Solving two varibles system equation above $\mathbb{C}$ A bit emmbarrassed to ask this newbie question:  
Let:
$$(1+i)x + y = 2$$
$$(1-i)x + iy = 0$$
Multiplying the first equation by $(-i)$ and summing the two equations, we have:
$$(2-2i)x + 2i = 0$$  
How to get the final result of: $$x = {1 \over 2} - {1\over 2}i$$?

If $(2-2i)x+2i=0$, then
$$x=\frac{-2i}{2-2i}=\frac{-i}{1-i}=\frac{-i(1+i)}{2}=\frac{1-i}{2}$$
Using the fact that $(1-i)(1+i)=|1-i|^2=2$
More generally, it's useful to remember
$$\frac{1}{a+ib}=\frac{a-ib}{a^2+b^2}$$
Or
$$\frac1z=\frac{\bar{z}}{|z|^2}$$
