Integration with variables Notes_Part_13

# Integration with variables Notes_Part_13 - Example 7.35 The...

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Example 7.35. The goal of this example is to construct an conformal map that takes a half disk D + = b | z | < 1 , Im z > 0 B (7 . 73) to the full unit disk D = {| ζ | < 1 } . The answer is not ζ = z 2 because the image of D + omits the positive real axis, resulting in a disk that has a slit cut out of it: {| ζ | < 1 , 0 < ph ζ < 2 π } . To obtain the entire disk as the image of the conformal map, we must think a little harder. The Frst observation is that the map z = ( w 1) / ( w + 1) that we analyzed in Example 7.24 takes the right half plane R = { Re w > 0 } to the unit disk. Moreover, it maps the upper right quadrant Q = { 0 < ph w < 1 2 π } to the half disk (7.73). Its inverse, w = z + 1 z 1 (7 . 74) will therefore map the half disk, z D + , to the upper right quadrant w Q . On the other hand, we just constructed a conformal map (7.72) that takes the upper right quadrant Q to the unit disk D . Therefore, if compose the two maps — replacing z by w in (7.72) and then using (7.74) — we obtain the desired conformal map: ζ = i w 2 + 1 i w 2 1 = i p z + 1 z 1 P 2 + 1 i p z + 1 z 1 P 2 1 = ( i + 1)( z 2 + 1) + 2( i 1) z ( i 1)( z 2 + 1) + 2( i + 1) z . The formula can be further simpliFed by multiplying numerator and denominator by i +1, and so ζ = i z 2 + 2 i z + 1 z 2 2 i z + 1 . The leading factor

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Integration with variables Notes_Part_13 - Example 7.35 The...

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