Use DeMoiver's Theorem to find (4cisπ/18)^3

Answer 1

To use DeMoiver's Theorem, we can rewrite \(4cis(\frac{\pi}{18})\) in rectangular form:

\(4cis(\frac{\pi}{18}) = 4(\cos(\frac{\pi}{18}) + i \sin(\frac{\pi}{18}))\)

Now, we can raise this complex number to the power of 3:

\((4cis(\frac{\pi}{18}))^3 = 4^3(\cos(\frac{\pi}{18}) + i \sin(\frac{\pi}{18}))^3\)

\(= 64(\cos(\frac{\pi}{18}) + i \sin(\frac{\pi}{18}))^3\)

Now, to find the cube of the complex number, we can use DeMoiver's Theorem. DeMoiver's Theorem states that for any complex number \(r cis(\theta)\), the power of n can be found using:

\((r cis(\theta))^n = r^n cis(n\theta)\)

Applying this theorem, we get:

\(64(\cos(\frac{\pi}{18}) + i \sin(\frac{\pi}{18}))^3\)

\(= 64(\cos(\frac{3\pi}{18}) + i \sin(\frac{3\pi}{18}))\)

\(= 64(\cos(\frac{\pi}{6}) + i \sin(\frac{\pi}{6}))\)

Now, we can convert this back to polar form:

\(64(\cos(\frac{\pi}{6}) + i \sin(\frac{\pi}{6}))\)

\(= 64 cis(\frac{\pi}{6})\)

Therefore, \((4cis(\frac{\pi}{18}))^3 = 64cis(\frac{\pi}{6})\)