Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

So I am about halfway through complex analysis (using Churchill amd Brown's book) right now. I began thinking some more about the nature and behavior of $i$ and ran into some confusion. I have seen the definition of $i$ in two different forms; $i = \sqrt{-1} $ and $i^2 = -1$. Now I know that these two statements are not equivalent, so I am confused as to which is the 'correct' definition. I see quite frequently that the first form is a common mistake, but then again Wolfram Math World says otherwise. So my questions are:

  1. What is the 'correct' definition of $i$ and why? Or are both definitions correct and you can view the first one as a principal branch?

  2. It seems that if we are treating $i$ as the number with the property $i^2 = -1$, it is implied that we are treating $i$ as a concept and not necessarily as a "quantity"?

  3. If we are indeed treating $i$ as a concept rather than a "quantity", how would things such as $i^i$ and other equations/expressions involving $i$ be viewed? How would such an equation have value if we treat $i$ like a concept?

I've checked around on the various imaginary number posts on this site, so please don't mark this as a duplicate. My questions are different than those that have already been asked.

share|improve this question
14  
To quote a professor of mine: "It is unfortunate that we call $i$ imaginary. What is a real number? A Dedekind cut or an equivalence class of Cauchy sequences. can you hold that in your hand?" The point, put a little less pithily is that these objects, while they may be defined in a way that seems very abstract and unlinked from reality, are in fact useful constructs with arguably as much reality as the reals. For example, the behavior of complex functions exerts tight control on the representation of real analytic functions as power series. –  JHance Aug 5 at 3:50
1  
Neither of those are correct definitions. The typical naive "construction" of the complex numbers is basically "set $i^2=-1$ and assume all other algebraic properties work as usual". That is not mathematics. It's too vague. What does "all other algebraic properties work as usual" mean, and how do you justify that that's "allowed"? It takes much more care to actually construct the complex numbers, see e.g. Chris Culter's answer for a start. –  Jack M Aug 6 at 1:03
2  
@JackM That is exactly what the complex numbers are, it takes theory to justify but not knowing the theory doesn't make the statement false. –  John Fernley Aug 6 at 21:32
    
@JohnFernley Sometimes vagueness is okay, because it's obvious how to fill in the gaps. Here though, it really isn't, which is why it took so long for complex numbers to be accepted. It's not clear that you can just make up a new number until you really sit down and think through what that means. –  Jack M Aug 6 at 21:35
1  
@JohnFernley The OP's confusion is in large part coming from not knowing that there's more to construction an algebraic structure than the simplistic approach that his teachers gave him. I know I spent a long time in the dark because I didn't realize that. –  Jack M Aug 6 at 21:50

8 Answers 8

The definition of complex number is given on page 1 of Churchill and Brown's book:

Complex numbers can be defined as ordered pairs $(x, y)$ of real numbers…

The definition of $i$ is given on page 2:

…let $i$ denote the pure imaginary number $(0,1)$…

So to answer your question, $i$ is not defined by the equation $i=\sqrt{-1}$, nor is it defined by the equation $i^2=-1$. Instead, it is defined as a particular ordered pair of real numbers, $i=(0,1)$. Then, given the definition of complex multiplication, one proves that $i^2=-1$.

share|improve this answer
1  
Of course, it could have used $i=(0,-1)$... but, this is a good answer. –  James S. Cook Aug 5 at 3:33
    
If $i$ is defined as an ordered pair, how do expressions such as $i^i$ make sense? Or is the only way to make sense of it by converting to polar coordinates? –  Exoyo Aug 5 at 3:36
11  
The definition of a complex number raised to a complex power is given in section 33, page 101 in the 8th edition: "When $z\neq0$ and the exponent $c$ is any complex number, the function $z^c$ is defined by means of the equation $z^c = e^{c\log z}$…". This definition, in turn, relies on the definitions of the complex exponential and logarithm functions, which are given in sections 29 and 30. –  Chris Culter Aug 5 at 3:41
    
(In general, these concepts aren't self-explanatory, so it's righteous to demand definitions for each one. The good news is that your textbook seems to do a pretty good job of providing those definitions in a reasonable order.) –  Chris Culter Aug 5 at 3:50
11  
$\Bbb C$ is $\Bbb R ^2$ just equiped with a special type of multiplication. –  Alizter Aug 5 at 4:22
  1. As you probably know, there are two solutions to $x^2+1=0$. We arbitrarily call one of them $i$ and the other $-i$, but this choice could have been made the other way (which is why complex conjugation is an automorphism of $\mathbb C$). So, both definitions are essentially correct, but of course, the first one is slightly less correct.

