Here's a step-by-step solution to the complex number problem: Part a) Express $z_1$ and $z_2$ in the form $r(\cos\theta + i\sin\theta)$ For a complex number $z = x + yi$, its polar form is $z = r(\cos\theta + i\sin\theta)$, where $r = \sqrt{x^2 + y^2}$ and $\theta = \arctan\left(\frac{y}{x}\right)$ adjusted for the correct quadrant, such that $-\pi < \theta \le \pi$. For $z_1 = 1 + \sqrt{3}i$: Step 1: Identify the real and imaginary parts. $x_1 = 1$, $y_1 = \sqrt{3}$. Step 2: Calculate the modulus $r_1$. $$r_1 = \sqrt{x_1^2 + y_1^2} = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2$$ Step 3: Calculate the argument $\theta_1$. Since $x_1 > 0$ and $y_1 > 0$, $\theta_1$ is in the first quadrant. $$\tan\theta_1 = \frac{y_1}{x_1} = \frac{\sqrt{3}}{1} = \sqrt{3}$$ $$\theta_1 = \arctan(\sqrt{3}) = \frac{\pi}{3}$$ This value is in the range $(-\pi, \pi]$. Step 4: Write $z_1$ in polar form. $$z_1 = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)$$ For $z_2 = -1 - i$: Step 1: Identify the real and imaginary parts. $x_2 = -1$, $y_2 = -1$. Step 2: Calculate the modulus $r_2$. $$r_2 = \sqrt{x_2^2 + y_2^2} = \sqrt{(-1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}$$ Step 3: Calculate the argument $\theta_2$. Since $x_2 < 0$ and $y_2 < 0$, $\theta_2$ is in the third quadrant. $$\tan\theta_2 = \frac{y_2}{x_2} = \frac{-1}{-1} = 1$$ The reference angle is $\frac{\pi}{4}$. For the third quadrant, $\theta_2 = -\left(\pi - \frac{\pi}{4}\right) = -\frac{3\pi}{4}$. This value is in the range $(-\pi, \pi]$. Step 4: Write $z_2$ in polar form. $$z_2 = \sqrt{2}\left(\cos\left(-\frac{3\pi}{4}\right) + i\sin\left(-\frac{3\pi}{4}\right)\right)$$ Part b) Evaluate $\left|\frac{z_1}{z_2}\right|^3$ and $\arg\left(\frac{z_1}{z_2}\right)^3$ Step 1: Find the polar form of $\frac{z_1}{z_2}$. If $z_1 = r_1(\cos\theta_1 + i\sin\theta_1)$ and $z_2 = r_2(\cos\theta_2 + i\sin\theta_2)$, then $$\frac{z_1}{z_2} = \frac{r_1}{r_2}(\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2))$$ Substitute the values from part a): $$\frac{r_1}{r_2} = \frac{2}{\sqrt{2}} = \sqrt{2}$$ $$\theta_1 - \theta_2 = \frac{\pi}{3} - \left(-\frac{3\pi}{4}\right) = \frac{\pi}{3} + \frac{3\pi}{4} = \frac{4\pi + 9\pi}{12} = \frac{13\pi}{12}$$ So, $$\frac{z_1}{z_2} = \sqrt{2}\left(\cos\left(\frac{13\pi}{12}\right) + i\sin\left(\frac{13\pi}{12}\right)\right)$$ Step 2: Evaluate $\left|\frac{z_1}{z_2}\right|^3$. Using De Moivre's Theorem, if $Z = r(\cos\theta + i\sin\theta)$, then $|Z^n| = r^n$. Here, $Z = \frac{z_1}{z_2}$, so $r = \sqrt{2}$ and $n=3$. $$\left|\frac{z_1}{z_2}\right|^3 = (\sqrt{2})^3 = 2\sqrt{2}$$ Step 3: Evaluate $\arg\left(\frac{z_1}{z_2}\right)^3$. Using De Moivre's Theorem, $\arg(Z^n) = n\theta$. Here, $\theta = \frac{13\pi}{12}$ and $n=3$. $$\arg\left(\frac{z_1}{z_2}\right)^3 = 3 \times \frac{13\pi}{12} = \frac{13\pi}{4}$$ The argument must be in the range $(-\pi, \pi]$. We adjust $\frac{13\pi}{4}$ by subtracting multiples of $2\pi$: $$\frac{13\pi}{4} - 2\pi = \frac{13\pi - 8\pi}{4} = \frac{5\pi}{4}$$ Since $\frac{5\pi}{4} > \pi$, we subtract $2\pi$ again: $$\frac{5\pi}{4} - 2\pi = \frac{5\pi - 8\pi}{4} = -\frac{3\pi}{4}$$ This value is in the range $(-\pi, \pi