find \( \left( -1 + \sqrt{3}i \right)^{10} \) by converting to trigonometric form and using the formula for raising powers (DeMoivre’s Theorem). Leave your answer in the form \( a + bi \). No decimals. No calculators for this problem.
#### Step-by-Step Solution:
1. **Convert to Trigonometric (Polar) Form**:
- Identify the magnitude (r) and argument (θ) of the complex number \( -1 + \sqrt{3}i \).
\[
r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2
\]
\[
\theta = \tan^{-1}\left(\frac{\sqrt{3}}{-1}\right)
\]
Since the complex number \( -1 + \sqrt{3}i \) lies in the second quadrant, where the tangent is positive but the x-component is negative, the angle is:
\[
\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}
\]
Thus, the trigonometric form is:
\[
-1 + \sqrt{3}i = 2 \left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right)
\]
2. **Apply DeMoivre’s Theorem**:
- DeMoivre's Theorem states that for a complex number in polar form \( r (\cos \theta + i \sin \theta) \),
\[
\left[r (\cos \theta + i \sin \theta)\right]^n = r^n \left(\cos (n\theta) + i \sin (n\theta)\right)
\]
Here, \( r = 2 \), \( \theta = \frac{2\pi}{3} \), and \( n = 10 \):
\[
[2 (\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}) ]^{10} = 2^{10} (\cos (\frac
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