  2. What is the difference between a "concept" and a "quantity"? This question seems more philosophical than mathematical...

  3. Well, $i^i$ is usually defined as $e^{i\log i}$, where $e^{a+bi}=e^a(\cos b+i\sin b)$ and $\log(re^{i\theta})=\ln r+i(\theta+2n\pi)$, where $n$ is any integer (the logarithm is a multivalued function since the exponential function is not injective)

share|improve this answer
1  
"All quantities are concepts, but not all concepts are quantities." (Which is to say: imaginary numbers may seem less 'real' than the real numbers, but the real numbers themselves are only real in a 'conceptual' sense.) –  Semiclassical Aug 5 at 3:28
    
@Nishant: 1. If the choice between which is $i$ and which is $-i$ is arbitrary and could be made either way, what cause that one to be slightly less correct? 2.) Guess I should have specified a bit. What I'm trying to get at here by a "concept" is that if $i$ is entirely different from real numbers (in that $i$ isn't assigned the same kind of values that you would find in the real numbers) how can functions involving complex numbers give you real values? –  Exoyo Aug 5 at 3:33
    
Hmm, I guess what I meant by "slightly less correct" is that you have to assume a principal value. It's like if you wrote $\sqrt{4}=-2$. This is correct in the sense that $(-2)^2=4$, but is not quite perfect since usually, square roots of nonnegative numbers are taken to be nonnegative. Well, you know that $\mathbb C$ naturally contains $\mathbb R$ as a subfield, so, even though in general, equations involving imaginary numbers give complex numbers, they can still be real. –  Nishant Aug 5 at 4:01
    
Slightly rambling comment towards point (2)... usually quantities (or objects, or values, etc) is a term used to describe an element of an equivalence class. So you might call $0 \pmod 2$ a quantity while you would call $\text{even}$ a concept (and many other things), even though they are 2 different ways of describing the same functionality. –  DanielV Aug 6 at 11:27

it is easy to become the captives of our preconceptions. if you have honed a very efficient machine $\sqrt{}$ for extracting the square roots of positive real numbers you will rightly rub your head in puzzlement if asked to operate it on $-1$. but rather than focus on the genesis of $i$, celebrate the vistas it opens up, with its quantum leap through the sign barrier which constrains the arithmetic of real numbers. that one magical property $i^2=-1$ has far-reaching consequences.

take, for example, how $i$ interacts with another very basic mathematical reality - the circular functions and their remarkable combinatorial properties. it can be shown on the basis of geometry that: $$ \def\c{\cos}\def\s{\sin} \c(x+y)=\c(x)\c(y)-\s(x)\s(y) $$ and $$ \s(x+y)=\s(x)\c(y)+\s(y)\c(x) $$ using these identities it is easy to give an inductive proof that for any integer $n \ge 0$ we have: $$ (\c(x) + i \s(x))^n = \c(nx) + i \s(nx) $$ this is easily extended to rational values of $n$ and Lo! suddenly you have the infinite family of roots of unity at your disposal. want a number whose fifth power is $1$? well, you could always have $1$! but now there are also $e^{\frac{2 \pi i}{5}},e^{\frac{4 \pi i}{5}},e^{\frac{6 \pi i}{5}}$ and $e^{\frac{8 \pi i}{5}}$. of course in order to win the right to describe them as exponentials you have to plug $x=i\theta$ into the McLaurin expansion for $e^x$ to see how this gives rise to the identity: $$ e^{i\theta}=\cos\theta +i\sin\theta $$ in short, learn to work with $i$, enjoy it! you can of course explore various dusty definitions, but $i$ is such a special number that you cannot really assimilate it to any previous mode of mathematical thinking. look at the transformations it has wrought in physics over the last 150 years.

