1. **Problem:** Solve the equation in $\mathbb{C}$: $$e^z = 3\sqrt{3} - 3i.$$ 2. **Formula and rules:** Recall that for any complex number $z = x + iy$, we have: $$e^z = e^{x+iy} = e^x (\cos y + i \sin y).$$ To solve $e^z = w$ for $w \in \mathbb{C}$, write $w$ in polar form: $$w = r (\cos \theta + i \sin \theta) = r e^{i \theta},$$ where $r = |w|$ and $\theta = \arg(w)$. Then the solutions are: $$z = \ln r + i (\theta + 2k\pi), \quad k \in \mathbb{Z}.$$ 3. **Intermediate work:** Calculate $r$: $$r = |3\sqrt{3} - 3i| = \sqrt{(3\sqrt{3})^2 + (-3)^2} = \sqrt{27 + 9} = \sqrt{36} = 6.$$ Calculate $\theta$: $$\theta = \arg(3\sqrt{3} - 3i) = \arctan\left(\frac{-3}{3\sqrt{3}}\right) = \arctan\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}.$$ 4. **Solution:** Thus, $$z = \ln 6 + i \left(-\frac{\pi}{6} + 2k\pi\right), \quad k \in \mathbb{Z}.$$ This is the complete set of solutions. **Final answer:** $$\boxed{z = \ln 6 + i \left(-\frac{\pi}{6} + 2k\pi\right), \quad k \in \mathbb{Z}}.$$