Solving two varibles system equation above $\mathbb{C}$

Question

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

Answer 1

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.

$\displaystyle \frac{-2i}{(2-2i)}\cdot \frac{(2+2i)}{(2+2i)}=\frac{-4i-4}{2^2+2^2}=\frac{-4i+4}{8}=\frac{1}{2}-\frac{1}{2}i$

Answer 2

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

$$\frac1z=\frac{\bar{z}}{|z|^2}$$