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Conformal transformation

  1. Jan 30, 2010 #1
    1. The problem statement, all variables and given/known data

    Use a sequence of simple conformal transformations to find the function f(z) mapping the line
    going trough the points {(-1,0),(0,i)} to the line given by the set {z=2+i*s, s is real}.

    2. Relevant equations



    3. The attempt at a solution

    let z=(-1,0) so when we put this into z=2+i*s, we get that s=3*i, when z=(0,i), we get s=1+2i.

    does this mean that it maps to the line joining the points (0,3i) and (1,2i)?

    I think this is wrong...
    any ideas?

    thank you.
     
  2. jcsd
  3. Jan 30, 2010 #2


    Yes this is wrong. You have tried to find your two original points, using the formula for the vertical line that you are supposed to en up with. But this is imposible, since the two points are not not the line {z=2+i*s, s is real}.

    You have to find conformal transformations that will "move" your two original points onto the vertical line given by {z=2+i*s, s is real}.

    So you need to check how each of the simple conformal transformations affect your two orignal points, and then try to find which conformal transformations that will actually move your points onto the vertical line.

    Torquil
     
  4. Jan 30, 2010 #3
    Said differently:

    The original line is given by

    { (-1,0) + r(1,i), r is real}

    If z -> f(z) is a conformal transformation, then this will be transformed into

    { f((-1,0) + r(1,i)), r is real}

    You need to find such a conformal transformation f, such that this set is the same as

    {2+i*s, s is real}

    So you need to check which types of f you are allowed to use, i.e. which f that represent conformal transformations. And when you know this, determine which of them that solves the problem.

    Good luck!

    Torquil
     
  5. Jan 30, 2010 #4
    Oh right, the problem is much clearer now, thanks.
    But, does this mean the line y=i(x+1) gets mapped onto y=2+is
    ?

    How do i know where the original points get mapped onto on the new line?
     
  6. Jan 30, 2010 #5
    Yes, but isn't it "y=i(x+1) gets mapped onto x=2"? You probably meant to say "z=2+is" and to think of s as a the parameter that defines the line when it takes different real values. In terms of only the x and y-variables (real and imag parts of z), you want to transform your original line y=i(x+1) into the line x=2.

    I don't know :-) Actually I think they can end up in many different places, because I think the solution is not unique.

    Torquil
     
  7. Jan 30, 2010 #6
    I see why it's the vertical line x=2.

    so the conformal transformations should be a rotation that makes the line y=i(x+1) vertical then a simple shift to the right.

    right?

    but now i have to figure out the equations of these tranformaitions
     
  8. Jan 30, 2010 #7
    Yes, that would be correct.
     
  9. Jan 30, 2010 #8
    for the rotation part, would it be:

    f(z) = ze^(-i*(pi/4))

    ?
     
  10. Jan 30, 2010 #9
    Since you have a minus in the exponent, wouldn't this result in a horizontal line? I would think that the sign should be positive to get a vertical line?
     
  11. Jan 30, 2010 #10
    Oh right.
    clockwise direction means negative so anticlockwise is positive...right?

    so would the final transformaiton be w=ze^(i*pi/4) + 3

    ?
     
  12. Jan 30, 2010 #11
    Yes that is probably right. Allthough you should test your result by applying it to those two points, and see if the x-value of both results is equal to 2.

    Torquil
     
  13. Jan 30, 2010 #12
    from the point (0,i)

    w = i(e^(i*pi/4))+3 = sqrt(2)/2 i + (3-sqrt(2)/2)

    :(

    so that means
    w=ze^(i*pi/4) + 3
    is wrong
    ?
     
  14. Jan 30, 2010 #13
    I think your +3 is incorrect. After having rotated the line to become vertical, it no longer intersects the x-axis at (-1,0). It will intersect at some other point. Therefore you need to find this point first, and after than you will know which you should use instead of +3.

    Torquil
     
  15. Jan 30, 2010 #14
    i can imagine it...the point will be closer to the origin.

    but how can i find this point?
     
  16. Jan 30, 2010 #15
    I can give you a hint for a simpler solution: First, translate so that your 45 degree line goes through the origin. Then rotate 45 degees. Then translate one more time to the right.

    Torquil
     
  17. Jan 30, 2010 #16
    to translate it to go through the origin, do i do: w=z+1 ? because this also wouldnt work
     
  18. Jan 30, 2010 #17
    I would say that the result should be

    Translate to intersect with the origin: z -> z+1
    Rotate 45 degrees: z -> z*e^(i*pi/4)
    Translate it to x=2: z -> z+2

    I think the composite transformation therefore is
    z -> (z+1)*e^(i*pi/4) + 2

    But I may be wrong of course, so you should check it using both of the points (-1,0) and (0,i). If it is wrong, then you could check each individual step of the transformation. E.g. check that you have a vertical line through the origin after step 2, and so on.

    Torquil
     
  19. Jan 30, 2010 #18
    yes, this works.
    sorry, i forgot to change the +3 to +2.

    you said that f(z) is not unique, how can i verify this?
     
  20. Jan 30, 2010 #19
    You can see this because you could have just as well rotated 180+45 degrees. Also it doesn't matter if you add any translation along the line in the beginning, or along the vertical line in the end. So e.g. it doesn't matter if you add anything proportional to i in the end result.

    Also, you could have also used any conformal scale transformation when your line was intersecting the origin. That would not have changed anything either, because a straight line through the origin is invariant under a scaling about the origin.

    EDIT: So from these examples, you see that the two original points that defined the original 45 degree line can end up anywhere in the line x=2. The scaling I mentioned would change the distance between them, a translation along the line will move them anywhere, and a 180 degree rotation will be able to change their order.

    Torquil
     
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