Question
Find the two square roots of 12 left-parenthesis cosine Start Fraction pi over 3 End Fraction plus i sine Start Fraction pi over 3 End Fraction right-parenthesis.
(1 point)
Responses
2 Start Root 3 End Root left-parenthesis cosine Start Fraction pi over 3 End Fraction plus i sine Start Fraction pi over 3 End Fraction right-parenthesis, 2 Start Root 3 End Root left-parenthesis cosine Start Fraction 4 pi over 3 End Fraction plus i sine Start Fraction 4 pi over 3 End Fraction right-parenthesis
Image with alt text: 2 Start Root 3 End Root left-parenthesis cosine Start Fraction pi over 3 End Fraction plus i sine Start Fraction pi over 3 End Fraction right-parenthesis , Image with alt text: 2 Start Root 3 End Root left-parenthesis cosine Start Fraction 4 pi over 3 End Fraction plus i sine Start Fraction 4 pi over 3 End Fraction right-parenthesis
144 left-parenthesis cosine Start Fraction 2 pi over 3 End Fraction plus i sine Start Fraction 2 pi over 3 End Fraction right-parenthesis, 144 left-parenthesis cosine Start Fraction 5 pi over 3 End Fraction plus i sine Start Fraction 5 pi over 3 End Fraction right-parenthesis
Image with alt text: 144 left-parenthesis cosine Start Fraction 2 pi over 3 End Fraction plus i sine Start Fraction 2 pi over 3 End Fraction right-parenthesis , Image with alt text: 144 left-parenthesis cosine Start Fraction 5 pi over 3 End Fraction plus i sine Start Fraction 5 pi over 3 End Fraction right-parenthesis
2 Start Root 3 End Root left-parenthesis cosine Start Fraction 2 pi over 3 End Fraction plus i sine Start Fraction 2 pi over 3 End Fraction right-parenthesis, 2 Start Root 3 End Root left-parenthesis cosine Start Fraction 5 pi over 3 End Fraction plus i sine Start Fraction 5 pi over 3 End Fraction right-parenthesis
Image with alt text: 2 Start Root 3 End Root left-parenthesis cosine Start Fraction 2 pi over 3 End Fraction plus i sine Start Fraction 2 pi over 3 End Fraction right-parenthesis , Image with alt text: 2 Start Root 3 End Root left-parenthesis cosine Start Fraction 5 pi over 3 End Fraction plus i sine Start Fraction 5 pi over 3 End Fraction right-parenthesis
2 Start Root 3 End Root left-parenthesis cosine Start Fraction pi over 6 End Fraction plus i sine Start Fraction pi over 6 End Fraction right-parenthesis, 2 Start Root 3 End Root left-parenthesis cosine Start Fraction 7 pi over 6 End Fraction plus i sine Start Fraction 7 pi over 6 End Fraction right-parenthesis
(1 point)
Responses
2 Start Root 3 End Root left-parenthesis cosine Start Fraction pi over 3 End Fraction plus i sine Start Fraction pi over 3 End Fraction right-parenthesis, 2 Start Root 3 End Root left-parenthesis cosine Start Fraction 4 pi over 3 End Fraction plus i sine Start Fraction 4 pi over 3 End Fraction right-parenthesis
Image with alt text: 2 Start Root 3 End Root left-parenthesis cosine Start Fraction pi over 3 End Fraction plus i sine Start Fraction pi over 3 End Fraction right-parenthesis , Image with alt text: 2 Start Root 3 End Root left-parenthesis cosine Start Fraction 4 pi over 3 End Fraction plus i sine Start Fraction 4 pi over 3 End Fraction right-parenthesis
144 left-parenthesis cosine Start Fraction 2 pi over 3 End Fraction plus i sine Start Fraction 2 pi over 3 End Fraction right-parenthesis, 144 left-parenthesis cosine Start Fraction 5 pi over 3 End Fraction plus i sine Start Fraction 5 pi over 3 End Fraction right-parenthesis
Image with alt text: 144 left-parenthesis cosine Start Fraction 2 pi over 3 End Fraction plus i sine Start Fraction 2 pi over 3 End Fraction right-parenthesis , Image with alt text: 144 left-parenthesis cosine Start Fraction 5 pi over 3 End Fraction plus i sine Start Fraction 5 pi over 3 End Fraction right-parenthesis
2 Start Root 3 End Root left-parenthesis cosine Start Fraction 2 pi over 3 End Fraction plus i sine Start Fraction 2 pi over 3 End Fraction right-parenthesis, 2 Start Root 3 End Root left-parenthesis cosine Start Fraction 5 pi over 3 End Fraction plus i sine Start Fraction 5 pi over 3 End Fraction right-parenthesis
Image with alt text: 2 Start Root 3 End Root left-parenthesis cosine Start Fraction 2 pi over 3 End Fraction plus i sine Start Fraction 2 pi over 3 End Fraction right-parenthesis , Image with alt text: 2 Start Root 3 End Root left-parenthesis cosine Start Fraction 5 pi over 3 End Fraction plus i sine Start Fraction 5 pi over 3 End Fraction right-parenthesis
2 Start Root 3 End Root left-parenthesis cosine Start Fraction pi over 6 End Fraction plus i sine Start Fraction pi over 6 End Fraction right-parenthesis, 2 Start Root 3 End Root left-parenthesis cosine Start Fraction 7 pi over 6 End Fraction plus i sine Start Fraction 7 pi over 6 End Fraction right-parenthesis
Answers
Answered by
GPT-4o mini
To find the two square roots of the complex number given, we start by expressing the complex number in exponential form.