share|improve this answer
    
This is beautiful. –  JHance Aug 5 at 4:30
    
Did you mean to use $c()$ and $s()$ throughout instead of $\cos()$ and $\sin()$? –  MJD Aug 6 at 17:29
    
they were just abbreviations. i was going to fill them in later, but got distracted by other things. thx for reminding me, i will do the edits –  David Holden Aug 6 at 17:46

I would like to add a bit to Chris Culter's excellent answer. It is possible that there is more than one possible correct definition for $i$. The definition Chris Culter quoted from your book is one such. One might ask if the definitions you quoted are also correct. The answer is that neither one is.

The expression “$\sqrt x$” is defined to mean the unique non-negative real number $y$ such that $y^2 = x$. This is a good definition when $x$ is a non-negative real number, but it is completely meaningless for other values of $x$, because for $x = -1$, say, there is no non-negative real number $y$ such that $y^2 = -1$. So an attempt to define $i$ by saying $i=\sqrt{-1}$ is an immediate failure; there is no such thing. The $\sqrt{}$ operator simply does not do that.

Going the other way, one can't define $i$ by simply saying that $i^2=-1$; that is not a definition. It describes the properties that we want $i$ to have, but it does not describe any object that actually has those properties, and there may not actually be one. It's very easy to assemble a list of properties that is not possessed by any object. For example, “the largest invisible purple water buffalo in Dubuque, Iowa” is a description of that sort; you can describe its properties, but there is no such thing. Or again, a recent question here asked for the volume of a polyhedron whose faces are five equilateral triangles. But there is no such polyhedron: every polyhedron with only triangular faces has an even number of them.

We have an idea of a number $i$ with $i^2=-1$, but in order to show that this idea is coherent, we must actually construct something that behaves that way. One way to do this is the construction in your book:

  1. We take the set of pairs of real numbers $\langle a,b\rangle$
  2. Define addition and multiplication operations on them
  3. Show that these operations have the properties one usually expects, such as $a\cdot(b+c) = (a\cdot b ) + (a\cdot c)$.
  4. Show that the resulting structure contains a proper substructure, namely the set of elements of the form $\langle a, 0\rangle$, that behaves just like the real numbers, so we can identify this substructure as the real numbers, and the element $\langle -1, 0\rangle$ is effectively the same as the real number $-1$
  5. Show that this structure contains an element, namely $i=\langle 0, 1\rangle$ (or $i=\langle 0,-1\rangle$ if you prefer), which has the property that $i\cdot i = -1$.

This is not the only way to define $i$; there are many structures we could invent that would behave the way we want the complex numbers to behave. In advanced algebra courses, one uses the same strategy as in the previous paragraph, but one constructs the complex numbers as classes of polynomials instead of as ordered pairs:

  1. Divide the polynomials into certain classes.
  2. Define addition and multiplication of these classes
  3. Show that these operations have the usual properties
  4. Show that some of the classes behave the same way that the real numbers behave, and in particular that there is a class $[-1]$ that corresponds to the real number $-1$
  5. Show that there is a certain class $[x]$ which has the property that $[x]\cdot[x] = [-1]$.

So there is more than one possible definition, but neither one of the suggestions you gave is enough to do it. Vladimirm's answer elsewhere in this thread shows yet another way to define $i$, again following the same basic strategy.

share|improve this answer

You could think of complex numbers as special linear transformations, of the form $\begin{bmatrix} a & -b \\ b & a\end{bmatrix}$ where this matrix represents the complex number $z = a + ib$, a transformation between points in $\mathbb{R}^2$. This matrix represents a rotation and uniform scale transformation, where you first rotate your point and then scale it. This can perhaps be better observed from the exponential representation, if $z = re^\theta$ then the corepsonding transformation matrix would be

$$ \begin{bmatrix} a & -b \\ b & a\end{bmatrix} = \begin{bmatrix} r & 0 \\ 0 & r\end{bmatrix}\begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta)\end{bmatrix} = \begin{bmatrix} r\cos(\theta) & -r\sin(\theta) \\ r\sin(\theta) & r\cos(\theta)\end{bmatrix} $$

You multiply two complex numbers by multiplying their coresponding matrices, and multiplication of matrices of this form is comutative. Then $0,1$ and $i$ are defined

$$ 0 = \begin{bmatrix} 0 & 0 \\ 0 & 0\end{bmatrix} \\ 1 = \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} \\ i = \begin{bmatrix} 0 & -1 \\1 & 0\end{bmatrix} $$

As you can see when complex numbers are defined this way there is nothing mystical about the number $i$. It's just another transformation, a $90^{\circ}$ counterclockwise rotation.

