complex numbers - correctly found the angle phi (arg z)? $left(frac{(sqrt2-isqrt2)(1-isqrt3)}{4}right)$

$$1) \left(\frac{(\sqrt2-i\sqrt2)(1-i\sqrt3)}{4}\right) = \left(\frac{(\sqrt2-\sqrt2\sqrt3)+i(\sqrt2*(-\sqrt3) - \sqrt2)}{4}\right) = \left(\frac{(\sqrt2-\sqrt6)+i(-\sqrt6-\sqrt2)}{4}\right) = \left(\frac{\sqrt2-\sqrt6}{4}+\frac{i(-\sqrt6-\sqrt2)}{4}\right)$$

$$2) tg \phi = \frac{y}{x}$$
$$3) tg\phi = \frac{-\sqrt6-\sqrt2}{\sqrt2-\sqrt6}=\frac{(-\sqrt6-\sqrt2)(\sqrt2+\sqrt6)}{2-6}= \frac{-(\sqrt6+\sqrt2)(\sqrt2+\sqrt6)}{-4} = \frac{(\sqrt6+\sqrt2)^2}{4}=\frac{6+2\sqrt{12}+2}{4}= \frac{8+2\sqrt{12}}{4}=\frac{8+4\sqrt3}{4} = 2+\sqrt3$$
$$4) tg\phi = 2+\sqrt3$$

$$5) \phi = arctg(2+\sqrt3)$$

Are all agreed right?

If not - tell me where the error.

And yet, how to find arg z ???