Showing that ${1 + i,−1 + i}$ is a basis for the vector space $\mathbb C$ over $\mathbb R$ I feel my justification for this is weak, and I'm seeking for improvement.
$$Span((1+i), (-1+i)) = a(1+i) + b(-1+i)$$
I have two conclusions from this:
1.
$$a(1+i) + b(-1+i) = a + ai -b + bi$$
$$a(1+i) + b(-1+i) = (a - b) + i (a +b)$$
And, other than just saying by setting $a$ and $b$ to any real number, any vector in $\mathbb C$ can be created.
2.
$a(1+i)$ is a vector that resembles the image of $y = x$ in that it can extend to any point in that direction.
$b(-1+i)$ is a vector that resembles the image of $y=-x$ in that it can extend to any point in that direction. 
Since these two vectors are orthogonal, any linear combination of the two can form any vector in $\mathbb C$.

Is this a valid proof? Or is there a better way, or proper way, to show this? Besides, the span of any two linearly independent vectors in $\mathbb C$ should be able to span all of $\mathbb C$, anyway. An example would be $(1+i)$ and $(3+2i)$.

Since the linear combination is
$(a-b)+(a+b)i$,
all you need to do
is to state that,
for any $c+di$,
you can make
$(a-b)+(a+b)i
=c+di$
by setting
$a=(c+d)/2$
and
$b = (d-c)/2$.
Nothing more.