EDIT

Also this way of thinking about complex numbers intuitively explains why the Cauchy-Riemann equations must hold for differentiable functions.

A complex function $f(z) = v(x,y) + iu(x,y)$ can be viewed as a simple $\mathbb{R}^2 \to \mathbb{R}^2$ transform (from "xy" coordinate sys. to "uv" ). So the best linear aproximation of the function $f$ near the point $z = \begin{bmatrix} x \\ y \end{bmatrix}$ (if the function is differentiable at z) is the Jacobian matrix $\begin{bmatrix} u_x & v_y \\ u_y & v_y \end{bmatrix}$ which is a generalization of the derivative in more than one dimension. $$f(z + \begin{bmatrix} \Delta x \\ \Delta y \end{bmatrix}) \approx f(z) + \begin{bmatrix} u_x & v_y \\ u_y & v_y \end{bmatrix} \begin{bmatrix} \Delta x \\ \Delta y \end{bmatrix}$$

But because $f$ is a complex function, differentiability at $z$ means

$$f( z + \Delta z ) = f(z) + f'(z)\Delta z +o(\Delta z)$$

Where $\Delta z = \Delta x + i\Delta y$. This means that the function is linear near $z$ ($\Delta z \to 0$), and therefore

$$f( z + \Delta z ) \approx f(z) + f'(z)\Delta z$$

where $f'(z)$ is a complex number which when represented by a linear transformation is actualy the Jacobian of this function. But because all complex numbers are special linear transformations of the form $\begin{bmatrix} a & -b \\ b & a\end{bmatrix}$ the Cauchy-Riemann equations follow, $u_x = v_y, v_x = -u_y$.

share|improve this answer
1  
One of your minus signs is in the wrong place. $\begin{bmatrix} a & -b \\ b & a \end{bmatrix}$ is correct, but this matches up with $\begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}$. –  Ian Aug 6 at 0:52
    
@Ian, yes you are right, I mistakenly wrote a clockwise rotation matrix, I fixed it, now its CCW. –  vladimirm Aug 6 at 5:24
    
Remark that said matrix is that associated to the linear map $\, z\mapsto (a+bi)z\,$ in the basis $\,[1,i],\,$ see this answer for the general idea (which works very widely). –  Bill Dubuque Aug 6 at 23:47

The remark "set $i^2=−1$ and assume all other algebraic properties work as usual" (quoted by @JackM in the comments) is indeed vague, but it's a good intuition to hold on to (the other answers here are more precise than what I'm about to say, but in their detail miss what I think is the key insight when teaching complex numbers).

You know about the real numbers: you can add, subtract, multiply and divide; you can therefore do squares and take the square roots of some numbers; and all this makes sense, or hangs together, in the sense that there are properties such as "if $a = 1/b$, then $a \cdot b = 1$", and so on.

You know about matrices: you can add, subtract and multiply these, and while there are properties such as "if $A = -B$ then $A+B=0$" for addition and subtraction, there's no way of defining division by a matrix in any way which respects the same sort of arithmetical rules that we have for Reals.

Now imagine there's a thing called $i$. We don't care (for the moment) what it 'is' nor how to represent it, but we decide that it has the property that $i \cdot i= -1$ (so it's clearly not a Real number). Can we do anything with this number?

The answer is yes, we can, and the elementary introduction to complex numbers consists of demonstrating that we can define addition, subtraction, multiplication and division of complex numbers in a way that results in these operations having the same properties that we find in the arithmetic of the Real numbers.

