Question
Given z subscript 1 baseline equals 12 left-parenthesis cosine Start Fraction pi over 3 End Fraction plus i sine Start Fraction pi over 3 End Fraction right-parenthesis and z subscript 2 baseline equals 3 left-parenthesis cosine Start Fraction 5 pi over 6 End Fraction plus i sine Start Fraction 5 pi over 6 End Fraction right-parenthesis, find Start Fraction z subscript 1 baseline over z subscript 2 baseline End Fraction where 0 less-than-or-equal-to theta less-than 2 pi.
(1 point)
Responses
Start Fraction z subscript 1 baseline over z subscript 2 baseline End Fraction equals 4 left-parenthesis cosine Start Fraction pi over 2 End Fraction plus i sine Start Fraction pi over 2 End Fraction right-parenthesis
Image with alt text: Start Fraction z subscript 1 baseline over z subscript 2 baseline End Fraction equals 4 left-parenthesis cosine Start Fraction pi over 2 End Fraction plus i sine Start Fraction pi over 2 End Fraction right-parenthesis
Start Fraction z subscript 1 baseline over z subscript 2 baseline End Fraction equals 9 left-parenthesis cosine Start Fraction 3 pi over 2 End Fraction plus i sine Start Fraction 3 pi over 2 End Fraction right-parenthesis
Image with alt text: Start Fraction z subscript 1 baseline over z subscript 2 baseline End Fraction equals 9 left-parenthesis cosine Start Fraction 3 pi over 2 End Fraction plus i sine Start Fraction 3 pi over 2 End Fraction right-parenthesis
Start Fraction z subscript 1 baseline over z subscript 2 baseline End Fraction equals 36 left-parenthesis cosine Start Fraction 7 pi over 6 End Fraction plus i sine Start Fraction 7 pi over 6 End Fraction right-parenthesis
Image with alt text: Start Fraction z subscript 1 baseline over z subscript 2 baseline End Fraction equals 36 left-parenthesis cosine Start Fraction 7 pi over 6 End Fraction plus i sine Start Fraction 7 pi over 6 End Fraction right-parenthesis
Start Fraction z subscript 1 baseline over z subscript 2 baseline End Fraction equals 4 left-parenthesis cosine Start Fraction 3 pi over 2 End Fraction plus i sine Start Fraction 3 pi over 2 End Fraction right-parenthesis
(1 point)
Responses
Start Fraction z subscript 1 baseline over z subscript 2 baseline End Fraction equals 4 left-parenthesis cosine Start Fraction pi over 2 End Fraction plus i sine Start Fraction pi over 2 End Fraction right-parenthesis
Image with alt text: Start Fraction z subscript 1 baseline over z subscript 2 baseline End Fraction equals 4 left-parenthesis cosine Start Fraction pi over 2 End Fraction plus i sine Start Fraction pi over 2 End Fraction right-parenthesis
Start Fraction z subscript 1 baseline over z subscript 2 baseline End Fraction equals 9 left-parenthesis cosine Start Fraction 3 pi over 2 End Fraction plus i sine Start Fraction 3 pi over 2 End Fraction right-parenthesis
Image with alt text: Start Fraction z subscript 1 baseline over z subscript 2 baseline End Fraction equals 9 left-parenthesis cosine Start Fraction 3 pi over 2 End Fraction plus i sine Start Fraction 3 pi over 2 End Fraction right-parenthesis
Start Fraction z subscript 1 baseline over z subscript 2 baseline End Fraction equals 36 left-parenthesis cosine Start Fraction 7 pi over 6 End Fraction plus i sine Start Fraction 7 pi over 6 End Fraction right-parenthesis
Image with alt text: Start Fraction z subscript 1 baseline over z subscript 2 baseline End Fraction equals 36 left-parenthesis cosine Start Fraction 7 pi over 6 End Fraction plus i sine Start Fraction 7 pi over 6 End Fraction right-parenthesis
