Substituting in and solving for w gives z i w iz 1

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Unformatted text preview: (independent of z) will suffice for the 3-dimensional 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 non-zero 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 z-plane 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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