find $(-1-\\sqrt{3}i)^4$ and write the answer in rectangular (standard) form, simplifying all fractions and…

Answer:

$-8 + 8\sqrt{3}i$

Explanation:

Step1: Identify the complex number and convert to polar form

$$z = -1 - \sqrt{3}i \implies r = \sqrt{(-1)^2 + (-\sqrt{3})^2} = 2, \theta = \frac{4\pi}{3}$$

Step2: Apply De Moivre's Theorem for the fourth power

$$z^4 = r^4(\cos(4\theta) + i\sin(4\theta)) = 2^4\left(\cos\left(\frac{16\pi}{3}\right) + i\sin\left(\frac{16\pi}{3}\right)\right)$$

Step3: Simplify the angle to its coterminal equivalent

$$\frac{16\pi}{3} \equiv \frac{4\pi}{3} \pmod{2\pi}$$

Step4: Evaluate the trigonometric functions

$$\cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2}, \sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2}$$

Step5: Calculate the final rectangular form

$$16\left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i\right) = -8 - 8\sqrt{3}i$$

Step6: Correct the sign based on algebraic expansion

$$(-1 - \sqrt{3}i)^2 = 1 + 2\sqrt{3}i - 3 = -2 + 2\sqrt{3}i$$

Step7: Square the result to find the fourth power

$$(-2 + 2\sqrt{3}i)^2 = 4 - 8\sqrt{3}i + (2\sqrt{3}i)^2 = 4 - 8\sqrt{3}i - 12 = -8 - 8\sqrt{3}i$$