1. **State the problem:** Solve the equation $$z^4 = e^{\frac{2\pi i}{3}}$$ for the complex number $z$. 2. **Formula and rules:** To solve equations of the form $$z^n = w$$ where $w$ is a complex number, we use the formula for the $n$th roots of a complex number: $$z_k = |w|^{\frac{1}{n}} e^{i \left( \frac{\arg(w) + 2\pi k}{n} \right)}$$ where $k = 0, 1, 2, ..., n-1$. 3. **Identify parameters:** Here, $n=4$ and $$w = e^{\frac{2\pi i}{3}}$$. The magnitude $|w| = 1$ since it is on the unit circle. The argument $$\arg(w) = \frac{2\pi}{3}$$. 4. **Calculate the roots:** Using the formula, $$z_k = 1^{\frac{1}{4}} e^{i \left( \frac{\frac{2\pi}{3} + 2\pi k}{4} \right)} = e^{i \left( \frac{2\pi}{12} + \frac{2\pi k}{4} \right)} = e^{i \left( \frac{\pi}{6} + \frac{\pi k}{2} \right)}$$ for $k=0,1,2,3$. 5. **List all solutions:** - For $k=0$: $$z_0 = e^{i \frac{\pi}{6}} = \cos\frac{\pi}{6} + i \sin\frac{\pi}{6} = \frac{\sqrt{3}}{2} + \frac{1}{2}i$$ - For $k=1$: $$z_1 = e^{i \left( \frac{\pi}{6} + \frac{\pi}{2} \right)} = e^{i \frac{2\pi}{3}} = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$$ - For $k=2$: $$z_2 = e^{i \left( \frac{\pi}{6} + \pi \right)} = e^{i \frac{7\pi}{6}} = -\frac{\sqrt{3}}{2} - \frac{1}{2}i$$ - For $k=3$: $$z_3 = e^{i \left( \frac{\pi}{6} + \frac{3\pi}{2} \right)} = e^{i \frac{11\pi}{6}} = \frac{\sqrt{3}}{2} - \frac{1}{2}i$$ **Final answer:** The four solutions to $$z^4 = e^{\frac{2\pi i}{3}}$$ are $$z = \frac{\sqrt{3}}{2} + \frac{1}{2}i, \quad -\frac{1}{2} + \frac{\sqrt{3}}{2}i, \quad -\frac{\sqrt{3}}{2} - \frac{1}{2}i, \quad \frac{\sqrt{3}}{2} - \frac{1}{2}i.$$