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Unformatted text preview: (independent of z)
will suffice for the 3dimensional solution. If, on the other hand, a and b above are linear
functions of z, then if we simply substitute them in the above formula, and notice that the
imaginary part is linear, we get uzz = 0 as well.
It would be nice not to have to rely so much on tables for our work, and for this, we
specialize to Linear Fractional Transformations. These have the form
az + b
w=
LFT
cz + d
where a, b, c, and d are complex constants. For it to be conformal, we need to ensure that
its derivative is nonzero and exists. This amounts to two conditions:
Condition 1: ad  bc ≠ 0
Condition 2: z ≠ d/c
We now look at what happens to regions of the zplane under these transformations. First
note that we can divide top & bottom by a or b (depending on which one is nonzero) and
thereby eliminate one of these constants. Thus there are only three constants in the
formula. This suggests that if we know where we want to map three points, we can plug
them in and solve for the constants uniquely. In other words, we can always find an LFT
that takes any three points to any other three points.
Note: S'pose we want the LFT to take z1Æw1, z2Æw2 and z3Æw3. Consider the LFT:
(w  w1)(w2  w3) (z  z1)(z2  z3)
=
(*)...
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 Fall '03
 StefanWaner
 Math, Algebra, Geometry, Complex Numbers

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