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We all know that $i$ doesn't have any sign: it is neither positive nor negative. Then how can people use $-i$ for anything?

Also, we define $i$ a number such that $i^2 = -1$. But it can also be seen that $(-i)^2 = -1$. Then in what way is $-i$ different from $i$? If it isn't, then why do we even have something like a $-i$?

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You are correct, $i$ doesn't have a sign, which is why we use the "-" sign to negate it. –  Ataraxia May 19 '13 at 14:33
    
$ni$ is for the vertical coordinate, where $n\geq 0$, if $n<0$ then $-ni.$ Look these links: $i$ and $-i$. –  Vladimir Putin May 19 '13 at 14:35
    
I've asked a related question here. I think you'll find some of the answers interesting. –  user1337 May 19 '13 at 14:49
    
@Panda The answer by Thomas is very nice. –  Parth Kohli May 19 '13 at 14:51
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@ADP, embodying "trust"/experience into the system, so that someone can't make a lot of accounts and go on a downvote frenzy, i.e. they have to prove themselves by asking or answering questions. –  dbaupp May 20 '13 at 7:05

16 Answers 16

up vote 77 down vote accepted

The vector $(1,-2)$ is also neither positive nor negative, yet there is value in having the operation $-(1,-2)=(-1,2)$.

Note: there is no natural meaning for "positive" for vectors, this is just an analogy.

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You must take care, even though $2^2 = (-2)^2 = 4$ this doesn't mean that $2 = -2$, indeed the problem here (thinking about reals for a while), is that if you consider negative numbers, then the function $f: \mathbb{R}\to\mathbb{R}$ given by $f(x) = x^2$ is not injective.

Since you have created $i$ to be $i=\sqrt{-1}$, then you are obviously allowed to create a new number $-i = -\sqrt{-1}$. Also, if you think on the function $f: \mathbb{C} \to \mathbb{R}^2$ given by $f(a+bi) = (a,b)$, this function identifies $\mathbb{C}$ with $\mathbb{R}^2$, and in this sense, a complex number is a point in the plane. Note that we have $f(i) = (0,1)$ and that $f(-i) = (0,-1)$, so there are two things to note: every number that can be written $ai$ for some $a \in \mathbb{R}$ can be identified with a point in the $y$ axis, and introducing $-i$ we have not just the upper $y$ axis, but the complete one.

EDIT: Because of the comments I thought it would be good to add this edit. Above I've provided a geometrical view of what happens when we introduce $-i$, now I'm just going to say a little about the algebraic point of view. Since every complex number $z \in \mathbb{C}$ is $z=a+ib$ for some $a,b \in \mathbb{R}$ if we do not allow things like $-i$ we would have some issues. For instance, consider quadratic equations with complex conjugate roots: one of them would be a well defined element of $\mathbb{C}$, but the other one wouldn't! Also, it wouldn't be possible to present for each $z$ some $-z$ with the nice property $z + (-z) = 0$ and it turns out that $\mathbb{C}$ wouldn't be a field. Also, if you use the standard operations of sum between complexes and you have $z = a+ib$ and you want the additive inverse $-z$ you can prove rather easily that you must have $-z = (-a) + i(-b)$, so that $-i$ would just really be $i(-1)$, so in practice, the element $-i$ (and any other negative complex in general) appears naturally when we define the complexes as we do and when we introduce the operations as we do.

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I don't think this answers the question. We can represent complex numbers as a pair of real numbers, but that doesn't mean that the $\mathbb{R}^2$ plane has the structure of the complex numbers. It could be the case that the complex numbers 'loop round' and $a + bi = a + (b+c)i$ for some real number $c$ for example. We can represent the rational numbers as a pair of integers, but we don't need to include both $\frac{1}{1}$ and $\frac{-1}{-1}$ to 'complete the plane'. –  jwg May 20 '13 at 7:54
    
