Would a negative imaginary number times an imaginary number be positive? I know that $\sqrt{-1} = i$
and that $ i^2 = -1$, and I'm sure $i^3 = -i$. So, would $i^4 = +1$?
I'm sort of confused about this. Can you smart guys explain?
 A: Yes that's correct $$i^4 = i^2\cdot i^2=(-1)\cdot (-1) = 1$$
A: Yes. You can see it yourself by 
$$
i^{4} = i^{3} \cdot i = -i \cdot i = - (\sqrt{-1})^{2} = -(-1) = 1. 
$$
A: Your powers of $i$ are correct.  $i$ is a fourth root of 1 together with $-1$, $-i$ and, of course, $1$ itself.  
It is not usual to talk of positive and negative with complex numbers.  It is not possible to define $>$ and $<$ for complex numbers and retain their familiar properties e.g. negative $\times$ negative $=$ positive.
So, $i$ is neither positive nor negative.  They are properties of real numbers and $i$ is not a real number.  
A: Indeed $i^4=1$. Since $$i^4=i^{2+2}=i^2\cdot i^2=(-1)(-1)=1$$
Your observation includes even more though:
$i^0=1;i^1=i;i^2=-1;i^3=-i;i^4=1$
and from here on the pattern repeats. The powers of $i$ repeat with a periodicity of $4$, so We can consider them modulo $4$: 
$i^n=i^{n\mod4}$
To prove this, write $n=4k+r$ with $r\in{\{0;1;2;3\}}$; then $i^n=i^{4k+r}=i^{4k}\cdot i^r=(i^4)^k\cdot i^r=1^k\cdot i^r=i^r$ and clearly $r\equiv n\mod4$.
And observing $4\equiv0\mod4$, $i^4=i^0=1$ is making sense anew.