This is a very Big Deal, not least because it is (historically) one of the first suggestions that those rules of arithmetic are not specific to the Real numbers, but a possible object of study themselves (and further down this road lies group theory, and the study of rings, and modules, and all that jazz).

Just by the way: One of my undergraduate epiphanies when studying complex analysis – which is the attempt to do calculus with complex numbers rather than just reals – is the point when I realised what the difference was between complex analysis and (2D) vector analysis: only in complex analysis can you define the derivative using plain old $\lim_{{\mathrm d}x\to 0} [f(x+{\mathrm d}x)-f(x)]/{\mathrm d}x$, where ${\mathrm d}x$ is a complex number, because only in the complex plane can you divide by ${\mathrm d}x$ in this way; differentiation has to be defined in a more roundabout way on the 2D plane, precisely because there's no operation of 'dividing by a vector'. It's that fact (well, that plus the Cauchy-Riemann condition) that gives the complex plane the huge (amazing) amount of structure that complex analysis reveals it has.

share|improve this answer

Complex numbers can be constructed in different ways. One such way (the most common) is to consider ordered pairs of real numbers and define operations on them. In this construction you expand $\Bbb R$ by embedding it in $\Bbb R^2$ (a real number $r$ becomes the pair $(r, 0)$), you call $i$ the element $(0,i)$ and, by defining appropriate operations on this set of pairs, it happens that $i^2=(-1,0)\equiv -1$.

Another construction (less common, but more explanatory of the "nature" of complex numbers and the imaginary number $i$) is to consider $\Bbb C$ as the splitting field of $x^2+1$ over $\Bbb R$. You take the set of real polynomials $\Bbb R[x]$ and consider $$\left.\Bbb R[x]\right/(x^2+1),$$that is the quotient of the domain $\Bbb R[x]$ with the ideal generated by the polynomial $x^2+1$. Now, our $\Bbb C$ is this set, equipped with its operations (induced from the operations on $\Bbb R[x]$). Here we call $i$ the element $(x^2+1)+x$, because this is the splitting element, i.e., a root of $x^2 +1 $. In this construction (equivalent up to isomorphism to the other) you can directly see the property of the element $i$, the fact that $i^2=-1$. In fact, in this construction, you take the field $\Bbb R$ and expand it by adding the roots of the polynomial $x^2+1$. Being $\Bbb C$ the splitting field of $(x^2+1)$ over $\Bbb R$ means that $\Bbb C$ is the smallest extension of $\Bbb R$ that contains the roots of $x^2 + 1$. So $\Bbb C= \Bbb R(i)$. This is why all elements of $\Bbb C$ can be written as $a+ib$ for real $a$ and $b$.

share|improve this answer
1  
$i$ is the element $\langle x^2+1\rangle + x$, not $\langle x^2+1\rangle + 1$. –  MJD Aug 6 at 17:30
    
@MJD Oh, sorry. Thank you :) –  Matteo Aug 6 at 21:11

The "nature" of a thing is what you can use it for. In your complex analysis class, you will learn how they help you solve algebraic equations, simplify trigonometry, understand power series, do computations in integral and differential calculus, do geometry, and other things. In other courses, you may learn how to use them to do Fourier analysis, compute with quantum mechanics, motivate ideas in algebraic geometry, or any number of things.

All of those things is the "nature" of the complex numbers.

It may be interesting that the very origins of the subject were "Huh, if I pretend I can take square roots of negative numbers and push symbols around as if they were actually numbers, I can find the correct solutions to cubic and quartic equations." This, incidentally, is why they were originally called "imaginary" numbers. (and, unfortunately, the term has stuck even though we've known better for centuries)

I vaguely recall that, except for finance, the use of negative numbers in in calculations was also novel at this time, and their use in solving cubic equations was a big part of helping those go mainstream too. (Someone correct me if this is wrong)

The point of a definition is for exposition -- to give you enough initial bits of information so that you can start using the tools to see what they can do and how they fit together. The fine details about what is actually a definition and how to set up the foundations of a subject are not important at all to using and understanding complex analysis.

However, understanding such things is useful as examples of how to go about construct your own mathematical objects when you need to do such things in the future. And possibly useful as a test case if you're interested in studying formal logic.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.