Rational numbers are not pairs of integers: you have to mod out by some equivalence relations. The plane $\Bbb R^2$ on the other hand can be identified with $\Bbb C$ as soon as you define properly a multiplication. Claiming that multiplying any $v\in\Bbb R^2$ by $(0,1)$ amounts to rotating $v$ counterclockwise by $\pi/2$ and extending by $\Bbb R$-linearity does the trick. –  Andrea Mori May 20 '13 at 10:54
    
@AndreaMori I know that the case for rational numbers and $\mathbb{Z}^2$ is not the same as for $\mathbb{C}$ and $\mathbb{R}^2$. My point is that the answer has to show this, rather than just say 'you can write a complex number using two reals, therefore you can negate one of the reals to get another complex number'. Your comment is a fine example of how to do this. –  jwg May 20 '13 at 11:08

The imaginary number $i$ was discovered/invented/revealed as a solution the equation $$ x^2+1=0\tag{$\ast$} $$ That equation, and that equation alone, is the link between $i$ and the reals. However, there is another solution to $(\ast)$: $-i=-1\times i$ also satisfies $(\ast)$. Basically, this is because $(-1)^2=1$.

Since the link between $i$ and the reals is the same equation that is satisfied by $-i$, one can easily wonder, "which is which?" When we choose a solution to $(\ast)$, do we get $i$ or $-i$? In a sense, it doesn't matter; both satisfy $(\ast)$ and $i^2=-1$ is the property we use to do math in $\mathbb{C}$.

On the other hand, complex conjugation, swapping $i\leftrightarrow-i$, is an important isomorphism of $\mathbb{C}$. For instance, any polynomial with real coefficients that has $x+iy$ as a root, must also have $x-iy$ as a root. There are many ways to show this, but one of the most basic is by swapping $i\leftrightarrow-i$. This isomorphism does not affect the real coefficients of the polynomial, but it swaps $x+iy\leftrightarrow x-iy$.

In the end, we call one root of $(\ast)$, $i$, and the other, $-i$; it really doesn't matter which. They both satisfy $(\ast)$ and that is what is important.

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I quickly scanned all the answers and didn't see anyone mention rotations, so I thought I would -- the simplest answer to your question is that $-i$ just represents a rotation of $\frac{\pi}{2}$ clockwise, whereas $i$ represents a rotation of $\frac{\pi}{2}$ counterclockwise. In other words, $i$ and $-i$ are inverses in the group $SO(2)\cong\left\{e^{i\theta}\,|\,\theta\in[0,2\pi)\right\}$ of rotations of the plane. From this perspective, $i$ and $-i$ are about as emphatically "different" as anyone could imagine.

EDIT: I thought I'd add that yet another way of saying what I've already said is that complex conjugation is just a way of swapping the orientation of the plane. (In particular, it is an orientation-reversing isometry.) The seemingly "sign-like" difference between $i$ and $-i$ is really just a difference in the choice of orientation.

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Three cheers for the SO(2) group. And you might add that the '^2' operation is really just a mapping from the complex plane using the vector norm to the real line and there must be an infinite number of such cases. –  DWin May 19 '13 at 19:09
    
I intended to add this answer if I didn't see it. Since I do, I'm giving you your tenth vote with my compliments. –  Ryan Reich Jun 2 '13 at 15:19
    
+1. Big fan of this answer!! I categorically promote the use of $i$ as a rotation operator! –  Eleven-Eleven Nov 21 '13 at 23:31

After defining a new symbol $i$ such that its square is the negative number $-1$, you can do algebra with it. So, you have an additive operation on the set of symbols like $a+bi$, with $a,b\in \mathbb{R}$. But, if you remember, any element must have an inverse additive and you denote it by $-i$.

Edit:

ps: I suggest you to read about the construction of complex numbers field. How is it defined?

ps1: when you talk about vectors $(a,b)$ (nothing more than ordered pairs) you are identifying $\mathbb{C}$ with $\mathbb{R}^2$.

