1. The problem is to express each complex number $z_k = x + yi$ in the form $\cos \theta + i \sin \theta$, where $\theta$ is the argument (angle) of the complex number. 2. Recall that for a complex number $z = x + yi$, the argument $\theta$ satisfies: $$\cos \theta = \frac{x}{r}, \quad \sin \theta = \frac{y}{r}$$ where $r = \sqrt{x^2 + y^2}$ is the modulus of $z$. Since all given points lie on the unit circle ($r=1$), we have $\cos \theta = x$ and $\sin \theta = y$. 3. We find $\theta$ by identifying the angle whose cosine and sine match the coordinates $(x,y)$. --- **1) For** $z_1 = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i$: - $x = \frac{\sqrt{2}}{2}$, $y = \frac{\sqrt{2}}{2}$ - This corresponds to $\cos \theta = \frac{\sqrt{2}}{2}$ and $\sin \theta = \frac{\sqrt{2}}{2}$ - The angle with these values is $\theta = \frac{\pi}{4}$ - So, $z_1 = \cos \frac{\pi}{4} + i \sin \frac{\pi}{4}$ **2) For** $z_2 = -\frac{1}{2} - \frac{\sqrt{3}}{2} i$: - $x = -\frac{1}{2}$, $y = -\frac{\sqrt{3}}{2}$ - $\cos \theta = -\frac{1}{2}$, $\sin \theta = -\frac{\sqrt{3}}{2}$ - This corresponds to $\theta = -\frac{2\pi}{3}$ (or equivalently $\theta = \frac{4\pi}{3}$) - So, $z_2 = \cos \left(-\frac{2\pi}{3}\right) + i \sin \left(-\frac{2\pi}{3}\right)$ **3) For** $z_3 = -\frac{\sqrt{3}}{2} + \frac{1}{2} i$: - $x = -\frac{\sqrt{3}}{2}$, $y = \frac{1}{2}$ - $\cos \theta = -\frac{\sqrt{3}}{2}$, $\sin \theta = \frac{1}{2}$ - This corresponds to $\theta = \frac{5\pi}{6}$ - So, $z_3 = \cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6}$ **4) For** $z_4 = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i$: - $x = -\frac{\sqrt{2}}{2}$, $y = -\frac{\sqrt{2}}{2}$ - $\cos \theta = -\frac{\sqrt{2}}{2}$, $\sin \theta = -\frac{\sqrt{2}}{2}$ - This corresponds to $\theta = -\frac{3\pi}{4}$ (or equivalently $\theta = \frac{5\pi}{4}$) - So, $z_4 = \cos \left(-\frac{3\pi}{4}\right) + i \sin \left(-\frac{3\pi}{4}\right)$ **5) For** $z_5 = i$: - $x = 0$, $y = 1$ - $\cos \theta = 0$, $\sin \theta = 1$ - This corresponds to $\theta = \frac{\pi}{2}$ - So, $z_5 = \cos \frac{\pi}{2} + i \sin \frac{\pi}{2}$ --- **Final answers:** $$\begin{aligned} z_1 &= \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \\ z_2 &= \cos \left(-\frac{2\pi}{3}\right) + i \sin \left(-\frac{2\pi}{3}\right) \\ z_3 &= \cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6} \\ z_4 &= \cos \left(-\frac{3\pi}{4}\right) + i \sin \left(-\frac{3\pi}{4}\right) \\ z_5 &= \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \end{aligned}$$