A $60^\circ$ rotation around the origin in the counter-clockwise direction is applied to $3 \sqrt{3} - 5i.$  What is the resulting complex number?
To rotate a complex number by an angle $\theta$ around the origin, I need to multiply it by $e^{i \theta}$, where $i$ is the imaginary unit. In this case, $\theta = 60^\circ$, so I need to multiply $3 \sqrt{3} - 5i$ by $e^{i 60^\circ}$. I recall that $e^{i 60^\circ} = \frac{1}{2} + \frac{\sqrt{3}}{2} i$, so I can write $e^{i 60^\circ} = \frac{1 + \sqrt{3} i}{2}$. I get $(3 \sqrt{3} - 5i)(\frac{1 + \sqrt{3} i}{2}) = \frac{3 \sqrt{3} - 5i + 9i - 5 \sqrt{3} i^2}{2}$. I know that $i^2 = -1$, so I can replace it and combine like terms. I get $\frac{3 \sqrt{3} - 5i + 9i + 5 \sqrt{3}}{2} = \frac{8 \sqrt{3} + 4i}{2}$. I can divide each term by 2 to get the final answer. I get $4 \sqrt{3} + 2i$.

