Is $\sqrt{-1}$ positive or negative? Do the concept of positive or negative make sense in this case? I remember that $\mathbb{R}^2$ has four quadrants thus ordered pairs of numbers could be $(+,+),(+,-),(-,-)(-,+)$, I presume that complex numbers have positivity or negativity  for both real and complex part, what if it's only $i$? Which is the ordered pair $(0,1)$, is it $(0,+)$?
 A: No. $\mathbb{C}$ is not an ordered field. 
If you are looking for more information about notions of positive and negative and how they can (or can't) be applied to other fields (and more broadly, sets), I suggest you look at these two links: 
http://en.wikipedia.org/wiki/Total_order
http://en.wikipedia.org/wiki/Ordered_field
They help generalize the notions you are discussing and as I understand it, the notion of positive and negative with respect to $\mathbb{R}$ falls out of these more general concepts.
A: There's no reason that you couldn't label the complex plane this way, but it's not generally done.
The reason is that when you say the square root of minus one, which one do you mean?  There are two complex numbers who's square is minus one $i$ and $-i$.
So if you answered that you meant $i$ not minus $i$ I could still ask you which one of the two solutions you meant by $i$.  You have defined $i$ as the solution to $i^2=-1$.  
In fact it doesnt mater which one you choose, if you set $\hat\imath = -i$ and went through a complex number calculation replacing all the $i$'s with $\hat\imath$ you'd still the same result, just with $i$ replaced by $\hat\imath$ in the answer.
It's in fact impossible to distinguish between them, which is why it makes no sense to label one of them as positive, because we can't know which one you mean by positive.  For the real line positive and negative numbers have different properties and it makes sense to distinguish between them.  For the complex numbers there's no natural way of making sense of this.  
A: Short answer: No, you cannot regard it to be positive.
Explanation: You could only possibly regard $\mathrm i$ to be positive either if …


*

*… you viewed it to be an element of any ordered field, or

*… there was any intrinsic preference of $\mathrm i$ over $-\mathrm i$.
However it turns out that you can’t view it to be an element of any ordered field and there is no intrinsic preference of $\mathrm i$ over $-\mathrm i$. Hence, you cannot regard the imaginary unit to be positive.
The imaginary numbers $\mathrm i$ and $-\mathrm i$ are the roots of $x^2 + 1$. In any scenario, this should always hold and be the defining property. So now consider this:


*

*In an ordered field you require $1$ to be positive,² the positive numbers to be multiplicatively closed and addition to preserve order. If $\mathrm i$ or $-\mathrm i$ were positive, so would be $\mathrm i·\mathrm i = (-\mathrm i)·(-\mathrm i) = -1$, which is not, as $-1 < 0 \iff 0 < 1$.

*You could define $\mathrm i$ to be positive since it has a positive real coefficient in the $ℝ$-base $(1,\mathrm i)$. But how does this differ from $-\mathrm i$ having a positive real coefficient in the $ℝ$-base $(1,-\mathrm i)$? The fact is, the complex plain is as a field axisymmetrical about the real axis. This symmetry is given by the complex conjugation $ℂ → ℂ,~z ↦ \overline z$. Visually, to put one root above the real axis in the complex plane, and the other below, is a choice we have to make. And the only reason $\mathrm i$ is chosen to be above the real axis, while $-\mathrm i$ is below it and expressed in terms of $\mathrm i$, is because we first want to go up and then want to go down. So it’s natural to first consider the root above the real axis and call it $\mathrm i$, and then look at the other root and refer to it as $-\mathrm i$ (since then it really is $-\mathrm i$, meaning the negative of $\mathrm i$).