Start Fraction z subscript 1 baseline over z subscript 2 baseline End Fraction equals 4 left-parenthesis cosine Start Fraction 3 pi over 2 End Fraction plus i sine Start Fraction 3 pi over 2 End Fraction right-parenthesis
Answers
GPT-4o mini
To find the quotient \( \frac{z_1}{z_2} \), where
\[
z_1 = 12 \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right)
\]
and
\[
z_2 = 3 \left( \cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6} \right),
\]
we will follow these steps:
1. **Convert to polar form:**
Given that \( z_1 = re^{i\theta} \) and \( z_2 = re^{i\theta} \):
- \( z_1 = 12 e^{i \frac{\pi}{3}} \)
- \( z_2 = 3 e^{i \frac{5\pi}{6}} \)
2. **Divide \( z_1 \) by \( z_2 \):**
\[
\frac{z_1}{z_2} = \frac{12 e^{i \frac{\pi}{3}}}{3 e^{i \frac{5\pi}{6}}} = \frac{12}{3} e^{i \left( \frac{\pi}{3} - \frac{5\pi}{6} \right)} = 4 e^{i \left( \frac{\pi}{3} - \frac{5\pi}{6} \right)}
\]
3. **Simplify the angle:**
Calculate \( \frac{\pi}{3} - \frac{5\pi}{6} \):
\[
\frac{\pi}{3} = \frac{2\pi}{6} \quad \text{and thus:}
\]
\[
\frac{2\pi}{6} - \frac{5\pi}{6} = -\frac{3\pi}{6} = -\frac{\pi}{2}
\]
4. **Convert the negative angle to a positive angle:**
Adding \( 2\pi \) to \( -\frac{\pi}{2} \):
\[
-\frac{\pi}{2} + 2\pi = \frac{3\pi}{2}
\]
5. **Reconstruct in exponential form:**
Thus,
\[
\frac{z_1}{z_2} = 4 e^{i \frac{3\pi}{2}}
\]
6. **Convert back to Cartesian form:**
Using \( \cos \frac{3\pi}{2} = 0 \) and \( \sin \frac{3\pi}{2} = -1 \),
\[
\frac{z_1}{z_2} = 4 \left( \cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2} \right) = 4(0 - i) = -4i
\]
This yields:
\[
\frac{z_1}{z_2} = 4 \left( \cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2} \right)
\]
**Answer:**
The correct response is:
\[
\frac{z_1}{z_2} = 4 \left( \cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2} \right)
\]
\[
z_1 = 12 \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right)
\]
and
\[
z_2 = 3 \left( \cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6} \right),
\]
we will follow these steps:
1. **Convert to polar form:**
Given that \( z_1 = re^{i\theta} \) and \( z_2 = re^{i\theta} \):
- \( z_1 = 12 e^{i \frac{\pi}{3}} \)
- \( z_2 = 3 e^{i \frac{5\pi}{6}} \)
2. **Divide \( z_1 \) by \( z_2 \):**
\[
\frac{z_1}{z_2} = \frac{12 e^{i \frac{\pi}{3}}}{3 e^{i \frac{5\pi}{6}}} = \frac{12}{3} e^{i \left( \frac{\pi}{3} - \frac{5\pi}{6} \right)} = 4 e^{i \left( \frac{\pi}{3} - \frac{5\pi}{6} \right)}
\]
3. **Simplify the angle:**
Calculate \( \frac{\pi}{3} - \frac{5\pi}{6} \):
\[
\frac{\pi}{3} = \frac{2\pi}{6} \quad \text{and thus:}
\]
\[
\frac{2\pi}{6} - \frac{5\pi}{6} = -\frac{3\pi}{6} = -\frac{\pi}{2}
\]
4. **Convert the negative angle to a positive angle:**
Adding \( 2\pi \) to \( -\frac{\pi}{2} \):
\[
-\frac{\pi}{2} + 2\pi = \frac{3\pi}{2}
\]
5. **Reconstruct in exponential form:**
Thus,
\[
\frac{z_1}{z_2} = 4 e^{i \frac{3\pi}{2}}
\]
6. **Convert back to Cartesian form:**
Using \( \cos \frac{3\pi}{2} = 0 \) and \( \sin \frac{3\pi}{2} = -1 \),
\[
\frac{z_1}{z_2} = 4 \left( \cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2} \right) = 4(0 - i) = -4i
\]
This yields:
\[
\frac{z_1}{z_2} = 4 \left( \cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2} \right)
\]
**Answer:**
The correct response is:
\[
\frac{z_1}{z_2} = 4 \left( \cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2} \right)
\]