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How do you know that the additive inverse of $i$ is not $i$ itself, or some other number such as $4 + 2i$? –  jwg May 20 '13 at 7:59
    
@jwg: You define $\Bbb C$ to be a finite field extension of $\Bbb R$ of degree $2$, hence by "forgetting" (part of) the multiplicative structure you get a real vector space isomorphic to $\Bbb R^2$. A basis of this space is given by $1$ and one of the roots of $X^2+1$, call it $i$. Then the additive inverse of $i$, which we denote by $-i$, is actually $(-1)*i$, and it is another (hence the other) root of $X^2+1$. –  A.P. Jun 16 '13 at 12:37
    
@jwg: To explicitly answer your question: It cannot be $i$ itself, since otherwise $i$ would be a torsion element of the abelian group $\Bbb R^2$, which is torsion free. To prove that it actually has to be $(-1)i$, first note (again) that this is the only root of $X^2+1$ different from $i$. Then observe that, since $i+(-i)=0$ by definition, we have $0=i0=i(i+(-i))=i^2+i(-i)$, so $i(-i)=1$. Moreover, $0=(i+(-i))^2=i^2+2i(-i)+(-i)^2=-1+2+(-i)^2$, hence $(-i)^2=-1$ and $-i$ is a root of $X^2+1$ different from $i$, thus equal to $(-1)*i$. –  A.P. Jun 16 '13 at 13:02
    
@A.P. I understand this. My question was rhetorical, pointing out that the answer is very incomplete. –  jwg Jun 26 '13 at 18:17
    
@jwg: I see, sorry about that. Anyway, I think I'll leave my comments... for the sake of completeness. –  A.P. Jun 27 '13 at 7:47

You are quite right! It is more or less an arbitrary convention that $i$ is represented by the point $(0,1)$ in the plane, and then necessarily $-i$ has to be represented by $(0,-1)$. It could have been the other way around! Regardless of the convention you choose, the PAIR $i, -i$ will always be the same as $(0,\pm 1)$.

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If you take moving along the usual number line as moving forward of backward, depending on the sign, and then extend this analogy to the complex plane, you could think of $\pm i$ as going either left or right, depending on your choice of convention. –  J. M. May 19 '13 at 16:34

In addition to another answer, I know a theorem in which we can extend the order of $\mathbb R$ to $\mathbb C$ such that two following are satisfied:

$P_1$: For all $a\in\mathbb R$ we have only one of $a>0,~a=0,-a>0$.

$P_2$: $a>0,b>0 \to a+b>0$.

In fact, if we let $\alpha=a+ib$ we can have $\alpha>0$ if and only if $a>0$ or $a=0,~b>0$. So for all $\alpha=a+ib$ we have just one of the following facts: $$a>0,a=0,-a>0$$ or $$a>0\to \alpha>0$$ or $$-a>0\to-\alpha>0$$ or if $a=0$ then $$b>0\to \alpha>0\\\\ b=0\to \alpha=0\\\\ -b>0\to-\alpha>0$$. Therefore, for any arbitrary $\alpha=a+ib\in\mathbb C$ we have just one of $$\alpha>0,~ \alpha=0,~-\alpha>0$$ I think this make a bit clear what we need here.

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This is a partial order and it doesn't respect the field operations, so it's not very useful. –  Ben Millwood May 20 '13 at 11:26
    
+1 Babak, dear friend! I saw your user page on Eng.Language.se! I'm "registered" there, but haven't posted anything yet! How are you? –  amWhy May 21 '13 at 1:25
    
@amWhy: Great! thanks God. My question there probably make me ashamed while you read them. Sorry for asking those ridiculous ones. I love English Writing Amy and up to now, have tried to be better in it. Recently, I took the IBT test and was evaluated Good in General Writing. –  B. S. May 21 '13 at 1:30
    
Oh no, no need to be ashamed about your questions! I really think there are many non-native English speakers who know English better than lots of Americans do! Good for you: $\large \bf GOOD$ in General Writing!! You don't give yourself enough credit. You do quite well with language! –  amWhy May 21 '13 at 1:32
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Thanks, Babak. You are so kind, and you are a poet, as well as a mathematician! –  amWhy May 21 '13 at 1:41

The signs "+" and "-" do not exactly correspond to positive and negative! The negative of a number (or any element) in a group is just the inverse of that element with respect to the group operation. We created $i$ as a solution to $x^{2} + 1 =0$ and extended the class of real numbers. However, this extended class had to be made into a group. So, $-i$ is just the additive inverse of $i$ in $\mathbb{C}$ with $0$ as the identity element.

