The expression in the image is:

$(3 + \sqrt{3}i)^5$

This is a complex number raised to the 5th power. To solve it, we can either:

1. Expand it using the binomial theorem, which can be time-consuming due to the powers involved.
2. Convert the complex number to polar form, use De Moivre’s Theorem, and then convert it back to rectangular form after raising it to the power.

The second approach is more efficient. Here are the steps to solve it:

Step 1: Convert to Polar Form

First, write $3 + \sqrt{3}i$ in polar form. The polar form of a complex number $z = a + bi$ is given by: $z = r(\cos \theta + i \sin \theta)$ where:

• $r = |z| = \sqrt{a^2 + b^2}$ is the modulus
• $\theta = \arg(z) = \tan^{-1} \left(\frac{b}{a}\right)$ is the argument.

For $3 + \sqrt{3}i$:

• $a = 3$
• $b = \sqrt{3}$

The modulus $r$ is: $r = \sqrt{3^2 + (\sqrt{3})^2} = \sqrt{9 + 3} = \sqrt{12} = 2\sqrt{3}$

The argument $\theta$ is: $\theta = \tan^{-1} \left(\frac{\sqrt{3}}{3}\right) = \tan^{-1} \left(\frac{1}{\sqrt{3}}\right) = 30^\circ = \frac{\pi}{6} \text{ radians}$

Thus, the polar form of $3 + \sqrt{3}i$ is: $3 + \sqrt{3}i = 2\sqrt{3} \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right)$

Step 2: Apply De Moivre's Theorem

De Moivre’s Theorem states that for any complex number in polar form: $z^n = r^n \left( \cos (n\theta) + i \sin (n\theta) \right)$

In this case, $n = 5$, $r = 2\sqrt{3}$, and $\theta = \frac{\pi}{6}$.

First, calculate $r^5$: $r^5 = (2\sqrt{3})^5 = 32 \sqrt{3}$

Now, calculate $5\theta$: $5\theta = 5 \times \frac{\pi}{6} = \frac{5\pi}{6}$

Thus, the result in polar form is: $(3 + \sqrt{3}i)^5 = 32\sqrt{3} \left( \cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6} \right)$

Step 3: Convert Back to Rectangular Form

Now, convert the polar form back to rectangular form. The values of the trigonometric functions for $\frac{5\pi}{6}$ are:

• $\cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2}$
• $\sin \frac{5\pi}{6} = \frac{1}{2}$

So: $(3 + \sqrt{3}i)^5 = 32\sqrt{3} \left( -\frac{\sqrt{3}}{2} + i \times \frac{1}{2} \right)$

Simplify: $(3 + \sqrt{3}i)^5 = 32\sqrt{3} \times \left( -\frac{\sqrt{3}}{2} + \frac{i}{2} \right)$ $= 32\sqrt{3} \times -\frac{\sqrt{3}}{2} + 32\sqrt{3} \times \frac{i}{2}$ $= -48 + 16i$

Final Answer:

$(3 + \sqrt{3}i)^5 = -48 + 16i$

Would you like further clarification or details on any step?

Here are 5 related questions:

1. How do you compute the modulus and argument of any complex number?
2. What is De Moivre's Theorem and how is it applied in solving powers of complex numbers?
3. How does the binomial theorem apply to complex numbers?
4. What are the advantages of using polar form for complex number operations?
5. How are trigonometric values derived from radians?

Tip: When dealing with complex numbers in high powers, converting to polar form often simplifies calculations dramatically.