1. **State the problem:** Write the complex number $z = \frac{1}{\sqrt{3}} + i$ in polar form and then use de Moivre's theorem to find $z^{20}$ in the form $a + bi$. 2. **Recall the polar form and de Moivre's theorem:** A complex number $z = x + yi$ can be written in polar form as $$z = r(\cos \theta + i \sin \theta)$$ where $r = \sqrt{x^2 + y^2}$ is the modulus and $\theta = \tan^{-1}(\frac{y}{x})$ is the argument. De Moivre's theorem states: $$z^n = r^n (\cos n\theta + i \sin n\theta)$$ 3. **Find $r$ and $\theta$ for $z$:** Given $x = \frac{1}{\sqrt{3}}$, $y = 1$, $$r = \sqrt{\left(\frac{1}{\sqrt{3}}\right)^2 + 1^2} = \sqrt{\frac{1}{3} + 1} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$$ $$\theta = \tan^{-1}\left(\frac{1}{\frac{1}{\sqrt{3}}}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$$ 4. **Write $z$ in polar form:** $$z = \frac{2}{\sqrt{3}} \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right)$$ 5. **Use de Moivre's theorem to find $z^{20}$:** $$z^{20} = \left( \frac{2}{\sqrt{3}} \right)^{20} \left( \cos \frac{20\pi}{3} + i \sin \frac{20\pi}{3} \right)$$ 6. **Simplify the angle:** Since $\cos$ and $\sin$ are $2\pi$ periodic, $$\frac{20\pi}{3} = 6\pi + \frac{2\pi}{3} = 3 \times 2\pi + \frac{2\pi}{3}$$ So, $$\cos \frac{20\pi}{3} = \cos \frac{2\pi}{3} = -\frac{1}{2}$$ $$\sin \frac{20\pi}{3} = \sin \frac{2\pi}{3} = \frac{\sqrt{3}}{2}$$ 7. **Calculate the magnitude:** $$\left( \frac{2}{\sqrt{3}} \right)^{20} = \frac{2^{20}}{(\sqrt{3})^{20}} = \frac{2^{20}}{3^{10}}$$ 8. **Write $z^{20}$ in rectangular form:** $$z^{20} = \frac{2^{20}}{3^{10}} \left( -\frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = \frac{2^{20}}{3^{10}} \times \left(-\frac{1}{2}\right) + i \frac{2^{20}}{3^{10}} \times \frac{\sqrt{3}}{2}$$ 9. **Simplify:** $$z^{20} = -\frac{2^{20}}{2 \times 3^{10}} + i \frac{2^{20} \sqrt{3}}{2 \times 3^{10}} = -\frac{2^{19}}{3^{10}} + i \frac{2^{19} \sqrt{3}}{3^{10}}$$ **Final answer:** $$z = \frac{2}{\sqrt{3}} \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right)$$ $$z^{20} = -\frac{2^{19}}{3^{10}} + i \frac{2^{19} \sqrt{3}}{3^{10}}$$