1. **State the problem:** Convert the complex number $\frac{\sqrt{6}}{2}(i-1)$ to polar form. 2. **Recall the polar form:** A complex number $z = x + yi$ can be written in polar form as $$z = r(\cos \theta + i \sin \theta) = r e^{i\theta}$$ where $r = |z| = \sqrt{x^2 + y^2}$ is the modulus and $\theta = \arg(z) = \tan^{-1}(\frac{y}{x})$ is the argument. 3. **Identify $x$ and $y$:** Given $z = \frac{\sqrt{6}}{2}(i - 1) = \frac{\sqrt{6}}{2}i - \frac{\sqrt{6}}{2}$, so $$x = -\frac{\sqrt{6}}{2}, \quad y = \frac{\sqrt{6}}{2}$$ 4. **Calculate modulus $r$:** $$r = \sqrt{x^2 + y^2} = \sqrt{\left(-\frac{\sqrt{6}}{2}\right)^2 + \left(\frac{\sqrt{6}}{2}\right)^2} = \sqrt{\frac{6}{4} + \frac{6}{4}} = \sqrt{\frac{12}{4}} = \sqrt{3}$$ 5. **Calculate argument $\theta$:** $$\theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{\frac{\sqrt{6}}{2}}{-\frac{\sqrt{6}}{2}}\right) = \tan^{-1}(-1)$$ Since $x < 0$ and $y > 0$, the point is in the second quadrant, so $$\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$$ 6. **Write the polar form:** $$z = r e^{i\theta} = \sqrt{3} e^{i \frac{3\pi}{4}}$$ **Final answer:** The polar form of $\frac{\sqrt{6}}{2}(i-1)$ is $$\boxed{\sqrt{3} e^{i \frac{3\pi}{4}}}$$