1. **State the problem:** Find the cube root of the complex number $81i$. 2. **Recall the formula:** The cube roots of a complex number $z = re^{i\theta}$ are given by $$\sqrt[3]{z} = \sqrt[3]{r} e^{i(\frac{\theta + 2k\pi}{3})} \quad \text{for } k=0,1,2.$$ 3. **Convert $81i$ to polar form:** - The modulus is $r = |81i| = 81$. - The argument $\theta$ is the angle with the positive real axis. Since $81i$ lies on the positive imaginary axis, $\theta = \frac{\pi}{2}$. 4. **Calculate the cube root of the modulus:** $$\sqrt[3]{81} = \sqrt[3]{3^4} = 3^{\frac{4}{3}} = 3 \times 3^{\frac{1}{3}} = 3 \sqrt[3]{3}.$$ 5. **Find the three cube roots:** For $k=0,1,2$, $$z_k = 3 \sqrt[3]{3} e^{i \frac{\frac{\pi}{2} + 2k\pi}{3}} = 3 \sqrt[3]{3} e^{i (\frac{\pi}{6} + \frac{2k\pi}{3})}.$$ 6. **Write the roots in rectangular form:** $$z_k = 3 \sqrt[3]{3} \left( \cos\left(\frac{\pi}{6} + \frac{2k\pi}{3}\right) + i \sin\left(\frac{\pi}{6} + \frac{2k\pi}{3}\right) \right).$$ 7. **Final answer:** The three cube roots of $81i$ are $$3 \sqrt[3]{3} \left( \cos\left(\frac{\pi}{6} + \frac{2k\pi}{3}\right) + i \sin\left(\frac{\pi}{6} + \frac{2k\pi}{3}\right) \right) \quad \text{for } k=0,1,2.$$