We are asked to use De Moivre's Theorem to express the given complex number in rectangular form.

The complex number provided is: $3\left(\cos\frac{\pi}{6} - i\sin\frac{\pi}{6}\right)^5$

Step 1: Apply De Moivre’s Theorem

De Moivre's Theorem states that for a complex number $z = r(\cos\theta + i\sin\theta)$, raised to the power $n$, we have: $z^n = r^n\left(\cos(n\theta) + i\sin(n\theta)\right)$

Here, $r = 3$, $\theta = -\frac{\pi}{6}$, and $n = 5$.

Step 2: Calculate the modulus and argument

First, calculate the modulus $r^5$: $r^5 = 3^5 = 243$

Now, calculate the argument $5\theta$: $5\theta = 5 \times \left(-\frac{\pi}{6}\right) = -\frac{5\pi}{6}$

Step 3: Use De Moivre's Theorem

We now have: $z^5 = 243\left(\cos\left(-\frac{5\pi}{6}\right) + i\sin\left(-\frac{5\pi}{6}\right)\right)$

Step 4: Evaluate trigonometric functions

$\cos\left(-\frac{5\pi}{6}\right) = -\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}$ $\sin\left(-\frac{5\pi}{6}\right) = -\sin\left(\frac{5\pi}{6}\right) = -\frac{1}{2}$

Step 5: Write the expression in rectangular form

Now substitute these values into the expression: $z^5 = 243\left(-\frac{\sqrt{3}}{2} - i\frac{1}{2}\right)$

Distribute the 243: $z^5 = 243 \times \left(-\frac{\sqrt{3}}{2}\right) - 243 \times \left(i\frac{1}{2}\right)$ $z^5 = -\frac{243\sqrt{3}}{2} - i\frac{243}{2}$

Thus, the rectangular form of the complex number is: $z^5 = -\frac{243\sqrt{3}}{2} - i\frac{243}{2}$

Let me know if you want more details or if you have any questions! Here are some related questions for practice:

1. What is the polar form of $z = 2 + 2i$?
2. How does De Moivre's Theorem relate to Euler's formula?
3. What are the real and imaginary parts of $z^5 = 1 + i$?
4. How do you convert from rectangular to polar form?
5. What happens to the magnitude of a complex number when raised to a power?

Tip: Always simplify trigonometric functions at key angles like $\frac{\pi}{6}$, $\frac{\pi}{4}$, etc., as they frequently appear in polar coordinates.