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Complex analysis - LFT

  1. May 2, 2006 #1
    Verify that the linear fractional transformation
    T(z) = (z2 - z1) / (z - z1)
    maps z1 to infinity, z2 to 1 and infinity to zero.

    ^^^ so for problems like these, do I just plug in z1, z2 and infinity in the eqn given for T(z) and see what value they give?
    in this case, do i assume 1/ 0 is infinity then?
    I just want to make sure I am taking the right approach

    (also ... for complex analysis in general, it is safe to assume 1/0 is infinity?)
     
  2. jcsd
  3. May 2, 2006 #2

    AKG

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    Yes, it's all correct. The thing you're asked to prove is only true when z1 and z2 are different, but I suppose the question intended that to be the case, so yes, just do what you said.
     
  4. May 3, 2006 #3

    matt grime

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    It is defined to be infinity, as is z/0 for any complex number z not equal to 0. 0/0 is not defined.
     
  5. May 3, 2006 #4
    what do i do with cases infinity/infinity and/or 0/0?
     
  6. May 3, 2006 #5

    matt grime

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    You cannot get LFTs which result in 0/0 or infinity/infinty. It is part of their definition: (az+b)/(cz+d) such that ad-bc=/=0
     
  7. May 27, 2006 #6
    in that case,in 0/0 or inf/inf i think u should take limits and use lh rule.
     
  8. May 27, 2006 #7

    shmoe

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    as already mentioned, 0/0 won't come up with a linear fractional transformation.

    If f(z)=(az+b)/(cz+d) with a,c both non zero you will get infinity/infinity if you try to evaluate f(infinity) by a straight substitution. Usualy linear fractional transformations will define f(infinity) either as a limit or, what amounts to the same thing, simple as f(infinity)=a/c.
     
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