1. **Problem statement:** We are given two complex numbers: $$z_1 = \sqrt{3} - 3i$$ and $$z_2 = 2 \sqrt[6]{e^{\frac{5\pi}{6}i}}$$ We need to: (a) Express $z_1$ in the form $re^{i\theta}$ where $r > 0$ and $-\pi < \theta \leq \pi$. (b) Find the cube roots $u$, $v$, and $w$ of $\frac{z_2}{z_1}$. --- 2. **Expressing $z_1$ in polar form $re^{i\theta}$:** - The polar form of a complex number $z = x + yi$ is $re^{i\theta}$ where: - $r = |z| = \sqrt{x^2 + y^2}$ (the modulus or length) - $\theta = \arg(z)$ (the angle with the positive real axis) - For $z_1 = \sqrt{3} - 3i$, we have: - $x = \sqrt{3}$ - $y = -3$ Calculate $r$: $$r = \sqrt{(\sqrt{3})^2 + (-3)^2} = \sqrt{3 + 9} = \sqrt{12} = 2\sqrt{3}$$ Calculate $\theta$: $$\theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{-3}{\sqrt{3}}\right) = \tan^{-1}(-\sqrt{3})$$ - $\tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}$ because tangent of $-\frac{\pi}{3}$ is $-\sqrt{3}$. - Since $x > 0$ and $y < 0$, the point is in the fourth quadrant, so $\theta = -\frac{\pi}{3}$ is correct and within $(-\pi, \pi]$. **Answer for (a):** $$z_1 = 2\sqrt{3} e^{-i\frac{\pi}{3}}$$ --- 3. **Simplify $z_2$ and find $\frac{z_2}{z_1}$:** Given: $$z_2 = 2 \sqrt[6]{e^{\frac{5\pi}{6}i}} = 2 e^{\frac{5\pi}{36}i}$$ - Because $\sqrt[6]{e^{i\alpha}} = e^{i\alpha/6}$. Now compute: $$\frac{z_2}{z_1} = \frac{2 e^{\frac{5\pi}{36}i}}{2\sqrt{3} e^{-i\frac{\pi}{3}}} = \frac{2}{2\sqrt{3}} e^{i\left(\frac{5\pi}{36} + \frac{\pi}{3}\right)} = \frac{1}{\sqrt{3}} e^{i\left(\frac{5\pi}{36} + \frac{12\pi}{36}\right)} = \frac{1}{\sqrt{3}} e^{i\frac{17\pi}{36}}$$ --- 4. **Find the cube roots $u$, $v$, and $w$ of $\frac{z_2}{z_1}$:** - The cube roots of a complex number $re^{i\theta}$ are given by: $$\sqrt[3]{r} e^{i\frac{\theta + 2k\pi}{3}} \quad \text{for } k=0,1,2$$ - Here: - $r = \frac{1}{\sqrt{3}}$ - $\theta = \frac{17\pi}{36}$ Calculate $\sqrt[3]{r}$: $$\sqrt[3]{\frac{1}{\sqrt{3}}} = \frac{1}{\sqrt[3]{\sqrt{3}}} = \frac{1}{(3^{1/2})^{1/3}} = \frac{1}{3^{1/6}}$$ Calculate the angles for $k=0,1,2$: - For $k=0$: $$\theta_0 = \frac{17\pi}{36} \times \frac{1}{3} = \frac{17\pi}{108}$$ - For $k=1$: $$\theta_1 = \frac{17\pi}{108} + \frac{2\pi}{3} = \frac{17\pi}{108} + \frac{72\pi}{108} = \frac{89\pi}{108}$$ - For $k=2$: $$\theta_2 = \frac{17\pi}{108} + \frac{4\pi}{3} = \frac{17\pi}{108} + \frac{144\pi}{108} = \frac{161\pi}{108}$$ So the cube roots are: $$u = \frac{1}{3^{1/6}} e^{i\frac{17\pi}{108}}, \quad v = \frac{1}{3^{1/6}} e^{i\frac{89\pi}{108}}, \quad w = \frac{1}{3^{1/6}} e^{i\frac{161\pi}{108}}$$ --- **Final answers:** (a) $$z_1 = 2\sqrt{3} e^{-i\frac{\pi}{3}}$$ (b) The cube roots of $\frac{z_2}{z_1}$ are: $$u = \frac{1}{3^{1/6}} e^{i\frac{17\pi}{108}}, \quad v = \frac{1}{3^{1/6}} e^{i\frac{89\pi}{108}}, \quad w = \frac{1}{3^{1/6}} e^{i\frac{161\pi}{108}}$$