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Complex numbers $\mathbb{C}$ is a set of ordered pairs $(a,b)$ where $a \in \mathbb{R}$ and $b \in \mathbb{R}$ with two arithmetic operations complex addition and complex multiplication defined by \begin{eqnarray} (a_1,b_1) + (a_2,b_2) & = & (a_1 + a_2,b_1 + b_2) \\ (a_1,b_1) \ \cdot \ (a_2,b_2) & = & (a_1 \cdot a_2 - b_1 \cdot b_2, a_1 \cdot b_2 + b_1 \cdot a_2) \ . \end{eqnarray} Here the new operations $+$ and $\cdot$ on the left hand side are defined using known operations $+$ and $\cdot$ on the right hand side defined on $\mathbb{R}$. Note that substraction is defined by the unique opposite element by \begin{equation} a-b = a+(-b) \ . \end{equation} It can be shown that $\mathbb{C}$ is a field, that is a set with two opeations addition and multiplication, write $(\mathbb{C}, +, \cdot)$, that satisfies the field axioms. Consider now the subset $F = \{(a,0)|a \in \mathbb{R}\} \subset \mathbb{C}$. It can be again shown that $F$ is a subfield of $\mathbb{C}$. In addition $F$ is isomorphic to $\mathbb{R}$, written $F \cong \mathbb{R}$ with the isomorphism mapping $\Phi$ defined by $\Phi(z) = \Re{(z)}$, where \begin{equation} \Re{((a,b))} = a \ . \end{equation} A field isomorphism mapping $\Phi$ is always a bijection and preserves all algebraic strucure, that is \begin{eqnarray} \Phi(z_1+z_2) & = & \Phi(z_1) + \Phi(z_2) \\ \Phi(z_1 \cdot z_2) & = & \Phi(z_1) \cdot \Phi(z_2) \end{eqnarray} for every $z_1$ and $z_2 \in F$. For the $\Phi$ defined above it can be shown using the definitions of complex addition and complex multiplication. It can be seen that $\Phi$ maps the unique neutral elements $0 = (0,0)$ and $1 = (1,0)$ of complex addition and multiplication correctly to the corresponding real elements.

The complex number $i$ is defined by \begin{equation} i = (0,1) \ . \end{equation} This is just a definition and follows Gauss' lead to assimilate $x+iy$ to the ordered pair $(x,y)$. We could of course define $i$ otherwise and still be able to calculate using the new definition. However, the current definition of $i$ is useful. It satisfies $i^2 = -1$, that makes multiplications of the numbers of the form $x+iy$ straightforward. The conventional definition of $i$ is also beautiful. It contains only neutral elements of the real field. There is also another alternative $(0,-1)$ for $i$, but it assimilates $x+iy$ to the ordered pair $(x,-y)$, that contains one additional calculation of opposite element. Hence we keep the conventional definition and calculate \begin{eqnarray} i^2 & = & i \cdot i = (0,1) \cdot (0,1) = (0 \cdot 0 - 1 \cdot 1,0 \cdot 1 + 1 \cdot 0) \\ & = & (-1,0) = (-1,-0) = -(1,0) = -1. \end{eqnarray} The number $1 = (1,0) \in \mathbb{C}$ is the unique neutral element of complex multiplication. Note that in the mapping $\Phi$ its image is $1 \in \mathbb{R}$, that is the neutral element of real multiplication. Note that there is an element $z' = (-a,-b)$ for every $z = (a,b) \in \mathbb{C}$ that satisfies the opposite element axiom: $\exists z'$ s.t. $z+z'=0$, where $0 \in \mathbb{C}$ is a neutral element of complex addition. The complex number $-i$ is now \begin{equation} -i = -(0,1) = (-0,-1) = (0,-1) \ . \end{equation}

