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A Question About the Complex Plane

  1. Mar 14, 2005 #1


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    I'm taking an introductory course in Complex Analysis. Close to the beginning of the term, we had a review of complex numbers, but we weren't expected to know any of this crazy number theory stuff, and as far as the reasons for this addition to the real numbers, my prof offered only this interesting tidbit (recalling from memory): [itex] \mathbb{C} [/itex] is closed under the operators [itex] \cdot \ \ and \ \ + [/itex] and is therefore a number "field" (whatever the heck that means). He sometimes spoke of [itex] \mathbb{R}^2 [/itex] instead of [itex] \mathbb{C} [/itex], during that discussion and explained that there was no reason for alarm, as they were essentially interchangeable.

    Okay, the point of my post is that I'm confused as to what that means, and why it should be the case. Isn't [itex] \mathbb{R}^2 [/itex] a vector space, not a "field"? Furthermore, my confusion was compounded when we went on to such things as complex valued functions, and calculus in the complex plane, which I admit seems to be developed in the same way, and have similar qualities as vector calculus. But if they're so interchangeable, why do the two forms of calculus exist independently. Why formulate things in terms of complex numbers if you can just use vectors? All my math classes are so frustrating because they're just a hodgepodge of branches of math that have interrelationships and coincidences of terminology that are interesting to ponder, but that I feel I will NEVER understand them because nobody outright states them. (eg. eigenvalues in DE's vs. Lin algebra, the term "singular" appearing all over the place etc). But let's not get sidetracked. I'm rambling because I'm confused. Let me offer a specific example of this confusion.

    A complex valued function f is a function of an independent complex variable z with real and imaginary parts such that:

    z = x + iy

    Therefore f: [itex] \mathbb{C} [/itex] ----> [itex] \mathbb{C} [/itex] (in general anyway. So the first [itex] \mathbb{C} [/itex] or some subset of it is the domain on which the values of z for which f is defined reside. And the second [itex] \mathbb{C} [/itex] is the range of f, ie in general it takes on complex values. But then we wrote f as:

    f(z) = u(x,y) + iv(x,y), where u and v are real valued functions of x and y that represent the real and imaginary parts of f respectively. So the domain of u is [itex] \mathbb{R}^2 [/itex], same with v, yet we seem to be claiming that that plane is the same as the plane [itex] \mathbb{C} [/itex] in which all the z's given by z = x + iy reside! After all, it's the same damn x and y that you're plugging in to u and v to evaluate them as the x and y you would plug into z to evaluate f(z) directly!!! So now I'm totally confused, because they are in fact NOT the same plane. We don't say u = u(z), because u is not a function of the complex variable x + iy, but of the two variables x and y SEPARATELY. You might argue that these planes are all just abstractions, so what difference does it make whether we label the y-axis with real or imaginary numbers? But I just don't GET it. It makes a big difference...or does it?

  2. jcsd
  3. Mar 14, 2005 #2
    Woah! A lot of stuff here. Frankly I haven't even taken any complex analysis yet, but I think I can still help.

    If a set [tex]F[/tex], with two operations, [tex]+[/tex] and [tex]\cdot[/tex], is such that the operations satisfy the following axioms:

    If [tex]a, \ b[/tex] and [tex]c[/tex] are in [tex]F[/tex], and denoting [tex]a \cdot b[/tex] by [tex]ab[/tex], then
    1. [tex]a+b = b+a[/tex] and [tex] ab = ba[/tex]
    2. [tex]a+(b+c) = (a+b)+c[/tex] and [tex] a(bc) = (ab)c[/tex]
    3. [tex]a(b+c) = ab + ac[/tex]
    4. [tex]\exists \ 0 \in F \ \mbox{s.t.} \; x + 0 = x \ \forall x \in F[/tex]
    5. [tex]\exists \ 1 \in F \ \mbox{s.t.} \; 1x = x \ \forall x \in F[/tex]
    6. [tex]a \neq 0 \Longrightarrow \exists a^{-1} \in F \ \mbox{s.t.} \ aa^{-1} = 1[/tex]
    7. [tex]\exists -a \in F \ \mbox {s.t.} \; a + (-a) = 0 [/tex]
    8. [tex] a\cdot b \in F[/tex] and [tex] a + b \in F[/tex]

    then [tex]F[/tex] is a field.

    On [tex]\mathbb{C}[/tex] these operations are just normal addition and multiplication. Similar operations can be defined on [tex]\mathbb{R}^2[/tex] (if [tex] (a, \ b), (c, \ d) \in \mathbb{R}^2, \ \mbox{define} \ \cdot \ \mbox{by} \ (a, \ b) \cdot (c, \ d) = (ac - bd, \ bc+ad), \mbox{and} \ + \ \mbox{by} \ (a, \ b) + (c, \ d) = (a+c, \ b+d)[/tex]).

    Evidently, every field is a vector space (but not the other way around).

