$\text{ Least positive argument of the }{4}^{\text{th }}\text{ root of the complex number }2-i\sqrt{12}\text{ is: }$

$\begin{array}{l}{z}^4=2(1-\sqrt{3}i)=4\left(\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)\\ =4\left[\cos \left(-\frac{\pi }{3}\right)+i\sin \left(-\frac{\pi }{3}\right)\right]\\ z=\sqrt{2}\left[\cos \frac{2m\pi -\left(\frac{\pi }{3}\right)}{4}+i\sin \frac{2m\pi -\left(\frac{\pi }{3}\right)}{4}\right]\\ For m=1, z=\sqrt{2}\left[\cos \left(\frac{5\pi }{12}\right)+i\sin \left(\frac{5\pi }{12}\right)\right]\\ \therefore The least positive argument of z is \frac{5\pi }{12}\end{array}$