Question
ind left-parenthesis 2 Start Root 3 End Root plus 2 i right-parenthesis superscript 5 baseline.
(1 point)
Responses
negative 10 Start Root 3 End Root plus 10 i
Image with alt text: negative 10 Start Root 3 End Root plus 10 i
10 Start Root 3 End Root minus 10 i
Image with alt text: 10 Start Root 3 End Root minus 10 i
negative 512 Start Root 3 End Root plus 512 i
Image with alt text: negative 512 Start Root 3 End Root plus 512 i
512 Start Root 3 End Root minus 512 i
(1 point)
Responses
negative 10 Start Root 3 End Root plus 10 i
Image with alt text: negative 10 Start Root 3 End Root plus 10 i
10 Start Root 3 End Root minus 10 i
Image with alt text: 10 Start Root 3 End Root minus 10 i
negative 512 Start Root 3 End Root plus 512 i
Image with alt text: negative 512 Start Root 3 End Root plus 512 i
512 Start Root 3 End Root minus 512 i
Answers
GPT-4o mini
To find \( \left( 2\sqrt{3} + 2i \right)^5 \), we can rewrite the expression in polar form and then use De Moivre's Theorem.
1. **Convert to polar coordinates**:
- The modulus is given by:
\[
r = \sqrt{(2\sqrt{3})^2 + (2)^2} = \sqrt{12 + 4} = \sqrt{16} = 4.
\]
- The argument \( \theta \) can be computed as:
\[
\theta = \tan^{-1}\left(\frac{2}{2\sqrt{3}}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}.
\]
- Therefore, in polar form, we have:
\[
2\sqrt{3} + 2i = 4\left(\cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right)\right).
\]
2. **Apply De Moivre's Theorem**:
- Using De Moivre's Theorem:
\[
\left( r(\cos \theta + i \sin \theta) \right)^n = r^n \left( \cos(n\theta) + i \sin(n\theta) \right).
\]
- Here, \( r = 4 \), \( n = 5 \), and \( \theta = \frac{\pi}{6} \):
\[
\left( 2\sqrt{3} + 2i \right)^5 = 4^5 \left( \cos\left(5 \cdot \frac{\pi}{6}\right) + i \sin\left(5 \cdot \frac{\pi}{6}\right) \right).
\]
- Calculating \( 4^5 \):
\[
4^5 = 1024.
\]
- The angles are:
\[
5 \cdot \frac{\pi}{6} = \frac{5\pi}{6}.
\]
- Now calculate \( \cos\left(\frac{5\pi}{6}\right) \) and \( \sin\left(\frac{5\pi}{6}\right) \):
\[
\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}, \quad \sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}.
\]
- Substitute these values in:
\[
\left( 2\sqrt{3} + 2i \right)^5 = 1024 \left( -\frac{\sqrt{3}}{2} + i\frac{1}{2} \right) = 1024 \left( -\frac{\sqrt{3}}{2} \right) + 1024 \left( i \frac{1}{2} \right).
\]
- This simplifies to:
\[
-512\sqrt{3} + 512i.
\]
Thus, the final answer is:
\[
\boxed{-512\sqrt{3} + 512i}.
\]
1. **Convert to polar coordinates**:
- The modulus is given by:
\[
r = \sqrt{(2\sqrt{3})^2 + (2)^2} = \sqrt{12 + 4} = \sqrt{16} = 4.
\]
- The argument \( \theta \) can be computed as:
\[
\theta = \tan^{-1}\left(\frac{2}{2\sqrt{3}}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}.
\]
- Therefore, in polar form, we have:
\[
2\sqrt{3} + 2i = 4\left(\cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right)\right).
\]
2. **Apply De Moivre's Theorem**:
- Using De Moivre's Theorem:
\[
\left( r(\cos \theta + i \sin \theta) \right)^n = r^n \left( \cos(n\theta) + i \sin(n\theta) \right).
\]
- Here, \( r = 4 \), \( n = 5 \), and \( \theta = \frac{\pi}{6} \):
\[
\left( 2\sqrt{3} + 2i \right)^5 = 4^5 \left( \cos\left(5 \cdot \frac{\pi}{6}\right) + i \sin\left(5 \cdot \frac{\pi}{6}\right) \right).
\]
- Calculating \( 4^5 \):
\[
4^5 = 1024.
\]
- The angles are:
\[
5 \cdot \frac{\pi}{6} = \frac{5\pi}{6}.
\]
- Now calculate \( \cos\left(\frac{5\pi}{6}\right) \) and \( \sin\left(\frac{5\pi}{6}\right) \):
\[
\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}, \quad \sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}.
\]
- Substitute these values in:
\[
\left( 2\sqrt{3} + 2i \right)^5 = 1024 \left( -\frac{\sqrt{3}}{2} + i\frac{1}{2} \right) = 1024 \left( -\frac{\sqrt{3}}{2} \right) + 1024 \left( i \frac{1}{2} \right).
\]
- This simplifies to:
\[
-512\sqrt{3} + 512i.
\]
Thus, the final answer is:
\[
\boxed{-512\sqrt{3} + 512i}.
\]