    It is a very astute observation that this might cause problems with writing [tex]f(z)[/tex] in that way, but in this case it is justified. Given [tex] \mathbb{C} \ni z = x + iy[/tex] we can always isolate [tex]x[/tex] and [tex]y[/tex]. Let [tex] \bar{z}[/tex] represent the complex conjugate of [tex]z[/tex], ie. [tex] \overline{a+bi} = a - bi[/tex] if [tex]a[/tex] and [tex]b[/tex] are real. Then [tex] \frac{z + \bar{z}}{2} = x \ \mbox{and} \ \frac{z - \bar{z}}{2i} = y [/tex], and so, since complex conjugation is certainly a valid operation to be used in the evaluation of [tex]f[/tex], we can justifiably write [tex]f(z) = u(x,y) + v(x,y)i[/tex].
    Last edited: Mar 15, 2005
  4. Mar 14, 2005 #3


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    You can look up what a field is at mathworld. Vector spaces and fields are related. For one, every vector space has an underlying field, so, if you recall, when vector spaces are said to be closed under scalar multiplication, the scalars involved come from the underlying field.
    Depending on how you want to look at it, it can be both. [itex]\mathbb{R}^2[/itex] is a vector space, normally thought to lie over the field [itex]\mathbb{R}[/itex], but you could have it lie over the rational field. It could not lie over the complex field. However, it can also be a field if multiplication between elements of R² is defined. Normally, (1, 2) x (8, -2.3) is not defined, but it can be defined such that R² satisfies the criteria of a field. Now, the complex numbers form a field, and in fact, the multiplication of complex numbers is one way to define multiplication on R². An example:

    (3 + 5i) x (4 + 2i) = (3 x 4) + (3 x 2)i + (5 x 4)i + (5 x 2)i² = 2 + 26i

    Now, if we have ordered pairs in R² (a, b) and (c, d), define multiplication on them by:

    (a, b) x (c, d) = (ac - bd, ac + bd). You will find that by this definition, multiplication meets all the requirements necessary for a field. Try to find the relationship between this multiplication and complex multiplication.

    Also, the complex numbers can be said to form a vector space. The complex numbers over the field of complex numbers is a 1-d vector space. The complex numbers over the real field is a 2-d vector space.
    Vector calculus is different. It deals with any vector space, and in most vector spaces, there is no defined multipilcation between elements. You'll recall that there is addition of elements, and then multiplication of elements with scalars, but not multiplication between elements. The complex field, on the other hand, is a special vector space, with a multiplication. However, whereas things like the cross product are usually most meaningful when talking about 3 dimensions, that is not the case with the complex field, so there is reason to treat complex analysis separately.
    Not some subset of C is the domain, all of C is the domain. And the second C is not the range, it is the co-domain. The range is a subset of the co-domain. In other words, f maps C only to values of C, but not necessarily to all values of C. for example, f: R --> R defined by f(x) = x² has R as its domain, so it takes all real values for x. Also, it is said to map to R, so it will only take x values to other real values. But clearly, it doesn't take R to all of R, since x² is always positive.
    z = x + iy. Now, with respect to a basis [itex]\beta[/itex] = {1, i} the co-ordinates of z can be expressed as (x, y), i.e.:

    [tex]z = x + iy[/tex]

    [tex][z]_{\beta} = (x, y)[/tex]

    So to say z = (x, y) is a little misleading. (x, y) are the co-ordinates of z:
    a) in the complex plane
    b) in the vector space [itex]\mathbb{C}(\mathbb{R})[/itex] with respect to basis {1, i}. Note [itex]\mathbb{C}(\mathbb{R})[/itex] denotes the vector space of complex numbers overlying the field R.

    In your example:

    f : C --> C
    u : R² --> R (edited, originally said co-domain was R²)
    v : R² --> R
    Last edited: Mar 14, 2005
  5. Mar 14, 2005 #4


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    It is common to "identify" various mathematical objects when they are "essentially the same". For example, "1" the natural number and "1" the real number are denoted the same way, although they are really different things. In general practice, using different notations for them would make notation more cumbersome, yet add no clarity. Thus, it's not used.

    Another example is symbols like R. R is used to denote a great many things, such as a set, a field, an order type, a topological space, a metric space, a vector space over itself, and more! However, all of these different aspects of the real numbers are "essentially the same", much like 1 the integer and 1 the real number are.

    One of the notations that bugs you is a common notation with functions that have vectors as inputs or outputs. If f is a function from R2 --> R, I might write it as [itex]f(\vec{x})[/itex], or I might write it as f(y, z). It is technically more accurate to say that f(y, z) would be a function from RxR --> R, but again going back to identifications, RxR is treated as the same thing as R2 anyways. (In fact, they might actually be the same thing!)

    Now, C is a vector space over R. (Or, if you prefer, that is one of the many aspects of the complex numbers) You can check directly by proving that C satisfies the vector space axioms. (Which might actually be a trivial task, depending on which definition of C you use)
  6. Mar 14, 2005 #5
    Actually in this case

    [tex] u: \mathbb{R}^2 \longrightarrow \mathbb{R}[/tex]
    [tex] v: \mathbb{R}^2 \longrightarrow \mathbb{R}[/tex]
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