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Conformal mapping definition

  1. Oct 14, 2015 #1
    "Definition: A map ƒ: A ⊂ ℂ→ ℂ is called conformal at z0, if there exists an angle θ ∈[0,2Pi) and an r > 0 such that for any curve γ(t) that is differentiable at t=0, for which γ(t)∈ A and γ(0)= z0, and that satisfies γ ' ≠0, the curve σ(t) = ƒ(γ(t)) is differentiable at t=0 and, setting u = σ'(0) and v = γ'(0), we have |u| = r |v| and arg(u)= arg(v)+ θ (mod 2π). A map is called conformal when it is conformal at every point."

    I am having a lot of trouble understanding this definition.
    First, what exactly is θ? Is it related v=γ'(0) or u=σ '(0)? Relative to what is θ being measured, or what two rays is θ being defined by? Also, how do I determine arg(u) and arg(v)?

    Why must γ(0) be z0?

    Why must we have σ(t) and γ(t) to begin with?

    I'm having a lot of trouble understanding this definition and what it exactly means. Could someone please explain and elaborate the statements made in this definition. I've read it plenty of times, but it just doesn't make any sense - I've been wrestling with this for sometime now. I looked at other sources, but they don't seem to be as detailed as this definition - which seems to offer more insight into the concept once understood.

    I kind of understand that the general point being made is that ƒ is a rotation through an angle θ and magnitude |ƒ ' (z0)|, however, I don't see how the definition shows this.I also don't understand where equation arg(u)= arg(v)+ θ comes from. Is ƒ a sort of matrix transformation with the properties that rotates and changes magnitudes?

    This definition was taking out of Marsden and Hoffman's Basic Complex Analysis, 3rd edition, definition 1.5.6 - in case you happen to have the book.
    Last edited: Oct 14, 2015
  2. jcsd
  3. Oct 14, 2015 #2


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    It means that the mapping treats tiny neighborhoods of every point z0 like a rotation by some angle θ and a stretching / shrinking by a factor of r in tiny image neighborhoods of f(z0). The rotation and stretching is identical in all directions from z0. The values of θ and r can be different at other z points. If you examine your definition carefully, you will see that they are trying to say that in a formal manor using arbitrary curves through z0 and the curve image mapped by f.
    Last edited: Oct 15, 2015
  4. Oct 14, 2015 #3
    What exactly is meant by "rotation"? What exactly is rotating and rotating relative to what? Are the curves themselves rotating? I'm not sure if I understand...
  5. Oct 14, 2015 #4


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    If you follow the mapping of the point z0 to f(z0) and points right around z0 to their images around f(z0), you will see that the surrounding point images have been rotated around f(z0) on the XY plane. A 90o rotation would map a point to the right (i.e. greater x, same y) of z0 above f(z0) (i.e. same x, greater y). A point above z0 (i.e. same x, greater y) would be mapped to the left of f(z0) (i.e. smaller x, same y), etc. etc..

    So any line through z0 will be mapped to a curve through f(z0) that is rotated in the XY plane by the angle θ (in the immediate neighborhood of f(z0)).
    Last edited: Oct 15, 2015
  6. Oct 14, 2015 #5
    [Apologies for length.] Suppose there is a mapping g: U ⊂ 22, where we assume U is some open set of 2.

    Before we can discuss whether g is conformal at some point p0 = (x0, y0), we need to know that g is differentiable at P.

    Definition: The function g is differentiable at p0 if there exists a linear mapping* L: 22 such that the expression

    ||(g(p0 + v) - g(p0) -L(v))|| / ||v|| approaches 0 as |v| approaches 0.

    If this is the case, the linear mapping L: 22 is called the the derivative of g at p0, and denoted by Dg(p0).

    Of course, this means that the expression

    g(p0) + L(v) = g(p0) + Dg(p0)(v)

    is a good approximation to the expression g(p0 + v), for small v.

    Assume the mapping g: U → 2 is differentiable at p0. Now we can discuss whether g is conformal at p0.

    Definition: The differentiable mapping g is conformal at p0 if the linear mapping g(p0) that is its derivative at p0 preserves angles. (To "preserve angles", a mapping must preserve both the size of the angle and its sense: clockwise or counterclockwise.)

    So, what kinds of linear mappings preserve angles? Certainly rotations do. So do uniform scalings. Also, if we compose a rotation with a uniform scaling, that linear mapping still preserves angles. (Note that any uniform scaling and any rotation can be composed in either order and the same mapping will result.)

    In fact, the composition of a rotation and a uniform scaling is the only kind of linear mapping L that preserves angles.

    As a matrix, the linear mapping L will be of the form
    $$\begin{bmatrix} r cos(θ) & -r sin(θ) \\ r sin(θ) & r cos(θ) \end{bmatrix}$$
    where θ is the rotation angle and r > 0 is the magnification factor of the uniform scaling.

    Therefore the most general form of such a matrix is this:
    $$\begin{bmatrix} a & -b \\ b & a \end{bmatrix}$$
    where a = r cos(θ) and b = r sin(θ). (Or conversely, r = √(a2+b2) and θ = atan2(b,a).)

    Finally: If you have an analytic function f: U → ℂ where U is an open set of ℂ, then we know that as a function freal:U → R2, f is differentiable. It is known that every analytic function is conformal at each point where its derivative is nonzero.

    In fact, its complex derivative c = f'(z0) at any point z0 of U is a single complex number c.

    Locally, the derivative linear mapping at z0 is just multiplication by c = a + ib (for some real a, b). This will produce the same effect on the complex number x + iy as will applying the matrix
    $$\begin{bmatrix} a & -b \\ b & a \end{bmatrix}$$
    to the vector
    $$\begin{pmatrix} x\\ y \end{pmatrix}$$
    (with the vector to the right of the matrix).

    * By a linear mapping here is meant a mapping of 2 to itself that is achieved by applying a 2×2 matrix to a vector.
  7. Oct 15, 2015 #6


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    IOW: you can see it as a map f so that if two curves ##C_1, C_2 ## intersect at an angle ##\theta ## at a point ##z_0##, then ##f(C_1), f(C_2) ## intersect at the same angle at the point ## f(z_0)##.
  8. Oct 15, 2015 #7


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    While your question has been completely answered already I found this helpful.

    The complex numbers can be described as all numbers of the form ##re^iθ## where r is the number's norm - i.e. its distance to 0 - and θ is its angle to the x-axis. Complex multiplication is

    ##re^iθ . se^iα = rse^i(θ+α)##

    If one multiplies that entire complex plane by the fixed number ##re^iθ## then the result is that each norm of every complex number has been multiplied by r and each angle has had θ added to it. This is a linear map that maps the plane into itself.

    A mapping of the complex plane is conformal at a point if its Jacobian is multiplication by a complex number. That is all that your book is saying.

    All the stuff about curves is a way to define the action of the Jacobian matrix on tangent vectors at the point z0. The tangent vectors are defined as the velocity vectors of curves. Instead of using the Jacobian directly, your book defines its action through the derivative vector of the curve f(γ(t)). It is the same thing.
    Last edited: Oct 16, 2015
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