Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Title: Applying DeMoivre's Theorem to Calculate Powers of Complex Numbers**
**Instructions:**
Use DeMoivre's Theorem to find the indicated power of the complex number. Write the result in standard (a + bi) form.
**Problem:**
Calculate \((1 + i)^3\).
**Explanation:**
DeMoivre's Theorem states that for any complex number \(z = r(\cos \theta + i \sin \theta)\), its \(n\)th power is given by:
\[z^n = r^n (\cos(n\theta) + i \sin(n\theta))\]
**Steps:**
1. Convert the complex number \(1 + i\) to polar form.
2. Use DeMoivre's Theorem to raise the complex number to the 3rd power.
3. Convert the result back to standard form (a + bi).
**Solution:**
1. **Convert to Polar Form:**
\[1 + i\]
The modulus \(r\) is:
\[r = \sqrt{1^2 + 1^2} = \sqrt{2}\]
The argument \(\theta\) is:
\[\theta = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4}\]
Therefore, in polar form:
\[1 + i = \sqrt{2} \left(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}\right)\]
2. **Apply DeMoivre's Theorem:**
\[(1 + i)^3 = \left(\sqrt{2}\right)^3 \left(\cos(3 \cdot \frac{\pi}{4}) + i \sin(3 \cdot \frac{\pi}{4})\right)\]
\[= 2\sqrt{2} \left(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}\right)\]
3. **Convert Back to Standard Form:**
\[\cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2}, \quad \sin \frac{3\pi}{4} = \frac{\sqrt{2}}{2} \]
\[(1 + i)^3 =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7fab864d-7419-4b8c-bc50-25971d273c3c%2F28dddc78-77ff-4dc9-85b1-2ee77ab466e0%2Frik0avp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Applying DeMoivre's Theorem to Calculate Powers of Complex Numbers**
**Instructions:**
Use DeMoivre's Theorem to find the indicated power of the complex number. Write the result in standard (a + bi) form.
**Problem:**
Calculate \((1 + i)^3\).
**Explanation:**
DeMoivre's Theorem states that for any complex number \(z = r(\cos \theta + i \sin \theta)\), its \(n\)th power is given by:
\[z^n = r^n (\cos(n\theta) + i \sin(n\theta))\]
**Steps:**
1. Convert the complex number \(1 + i\) to polar form.
2. Use DeMoivre's Theorem to raise the complex number to the 3rd power.
3. Convert the result back to standard form (a + bi).
**Solution:**
1. **Convert to Polar Form:**
\[1 + i\]
The modulus \(r\) is:
\[r = \sqrt{1^2 + 1^2} = \sqrt{2}\]
The argument \(\theta\) is:
\[\theta = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4}\]
Therefore, in polar form:
\[1 + i = \sqrt{2} \left(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}\right)\]
2. **Apply DeMoivre's Theorem:**
\[(1 + i)^3 = \left(\sqrt{2}\right)^3 \left(\cos(3 \cdot \frac{\pi}{4}) + i \sin(3 \cdot \frac{\pi}{4})\right)\]
\[= 2\sqrt{2} \left(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}\right)\]
3. **Convert Back to Standard Form:**
\[\cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2}, \quad \sin \frac{3\pi}{4} = \frac{\sqrt{2}}{2} \]
\[(1 + i)^3 =
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