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**Unformatted text preview: **on the unit disk D for an impulse concentrated at the origin; see Section 6.3 for details. How do we obtain the corresponding solution when the unit impulse is concentrated at another point = + i D instead of the origin? According to Example 7.25, the linear fractional transformation w = g ( z ) = z z 1 , where | | &lt; 1 , (7 . 104) maps the unit disk to itself, moving the point z = to the origin w = g ( ) = 0. The logarithmic potential U = 1 2 log | w | will thus be mapped to the Greens function G ( x, y ; , ) = 1 2 log vextendsingle vextendsingle vextendsingle vextendsingle z 1 z vextendsingle vextendsingle vextendsingle vextendsingle (7 . 105) at the point = + i . Indeed, by the properties of conformal mapping, since U is harmonic except at the singularity w = 0, the function (7.105) will also be harmonic except at the image point z = . Furthermore, as you are asked to prove in Exercise , the conformal mapping preserves the delta function singularity in G . Finally, since the conformal map does not alter the boundary | z | = 1, the function (7.105) continues to satisfy the homogeneous Dirichlet boundary conditions there. Formula (7.105) reproduces the Poisson formula (6.131) for the Greens function that we previously derived by the method of images. This identification can be verified by substituting z = re i , = e i , or, more simply, by noting that the denominator in the logarithmic fraction gives the potential due to a unit impulse at z = , while the numerator represents the image potential at z = 1 / required to cancel out the effect of the interior...

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