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Conformal mapping

  1. Apr 29, 2014 #1
    1. The problem statement, all variables and given/known data
    Find function that maps area between ##|z|=2## and ##|z+1|=1## on area between two parallel lines.


    2. Relevant equations



    3. The attempt at a solution

    I don't know how to check if my solution works for this problem?

    I used Möbious transformation:

    ##f(z)=\frac{az+b}{cz+d}## where I decided that ##f(-2i)=\infty ## and ##f(2i)=0## and ##f(2)=2i##.

    Using this I find out that ##f(z)=\frac{(z-2i)(i-4)}{\frac{1-i}{2}z+i+1}##

    Now I am almost 100% if the circles would both have center in point 0. However, they have different origins and I somehow don't know how to deal with this kind of problems. :(
     
  2. jcsd
  3. Apr 29, 2014 #2

    pasmith

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    Homework Helper

    The circles [itex]|z| = 2[/itex] and [itex]|z + 1| = 1[/itex] intersect only at [itex]z = -2[/itex], so you must have [itex]f(-2) = \infty[/itex].

    Also, the first choice you need to make is the parallel lines bounding the image of the area in question.

    I think I would work backwards, and map the parallel lines [itex]\mathrm{Re}(z) = 0[/itex] and [itex]\mathrm{Re}(z) = 1[/itex] to the circles [itex]|z + 1| = 1[/itex] and [itex]|z| = 2[/itex] respectively by a Mobius map [itex]g[/itex] (these parallel lines are cunningly chosen so that I can take [itex]g(0) = 0[/itex] and [itex]g(1) = 2[/itex]). I would then need to check that [tex]
    |g(iy) + 1| = 1[/tex] and [tex]
    |g(1+ iy)| = 2[/tex] for all [itex]y \in \mathbb{R}[/itex] (which seems easier than checking that [itex]f(z)[/itex] lies on a particular straight line when [itex]|z + 1| = 1[/itex] and so forth) and that if [itex]\mathrm{Re}(z) \in (0,1)[/itex] then [itex]g(z)[/itex] lies between the two circles.

    (Finally, of course, I must compute [itex]f = g^{-1}[/itex] which what I'm actually asked to find.)
     
  4. Apr 30, 2014 #3
    Hmm, ok, I found out that for ##g(0)=0##, ##g(1)=2## and ##g(\infty )=-2## the Mobius transformation is ##g(z)=\frac{z}{1-\frac{1}{2}z}##

    Now I don't understand how you came up with following conditions:

    ##|g(iy) + 1| = 1## and ##|g(1+ iy)| = 2##

    For example, the first one:
    ##|\frac{iy}{1-\frac{1}{2}iy}+\frac{1-\frac{1}{2}iy}{1-\frac{1}{2}iy}|=|\frac{1+\frac{1}{2}iy}{1-\frac{1}{2}iy}|=1##

    ##|\frac{(1+\frac{1}{2}iy)^2}{1-\frac{1}{4}y^2}|=1##

    ##|(1+\frac{1}{2}iy)^2|=1-\frac{1}{4}y^2##

    ##2iy=0##

    Now it is possible that I made a mistake.. ?
     
  5. Apr 30, 2014 #4

    pasmith

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    Homework Helper

    We need the image of [itex]\mathrm{Re}(z) = 0[/itex] to be [itex]|z + 1| = 1[/itex]. If [itex]\mathrm{Re}(z) = 0[/itex] then [itex]z = iy[/itex] for some [itex]y \in \mathbb{R}[/itex]. Hence we need [itex]|g(iy) + 1| = 1[/itex] for all [itex]y \in \mathbb{R}[/itex].

    You are checking whether the condition [itex]|g(iy) + 1| = 1[/itex] is true for all [itex]y \in \mathbb{R}[/itex], so you can't assume this in your calculations. Thus you should start with [tex]
    |g(iy) + 1| = \left|\frac{iy}{1-\frac{1}{2}iy}+\frac{1-\frac{1}{2}iy}{1-\frac{1}{2}iy}\right|=\left|\frac{1+\frac{1}{2}iy}{1-\frac{1}{2}iy}\right| =
    \left|\frac{(1 + \frac12 iy)^2}{1 + \frac14y^2}\right| = \frac1{1 + \frac14y^2} \left|(1 + \tfrac12 iy)^2\right|
    [/tex]
    and you need to show that this is equal to 1. You have now to calculate [tex]\left|(1 + \tfrac12 iy)^2\right| = \left|1 - \tfrac14y^2 + iy\right|
    = \sqrt{(1 - \tfrac14y^2)^2 + y^2}.
    [/tex] Fortunately, the expression under the square root turns out to be a perfect square.
     
  6. Apr 30, 2014 #5
    Aaa ok I see it now. We are trying to prove that both of the lines map where they are supposed to; vertical line where ##Re(z)=0## into ##|z+1|=1## and similar for the line ##Re(z)=1##.

    Second condition:

    ##|g(1+iy)|=2##

    ##|\frac{1+iy}{1-\frac{1}{2}(1+iy)}|=2##

    After some calculation, we find out that: ##|\frac{1+iy}{1-\frac{1}{2}(1+iy)}|=\frac{4}{2}\frac{1+y^2}{1+y^2}=2##

    Second condition is also good. Therefore yes, the area between two parallel lines is mapped to the area between the desired circles with this Mobius transformation: ##g(z)=\frac{z}{1-\frac{1}{2}z}##

    Of course, I need inverse of that, which is ##f(z)=\frac{z}{\frac{1}{2}z+1}##

    pasmith, thank you very much!
     
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