Let's solve the expression $(-2 \times \sqrt{3} + 2i)^3$.

This expression represents a complex number in the form $a + bi$, where $a = -2\sqrt{3}$ and $b = 2$.

Step 1: Express in Polar Form

To express the complex number in polar form $re^{i\theta}$, we first calculate the magnitude $r$ and the argument $\theta$:

1. Magnitude $r$: $r = \sqrt{a^2 + b^2} = \sqrt{(-2\sqrt{3})^2 + 2^2} = \sqrt{12 + 4} = \sqrt{16} = 4$

2. Argument $\theta$: $\theta = \tan^{-1}\left(\frac{b}{a}\right) = \tan^{-1}\left(\frac{2}{-2\sqrt{3}}\right)$ Simplifying: $\theta = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = \tan^{-1}(-\frac{\sqrt{3}}{3}) = -\frac{\pi}{6}$ Since the complex number is in the second quadrant (negative real part and positive imaginary part), we adjust the angle: $\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$

Step 2: Apply De Moivre's Theorem

Using De Moivre's Theorem to find $(re^{i\theta})^3$: $(re^{i\theta})^3 = r^3 \cdot e^{i \cdot 3\theta}$

Substitute the values: $r^3 = 4^3 = 64$ $3\theta = 3 \times \frac{5\pi}{6} = \frac{15\pi}{6} = \frac{5\pi}{2}$

Step 3: Convert Back to Rectangular Form

We now convert the polar form back to rectangular form: $64 \times \left(\cos\left(\frac{5\pi}{2}\right) + i\sin\left(\frac{5\pi}{2}\right)\right)$ Since $\frac{5\pi}{2}$ is equivalent to $\frac{\pi}{2}$ (adding $2\pi$ to get within the standard interval): $64 \times \left(\cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right)\right)$ $= 64 \times (0 + i \times 1) = 64i$

Thus, $(-2 \times \sqrt{3} + 2i)^3 = 64i$.

Would you like any further details or have any questions?

Related Questions:

1. What is the significance of the argument $\theta$ in complex numbers?
2. How do you convert a complex number from rectangular to polar form?
3. Can you explain De Moivre's Theorem and its applications?
4. What are the properties of powers of complex numbers?
5. How does the magnitude of a complex number affect its powers?

Tip:

When dealing with complex numbers in powers, converting to polar form often simplifies the computation significantly.