Given:
\[ 12 \left( \cos\frac{\pi}{3} + i \sin\frac{\pi}{3} \right) \]
Using Euler's formula, we can rewrite this as:
\[ 12 e^{i \frac{\pi}{3}} \]
Next, we need to find the square roots of this expression. The general formula for the square roots of a complex number in polar form \( r e^{i\theta} \) is given by:
\[ \sqrt{r} e^{i \frac{\theta}{2}} \quad \text{and} \quad \sqrt{r} e^{i \left( \frac{\theta}{2} + \pi \right)} \]
1. **Calculate \( \sqrt{r} \)**:
\[
r = 12 \implies \sqrt{r} = \sqrt{12} = 2\sqrt{3}
\]
2. **Calculate \( \frac{\theta}{2} \)**:
\[
\theta = \frac{\pi}{3} \implies \frac{\theta}{2} = \frac{\pi}{6}
\]
Using these, we find the two roots:
1. The first root:
\[
2\sqrt{3} e^{i \frac{\pi}{6}} = 2\sqrt{3} \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right)
\]
2. The second root:
\[
2\sqrt{3} e^{i \left( \frac{\pi}{6} + \pi \right)} = 2\sqrt{3} e^{i \frac{7\pi}{6}} = 2\sqrt{3} \left( \cos \frac{7\pi}{6} + i \sin \frac{7\pi}{6} \right)
\]
Thus, the two square roots are:
1. \( 2\sqrt{3} \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right) \)
2. \( 2\sqrt{3} \left( \cos \frac{7\pi}{6} + i \sin \frac{7\pi}{6} \right) \)
Now we can match this with the provided options:
- The correct response is:
\[
2\sqrt{3} \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right), \quad 2\sqrt{3} \left( \cos \frac{7\pi}{6} + i \sin \frac{7\pi}{6} \right)
\]
This matches the last option:
**Response:**
\( 2\sqrt{3} \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right), \quad 2\sqrt{3} \left( \cos \frac{7\pi}{6} + i \sin \frac{7\pi}{6} \right) \)
Given:
\[ 12 \left( \cos\frac{\pi}{3} + i \sin\frac{\pi}{3} \right) \]
Using Euler's formula, we can rewrite this as:
\[ 12 e^{i \frac{\pi}{3}} \]
Next, we need to find the square roots of this expression. The general formula for the square roots of a complex number in polar form \( r e^{i\theta} \) is given by:
\[ \sqrt{r} e^{i \frac{\theta}{2}} \quad \text{and} \quad \sqrt{r} e^{i \left( \frac{\theta}{2} + \pi \right)} \]
1. **Calculate \( \sqrt{r} \)**:
\[
r = 12 \implies \sqrt{r} = \sqrt{12} = 2\sqrt{3}
\]
2. **Calculate \( \frac{\theta}{2} \)**:
\[
\theta = \frac{\pi}{3} \implies \frac{\theta}{2} = \frac{\pi}{6}
\]
Using these, we find the two roots:
1. The first root:
\[
2\sqrt{3} e^{i \frac{\pi}{6}} = 2\sqrt{3} \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right)
\]
2. The second root:
\[
2\sqrt{3} e^{i \left( \frac{\pi}{6} + \pi \right)} = 2\sqrt{3} e^{i \frac{7\pi}{6}} = 2\sqrt{3} \left( \cos \frac{7\pi}{6} + i \sin \frac{7\pi}{6} \right)
\]
Thus, the two square roots are:
1. \( 2\sqrt{3} \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right) \)
2. \( 2\sqrt{3} \left( \cos \frac{7\pi}{6} + i \sin \frac{7\pi}{6} \right) \)
Now we can match this with the provided options:
- The correct response is:
\[
2\sqrt{3} \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right), \quad 2\sqrt{3} \left( \cos \frac{7\pi}{6} + i \sin \frac{7\pi}{6} \right)
\]
This matches the last option:
**Response:**
\( 2\sqrt{3} \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right), \quad 2\sqrt{3} \left( \cos \frac{7\pi}{6} + i \sin \frac{7\pi}{6} \right) \)