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: $$x = {1 \over 2} - {1\over 2}i$$?

Okay solve for $x$
$\displaystyle x=\frac{-2i}{2-2i}$
The way we divide by a complex number and get in standard form is to multiply the numerator and denominator by the complex conjugate of the denominator.
$\displaystyle \frac{-2i}{(2-2i)}\cdot \frac{(2+2i)}{(2+2i)}=\frac{-4i-4}{2^2+2^2}=\frac{-4i+4}{8}=\frac{1}{2}-\frac{1}{2}i$
