$$z^4+2(\sqrt{12}-2i)z^2+8-4\sqrt{12}\cdot i = 0\\ \boxed {v=z^2}\\ v^2+2(\sqrt{12-2i})v+8-4\sqrt{12}\space i=0\\ ⇒\text{pq-Formel}\\ v_{1,2}=-(\sqrt{12}-2x)\pm \sqrt{(2i-\sqrt{12})^2-8+4\sqrt{12}i}\\ v_{1,2}=-(\sqrt{12}-2x)\pm \sqrt{8-8i\sqrt{3}-8+4\sqrt{4}\sqrt{3}i}\\ v_{1,2}=-(\sqrt{12}-2x)\pm \sqrt{0}\\ v_{1,2}=-\sqrt{4}\sqrt{3}+2i\\ v_{1,2}=-2\sqrt{3}+2i\\ ⇒v=z^2=-2\sqrt{3}+2i\\ z_{1,2}=\pm \sqrt{-2\sqrt{3}+2i}\\ [\text{Polarform: z=r (cos(2) + i sin(2)})]\phantom{4}\text {allgemein}\\ ⇒-2\sqrt{3}+2i\\ \text{Betrag}=\sqrt{(-2\sqrt{3})^2+2^2}=4\\ \text{Winkel(=)}tanφ=\frac{Imaginärteil}{Realteil}=\frac{2}{-2\sqrt{3}}=-\frac{\sqrt{3}}{3}\\ φ=\frac{5}{6}π\phantom{4}\text{(2. Quadrant)}\\ ⇒\\ z=\pm \sqrt{4\cdot e^{i\frac{5}{6}π}}=\pm (4\cdot e^{i\frac{5}{6}π})^\frac{1}{2}\\ z=\pm 2\cdot e^{i\frac{5}{12}π}\\ z=\pm 2\bigl[cos(\frac{5}{12}\cdot π)+i\space sin(\frac{5}{12}\cdotπ)\bigr]$$