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You can also calculate –  Juha-Matti Vihtanen Jun 2 '13 at 22:34
    
$$\begin{eqnarray} \sqrt{-1} & = & (-1)^\frac{1}{2} = (e^{i\pi+2\pi i k})^\frac{1}{2} = e^{(i\pi + 2\pi i k)\frac{1}{2}} \\ & = & e^{i \pi k + i \pi \frac{1}{2}} = e^{i\pi k} e^{i\pi \frac{1}{2}} = (e^{i\pi})^k (e^{i\pi})^\frac{1}{2} \\ & = & (-1)^k (-1)^\frac{1}{2} = (-1)^k \sqrt{-1} \end{eqnarray}.$$ Hence the complex square root should not be unique. –  Juha-Matti Vihtanen Jun 2 '13 at 22:54

Most of the answers already take the whole structure of the complex numbers as given. Therefore let me go to a more elementary level, where we don't yet assume anything about the complex numbers except that they contain the real numbers, and at least one additional number $i$ which solves the equation $x^2+1=0$.

Now consider the equation $x+i=0$. I think you would agree that a set of numbers where we could not solve this equation would not be terribly useful. By definition, we denote the solution of $x+a=0$ with $-a$, so the solution here is $-i$. However that's up to now only a notation; for example, the same way we get $-0$, which we know is the same as $0$. Therefore the question is: What is $-i$?

First, we note that $-i$ cannot be a real number, because otherwise $i$ would be its negative (also a real number), and we already know $i$ is not a real number. So there remain two possibilities: Either $-i=i$, or $-i$ is a different non-real number than $i$.

We also want that the distributive law continues to hold.

Now assume $-i=i$, that is, $i+i=0$. Then we have $-2 = (-1) + (-1) = i^2 + i^2 = i(i+i) = i\cdot 0 = 0$. Now that's obviously wrong, therefore we cannot have $-i=i$. Thus $-i$ is a distinct number from $i$.

Edit: On A.P.'s request, I also add the proof that $-i=(-1)i$.

To prove that $-i=(-1)i$ means to prove that $i+(-1)i=0$, because that's how $-i$ is defined.

We obviously have, using the fact that $x=1x$ for all $x$: $$i +(-1)i = 1i + (-1)i =(1+(-1))i = 0i = 0$$

Note that you can prove essentially the same way that $i(-1) = -i$ (in case you take into account that multiplication might not turn out to be commutative, which actually would not be that unusual, although for complex numbers multiplication actually is commutative)- Therefore you also get $$(-i)^2+1 = i(-1)(-1)i+1 = i^2+1 = 0$$

Now for the proof of $(-1)i=-i$ as well as above when proving $-i\ne i$, I've used $0i=0$, which seems obvious, but to be sure, let's prove this as well. We have $$0i = (0+0)i = 0i+0i$$ Now we add $-(0i)$ to both sides of the equation (remember, the demand that for each $x$ we have a $-x$ which fulfils $x+(-x)=0$ was how we arrived at $-i$ at the first place), to get for the left hand side $0i + (-(0i)) = 0$, and for the right hand side $0i + 0i + (-(0i)) = 0i + 0 = 0i$ (note that in this, I've made another assumption, namely that addition is still associative even if expressions with $i$ are involved; but that's also a natural assumption which you'd not give up unless necessary).

Note that ultimately, all I needed was

  • the rules for real numbers continue to hold unchanged
  • the fact that $i$ solves the equation $x^2+1=0$
  • the associativity of addition
  • $x+0=0$ for all $x$
  • the existence of an additive inverse $-x$ for any number $x$
  • the associativity of multiplication
  • $1x=x$ for all $x$
  • the distributive law

The first two are given by the problem, while the rest are very fundamental assumptions which you wouldn't give up lightly.

