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Cz 1 then i claim that r maps the unit disc onto

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Unformatted text preview: ÆH therefore gives us a complex potential DÆC , and hence a harmonic I function ∞:DÆI . This is exactly what we were doing two sections ago, and now we do R it some more, this time in the context of complex potentials. Examples (A) Potential between two semicircular plates Consider the following scenario: 32 3 kV insulation –3 kV We would like a conformal mapping sending the disk to H . Without being too demanding, let us go back to Example (B) on p. 23 of these notes, where we saw that 1+z w= 1-z takes the above disc onto the right-hand half-plane as shown: 3 kV 3 kV –3 kV –3 kV So what we need now is a nice potential for the right-hand region. But this is an angular one, so our potential is given by (see the last section) ∞ = Aø + B 6 = πArg(z) (Recall that Arg is fine for the right-hand plane; -π < Arg(z) ≤ π) 6 6 This is the imaginary part of πLn(z) or the real part of -iπLn(z). Thus ∞ is the real part of 6 F(z) = -iπLn(z) Transferring this over the left-h...
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