Proving that a complex manifold is orientable.

Question

I am in the process of proving that a complex manifold is orientable. Consider the case $m=1$ so that in some chart, the usual coordinates of $p\in M$ are $(x,y)$. In some overlapping chart, let the coordinates of $p$ be denoted $(u,v)$. The manifold by assumption is analytic and so the Cauchy Riemann equations hold for the transition maps. For this case the proof is straightforward (as explained in one of the related posts). One need only show that the determinant of the Jacobian matrix comes out positive. And for this case, it does indeed (by use of Cauchy Riemann equations):$$\begin{vmatrix}u_x&u_y\\v_x&v_y \end{vmatrix}=u_x^2+v_x^2>0.$$ The problem arises when I try to extend this argument. Consider the case $m=2$. Then the coordinates of $p\in M$ in the overlapping charts are $(x^1, y^1,x^2,y^2)$ and $(u^1, v^1,u^2,v^2)$. The Cauchy Riemann equations read: $$u_{x^1}^1=v_{y^1}^1,v_{x^1}^1=-u_{y^1}^1,u_{x^1}^2=v_{y^1}^2,v_{x^1}^2=-u_{y^1}^2,u_{x^2}^1=v_{y^2}^1,v_{x^2}^1=-u_{y^2}^1,u_{x^2}^2=v_{y^2}^2,v_{x^2}^2=-u_{y^2}^2.$$ Putting these to use I work out the Jacobian determinant as:$$\begin{vmatrix}u_{x^1}^1&u_{y^1}^1&u_{x^2}^1&u_{y^2}^1\\v_{x^1}^1&v_{y^1}^1&v_{x^2}^1&v_{y^2}^1\\u_{x^1}^2&u_{y^1}^2&u_{x^2}^2&u_{y^2}^2\\v_{x^1}^2&v_{y^1}^2&v_{x^2}^2&v_{y^2}^2\end{vmatrix}$$$$=(u^1_{x^1}u^2_{x^2}+v^1_{x^2}v^2_{x^1})^2+(u^1_{x^1}v^2_{x^2}-u^2_{x^1}v^1_{x^2})^2+(u^1_{x^2}u^2_{x^1}+v^1_{x^1}v^2_{x^2})^2+(u^2_{x^2}v^1_{x^1}-u^1_{x^2}v^2_{x^1})^2-2u^1_{x^1}u^1_{x^2}v^2_{x^1}v^2_{x^2}-2u^1_{x^1}u^1_{x^2}u^2_{x^1}u^2_{x^2}-2u^2_{x^1}u^2_{x^2}v^1_{x^1}v^1_{x^2}-2v^1_{x^1}v^1_{x^2}v^2_{x^1}v^2_{x^2}.$$ I am not being able to make progress beyond this point. The squared terms are obviously not the problem, rather it is the last four terms that are not allowing me to deduce that the determinant be positive. Kindly guide me.