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What is a complex number?

  1. Oct 4, 2013 #1
    I have for a while been trying to really understand the notion of a complex number and the construction of the complex number system. My knowledge of mathematics so far is very limited and spans mostly linear algebra (no pun intended), discrete mathematics (where I have yet to see complex numbers be used) and real analysis.

    Historically the complex numbers have been treated as a mysterious object, due to there being polynomials over the real field having no real roots. Thus mathematicians were inclined to define roots of these polynomials as if they existed on some space outside the real field. But many standard algebraic operations such as exponentiation and logarithms do not work the same way they do on complex numbers as they do with reals.

    In one sense, the complex number is an algebraic object that allows us to form an algebraically closed field, that is every polynomial over the complex field must split over the complex field. This has important applications in linear algebra, such as guaranteeing the existence of a Jordan normal form for matrices over the complex field.

    Is it is a theorem of algebra that the complex field cannot be ordered? Is it a theorem of algebra that the complex field is the smallest algebraically closed field containing the reals? What fields/vector spaces is the complex field isomorphic to?

    And how are the complex numbers really axiomatized? As algebraic objects, or as geometric objects?

    The reason I ask these questions is that complex numbers are immensely useful in the field of signal processing, but because the field is algebraically closed, but more because trigonometric functions can be written in the form of complex exponentials and that computations on complex exponentials are much faster. Most of the use of complex numbers in signal processing is from the Euler identity combined with the Fourier transform.

    So far, most textbooks I have seen define complex numbers as a two dimensional vector space over the reals with an embedded product such that (0,1)*(0,1) = (-1,0). This geometric interpretation is handy, as it reduces the mystery of the complex field to a vector space isomorphic to ##ℝ^{2}## with some additional structure, but if this is the direction we take, how can we develop the link between such a geometric interpretation and the inevitable powerful algebraic properties of the complex field and reconcile the two?

    And then there's Euler's identity, and calculus on complex-valued functions. How does that work?

    A recommendation on a textbook, but more importantly, a brief explanation (assuming one exists) in your own words that may resolve my questions is deeply appreciated!

  2. jcsd
  3. Oct 4, 2013 #2


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    Yes, the most rigorous way to define the complex numbers does NOT start with "i". Instead define the comples numbers to be ordered pairs of real numbers, (x, y), with addition defined "coordinate-wise", (a, b)+ (c, d)= (a+ c, b+ d), multiplication defined by (a, b)(c, d)= (ac- bd, ad+ bc). Notice that does NOT directly mention an "i" such that "[itex]i^2= -1[/itex]". We don't need to simply "assert" such a thing.

    It is easy to show that the usual field properties, commutativity, existance of additive and multiplicative identities, additive inverses, and multiplicative inverses, etc. still hold. The additive identity is (0, 0), the multiplicative identity is (1, 0), the additive inverse of (a, b) is (-a, -b), and the multiplicative inverse is [itex](a/(a^2+ b^2), -b/(a^2+ b^2)[/itex] which exist as long as [itex]a^2+ b^2\ne 0[/itex]- that is, as long as (a, b) is not (0, 0), the additive identity. We can think of the real numbers as a subfield of the complex numbers by associating the real number x with the complex number (x, 0).

    The critical point is that [tex](0, 1)(0, 1)= (0(0)- 1(1), 0(1)+ 1(0))= (-1, 0) which we associate, as above, with -1. That is, defining "i" to be the complex number (0, 1), we have [itex]i^2= -1[/itex] as a result of these definitions.

    As for "Euler's identity", do you know the Taylor's series for [itex]e^x[/itex], [itex]sin(x)[/itex], and [itex]cos(x)[/itex]? They are
    [tex]\sum_{n=0}^\infty \frac{1}{n!}x^n[/tex]
    [tex]\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}[/tex]
    [tex]\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}[/tex]

    Notice that, except for the "[itex](-1)^n[/itex]", the sine and cosine are the even and odd power parts of [itex]e^x[/itex]. Now, [itex]i^0= 1[/itex], [itex]i^2= -1[/itex], [itex]i^3= i^2(i) -1(i)= -i[/itex], [itex]i^4= (i^2)(i^2)= (-1)(-1)= 1 and then every thing repeats: [itex]i^5= i[/itex], [itex]i^6= -1[/itex], [itex]i^7= -i[/itex], [itex]i^8= 1[/itex], etc. If we replace the "x" in the series formula for [itex]e^x[/itex] with "ix" we get
    [tex]\sum_{n=0}^\infty \frac{1}{n!}(ix)^n= 1+ i+ \frac{1}{2!}(ix)^2+ \frac{1}{3!}(ix)^3+\frac{1}{4!}(ix)^4+ \cdot\cdot\cdot[/tex]
    [tex]= 1+ ix+ \frac{1}{2!}(-x^2)+ \frac{1}{3!}(-ix^3)+ \frac{1}{4!}(x^4)+ \frac{1}{5!}ix^5+ \cdot\cdot\cdot[/tex]
    Separating "real" and "imaginary" parts, we have
    [tex](1- \frac{1}{2!}x^2+ \frac{1}{4!}x^4+\cdot\cdot\cdot)+ i(x- \frac{1}{3!}x^3+ \frac{1}{5!}x^5+ \cdot\cdot\cdot[/tex]
    so that the odd powers have a factor of "i" and the even powers do not. But those even and odd power series are precisely the Taylor's series for cosine and sine. That is
    [itex]e^{ix}= cos(x)+ i sin(x)[/itex], "Euler's formula".

    As for "calculus on complex-valued functions" it is essentially the same as calculus for real valued functions except that it is "two- dimensional". If you have taken "Calculus of functions of two variables" you might remember that instead of having limits "from the right" and "from the left", in order that a limit, at a given point, exist, it must be the same as we approach that point from any direction. That makes "limits" and "derivatives" a lot more restrictive.

    Also, because the fundamental property of complex numbers is "[itex]i^2= -1[/itex]", the simplest way to handle more complicated functions is to write them as "power series", as with [itex]e^x[/itex], sin(x), and cos(x), above. In fact, it can be shown that a function of a complex variable is differentiable at every point of a neighborhood if and only if it is infinitely differentiable on that set (so we can define it Taylor's series on the set) and that series converges to the function on the set. Such functions are called "analytic" functions and the study of "calculus of functions of a complex variable" is largely the study of "analytic functions".
  4. Oct 6, 2013 #3
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