Question

Use DeMoivre’s Theorem to find left parenthesis 4 cis start fraction pi over 18 end fraction right parenthesis cubed.
(1 point)
Responses

32 plus 32 start root 3 end root lower i
Image with alt text: 32 plus 32 start root 3 end root lower i

32 start root 3 end root plus 32 lower i
Image with alt text: 32 start root 3 end root plus 32 lower i

6 plus 6 start root 3 end root lower i
Image with alt text: 6 plus 6 start root 3 end root lower i

6 start root 3 end root plus 6 lower i

Answer 1

To find \((4 \text{ cis } \frac{\pi}{18})^3\) using DeMoivre's Theorem, we start by applying the theorem.

DeMoivre's Theorem states that for any complex number in polar form \( r \text{ cis } \theta \):
\[
(r \text{ cis } \theta)^n = r^n \text{ cis } (n\theta)
\]

In our case, we have:
- \(r = 4\)
- \(\theta = \frac{\pi}{18}\)
- \(n = 3\)

Now we can compute:
1. The modulus raised to the power:
\[
r^3 = 4^3 = 64
\]

2. The angle multiplied by \(n\):
\[
n\theta = 3 \cdot \frac{\pi}{18} = \frac{3\pi}{18} = \frac{\pi}{6}
\]

Putting these together using DeMoivre's Theorem:
\[
(4 \text{ cis } \frac{\pi}{18})^3 = 64 \text{ cis } \frac{\pi}{6}
\]

Now, we can express \(64 \text{ cis } \frac{\pi}{6}\) in rectangular form. Recall that:
\[
\text{cis } \frac{\pi}{6} = \cos \frac{\pi}{6} + i \sin \frac{\pi}{6}
\]

Calculating the values:
\[
\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}, \quad \sin \frac{\pi}{6} = \frac{1}{2}
\]

So:
\[
\text{cis } \frac{\pi}{6} = \frac{\sqrt{3}}{2} + i \cdot \frac{1}{2}
\]

Now multiply by \(64\):
\[
64 \text{ cis } \frac{\pi}{6} = 64 \left( \frac{\sqrt{3}}{2} + i \cdot \frac{1}{2} \right) = 32\sqrt{3} + 32i
\]

Thus, the final result is:
\[
\boxed{32\sqrt{3} + 32i}
\]