The Imaginary Unit $i$ as an Operator Let us consider the real line, and a point $n$ on it which corresponds to the the real number $n$. Let us also imagine a line segment that represents this number $n$ by possessing the length equal to $n$ units to the right of $0$. If we multiply the number $n$ by the imaginary unit $i$, we have the number $ni$. Here we consider $i$ as an operator. Graphically, multiplying $n$ by $i$ corresponds to rotating the line segment through an angle of $90 ^{\circ}$. However, if we multiply the number $n$ by $-i$, then graphically it is thought of  as rotating the line segment through an angle of $-90^{\circ}$ or through an angle of $90^{\circ}$ in the clockwise direction.

Question: Should we think of $i$ and $-i$ as two different and separate operators in the context of graphical representation? 

I like to think of $i$ as the only operator which may act on both positive and negative reals. When it acts on negative reals, it corresponds to rotating the line that represents this number through an angle of $90^{\circ}$. For example:
Let $a$ be some real number and $a>0$, and let there be a line segment $l_1$ that represents this number $a$ and it has the length $a$ units to the right of $0$. If we multiply $a$ by $-i$ (or if the operator $-i$ acts on $a$), then graphically it means rotating the line segment $l_1$ through an angle of $-90^{\circ}$. However, if we think of it in this way that the operator $i$ acts on $-a$, then we may say that the line, let us say $l_2$, that represents $-a$ has been rotated $90^{\circ}$, and in both cases the extremities of our lines end up on the same locations. In this way we just have to think about $i$ and not $-i$. But maybe there is a flaw in my thinking.

 A: If you think of them as functions $\mathbb{C} \rightarrow \mathbb{C}$, then notice that $-i=i^{-1}$, i.e, they are inverses of each other.
The only element of the field $\mathbb{C}$ identified with It's inverse is $1$ (Which is unique)! If that's not enough you can tell they are not the same function, exactly by looking at their values in $\mathbb{R}^2$ space with basis $\{1,i\}$.  
Note: 
Nothing is really rotating anywhere. You cannot really visualize the field $\mathbb{C}$ like you can(?) with $\mathbb{R}$. What you can do is build a vector space (For example the one above; The one you mistakenly identify with  $\mathbb{C}$) which is isomorphic to $\mathbb{C}$, and only then the rotations may exist, and the direction is dependent on the basis you choose. 
A: There are exactly three linear associative algebras of order two:
I J K L
J I L K
K L I J
L K J I

I J K L
J K L I
K L I J
L I J K

I J K L
J 0 L 0
K L I J
L 0 J 0

In the second of these tables, J is the imaginary unit +i and L is -i. It is only a historical accident that +/- are used, with the + typically being supressed. The operators +i and -i are two distinct elements of the cyclic group of order four.  The first table is the Klein viergruppe. In it, J is +1 and L is -1. In the third table, J is the differential operator d and J is the operator -d. I is the identity element in all three tables. K is the inverse identity element in all three tables.
