Writing ln[(x + jy + a)/(x + jy a)] = u + jv, show that (a)
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Writing ln[(x + jy + a)/(x + jy – a)] = u + jv, show that
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(a) x + y - 2ax coth u + a = 0 (b) x = a sinh ul(cosh u cos v) (c) |x + jy| = a(cosh u + cos v)/(cosh u - cos v)
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