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Conformal Maps

  1. Oct 6, 2006 #1
    This issue of infinity (undefined?) keeps coming up in the following problems.

    For example, the following question:

    Computer the image of the sector [itex]0 \leq r \leq 1, 0 \leq \theta \leq \pi[/itex], under the map ln(z).

    So I first graphed this thing in the x,y (z-plane) and obviously we get a half circle with radius 1, above/including x-axis.

    Then I looked at the following points. A=0, B=1, C=i, D=-1. Then if we map them under [itex]w= ln(z)[/itex] we get,
    A' = w(A)= ln(A) = ln(0) = ?
    B' = ln(1) = 0
    C' = ln(i) = [tex]i \frac{\pi}{2}[/tex]
    D' = ln(-1) = [tex]i \pi[/tex]

    If we then look at each segment, and map them under w, well I know what to do with everything but the parts that involve, or go though A. But the whole infinity, or undefined, issue bugs me.

    Any ideas? Thanks.
    Last edited: Oct 6, 2006
  2. jcsd
  3. Oct 6, 2006 #2


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    Staff Emeritus
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    Gold Member

    Short answer:

    The map

    f(z) := \begin{cases}
    \ln z & z \neq 0 \\
    \infty & z = 0

    is a continuous map from your half-disk to the Riemann sphere. It's probably even complex-analytic.

    Long answer:

    It's sort of like the function

    [tex]f(x) = \frac{x}{x}[/tex]

    It's not defined at x = 0, but that's more of a technicality rather than an important property of the function.

    In many geometric contexts, we will (implicitly) continuously extend our functions so that they will be defined at such removable discontinuities.

    It's often fruitful to think not of the complex plane, but instead to think of the Riemann sphere. So what you really want to do is to consider your map as a map from the half-disk into the Riemann sphere. Your map has a removable discontinuity at the origin, so it's fruitful to work with its continuous extension instead.
  4. Oct 6, 2006 #3
    Have you tried looking at more points at some radius between 1 and 0 on that semicircle? Looking at 0+0i was probably just a bad choice.
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