Certainly! Let's go through the problem step-by-step to determine the new modulus and argument after raising the given complex number to the 6th power.
Given:
- Initial modulus: [tex]\(\sqrt{3}\)[/tex]
- Initial argument: [tex]\(\frac{\pi}{18}\)[/tex]
We need to raise the complex number defined by these values to the 6th power. We will utilize De Moivre's Theorem, which states:
[tex]\[
(r[\cos \theta + i \sin \theta])^n = r^n[\cos (n\theta) + i \sin (n\theta)]
\][/tex]
Where:
- [tex]\(r\)[/tex] is the modulus (in this case, [tex]\(\sqrt{3}\)[/tex])
- [tex]\(\theta\)[/tex] is the argument (in this case, [tex]\(\frac{\pi}{18}\)[/tex])
- [tex]\(n\)[/tex] is the power to which the expression is raised (in this case, [tex]\(6\)[/tex])
### Step 1: Calculate the new modulus
According to De Moivre's Theorem, the new modulus [tex]\(r'\)[/tex] will be:
[tex]\[
r' = (\sqrt{3})^6
\][/tex]
Calculating [tex]\((\sqrt{3})^6\)[/tex]:
[tex]\[
(\sqrt{3})^6 = (3^{1/2})^6 = 3^{(1/2) \cdot 6} = 3^3 = 27
\][/tex]
### Step 2: Calculate the new argument
According to De Moivre's Theorem, the new argument [tex]\(\theta'\)[/tex] will be:
[tex]\[
\theta' = 6 \cdot \frac{\pi}{18}
\][/tex]
Simplifying [tex]\(6 \cdot \frac{\pi}{18}\)[/tex]:
[tex]\[
6 \cdot \frac{\pi}{18} = \frac{6\pi}{18} = \frac{\pi}{3}
\][/tex]
### Conclusion
After raising the given complex number [tex]\((\sqrt{3})[\cos \frac{\pi}{18} + i \sin \frac{\pi}{18}]\)[/tex] to the 6th power:
- The new modulus is: [tex]\(27\)[/tex]
- The new argument is: [tex]\(\frac{\pi}{3}\)[/tex]
Thus, the correct answer is:
[tex]\[
\boxed{\text{modulus } = 27 ; \text{ argument } = \frac{\pi}{3}}
\][/tex]
