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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.

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Samama Fahim
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  • What do you mean by operator, and what do you mean by "should we think"? – Jack M Feb 27 '14 at 17:02
  • By operator I simply mean what is meant by the addition or subtraction operators $+$ or $-$. $i$ graphically tells us to change the position of $n$ through $90^{\circ}$ with respect to its previous position in the same way that addition or subtraction operations tell us to add or subtract something to or from some number. "Should we think" mean what I should take them as? Are they two different operators or is there only $i$? – Samama Fahim Feb 27 '14 at 17:08
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    Of course they are different - they do different things. One rotates clockwise, the other rotates counterclockwise. Is that what you meant? – Jack M Feb 27 '14 at 17:09
  • So, can we not say that the operator $i$ may graphically rotate a number in either clockwise or anticlockwise direction because $-i$ is supposed to rotate the number through clockwise direction? – Samama Fahim Feb 27 '14 at 17:14
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    $i$ rotates things counter clockwise, $-i$ rotates clockwise. That isn't unusual because $i\neq-i$. Your confusion seems to be that you think $i$ simultaneously does both directions of rotation, but it doesn't - $i$ does one direction, and $-i$ does the other. – Jack M Feb 27 '14 at 18:17

2 Answers2

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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.

user76568
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  • In the book I'm reading it is written that the operator $i$ (if we think of it as an operator) may graphically rotate a number in either direction (clockwise or counterclockwise). How can $i$ rotate a number in both the directions? I think that you need $-i$ to rotate it in the clockwise direction. Is this a misstatement in the book or may be I'm misinterpreting the statement? (This is how my question arises). – Samama Fahim Feb 27 '14 at 17:23
  • If you switch the order of the basis, It'll rotate in the other direction :-). – user76568 Feb 27 '14 at 17:25
  • I do not understand the concept of basis in the context you refer to it. Can you refer me to a link to some definition of basis? – Samama Fahim Feb 27 '14 at 17:54
  • http://en.wikipedia.org/wiki/Complex_plane – user76568 Feb 27 '14 at 18:22
  • http://en.wikipedia.org/wiki/Basis_(linear_algebra) – user76568 Feb 27 '14 at 18:24
  • Do you mean that if we change the convention that "angles in counterclockwise direction are positive and angles in clockwise direction are negative" to the convention which is the other way, then $i$ may also suggest rotation through clockwise direction? – Samama Fahim Feb 27 '14 at 19:49
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    That's not really what I meant, but is isomorphic to it ;-) – user76568 Feb 27 '14 at 19:52
  • $1$ is not the only element of $\mathbb{C}$ that is equal to its inverse. $-1$ is also equal to its inverse. But I don’t see how the topic of inverses is useful or necessary for the proof that $i\neq{-i}$. Also, why do you say that $\mathbb{R}$ can be visualized but $\mathbb{C}$ cannot be visualized? A plane can be visualized as easily as a line can. – Radial Arm Saw Jul 07 '20 at 16:05
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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.

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