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Proof by contradiction -- nice. –  Lambda Fairy Jun 9 '13 at 3:51
    
You proved that $-i$ is different from $i$, which is nice, but not that it is equal to $(-1)*i$ (or, equivalently, that it is the other root of $X^2+1$. –  A.P. Jun 16 '13 at 12:56
    
OK, I've added the proof, although it wouldn't have been necessary for answering the question. I also added a proof that $0i=0$ which you didn't request, but which might be seen as an omission anyway. I also listed explicitly all the needed assumptions. –  celtschk Jun 16 '13 at 19:45

The imaginary unit $i$ is just a unit, like $m$ (meters), $s$ (seconds), or $\hat{u}$ (a unit vector). Any unit can be multiplied by a scalar, so $-i$ is jus the result of multiplying the imaginary unit by $-1$, no different than $4i$, $6i$, or $2.5i$.

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I disagree. $i$ is not a unit like meters, and it is not like those units either. Dimensional analysis says that $i$ must be a scalar quantity. –  MJD May 19 '13 at 15:05
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@MarkHurd doesn't that contradict the above claim that $i$ has to be a scalar quantity? –  Ataraxia May 19 '13 at 15:19
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Dimensionally, or perhaps geometrically, $i$ is in fact a unit vector in the complex plane. It is multiplied by -1 exactly how a vector would be. –  Kaz May 19 '13 at 15:50
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The name "imaginary unit" for i, which is a very common one, suggests already that there is some truth in @ZettaSuro 's interpretation. Of course its not the only possible interpretation... –  alexlo May 19 '13 at 16:12
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This way of looking at $i$ only works if you want to add complex numbers and multiply them by reals. It breaks down if you consider that they can be multiplied by each other and that $i^2 = -1$. –  jwg May 20 '13 at 8:05

You need to impose an orientation on the plane. Since we mathematicans like counterclockwise to be the positive direction, we think of $i$ as being "one-up" from zero. You could equally well work with the opposite. It's a "tom(ay)to vs. tom(ah)to" situation. I refer all readers to this (humorously)....

http://www.youtube.com/watch?v=zZ3fjQa5Hls

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If we say that $i^2$ = -1, then $(-i)^2 = (-1)^2(i)^2 = 1(-1) = -1$ as well. i can't be zero, since $0^2 = 0$, not -1. So, $i$ and $-i$ are distinct. You can't have one without the other, and they are defined identically. However, they are in effect defined as a pair: There are two numbers which squared give you -1; we call one of them $i$, and that makes the other $-i$.

Make your choice about which is which and stick to it for your equation or expression, and all will be well. If you like, you can change your mind about which is which every alternate Thursday (except on leap day), and as long as you're consistent within an equation that's just fine. Just don't tell anyone you did it; they'll only be confused.

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Comparing $\mathbb{C}$ with vectors in $\mathbb{R}^2$ the $i$ can be viewed as describing a direction in 2D space (vertical, if you want), and it gets its magnitude and orientation (i.e. sign) from the coordinate that is a real number. It turns out that defining $i$ as the $\sqrt{-1}$ (which doesn't exist in $\mathbb{R}$), we get a field extension of $\mathbb{R}$ that is algebraically very useful and has the "geometrical properties" of the usual $\mathbb{R}^2$ vectorspace.

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You seem to be explaining what the complex numbers are, rather than answering the question? –  jwg May 20 '13 at 11:11
    
@jwg Why are you saying this? Read it again, think about it, and you will see that's not true. The interpretation for $i$ that I am offering is "...$i$ can be viewed as describing a direction in 2D space...",i.e. a unit vector, and we have $i$ and $-i$ to be able to specify its orientation. Furthermore requiring that $i=\sqrt{-1}$ is helpful to be able to do algebra with this "new plane". Of course this also describes something about the complex numbers. What else did you expect? –  alexlo May 20 '13 at 15:21
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The question was about $-i$. You seem to be explaining $i$ but not saying anything about $-i$. –  jwg May 20 '13 at 21:49

The complex number $i$ is the representation of an angle of $\pi/2$ between the real axis and imaginary axis and $-i$ mean that the angle is $-\pi/2$. So there is a huge difference between $i$ and $-i$ as they have $\pi$ radians phase difference.

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Before you spend too much effort mulling over the "meaning" of $-i$, you should consider the fact that if $x = -2$, then the "meaning" of $-x$ is something that has the value $2$.

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