If we want to make the ln continuous we remove that

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Unformatted text preview: ate w = ln z as follows: First write z in the form z = reiø. Now let w = u+iv. Then ew = z gives eu+iv = z = reiø. Thus, eu eiv = reiø. Equating magnitudes and arguments, eu = r, v = ø, or u = ln r, v = ø. Thus, Formula for ln z ln z = ln r + iø, r = |z|, ø = arg(z) 3. If ø is chosen as the principal value of arg(z), that is, -π < ø ≤ π, then we get the principal value of ln z, called Ln z. Thus, Formula for Ln z Ln z = ln r + iø, r = |z|, ø = Arg(z) Also ln z = Ln z + i(2nπ); n = 0, ±1, ±2, ... What about the domain of the function Ln? Answer: Ln: C -{0}ÆC . However, Ln is discontinuous everywhere along the negative I I x-axis (where Arg(z) switches from π to numbers close to -π. If we want to make the Ln continuous, we remove that nasty piece from the domain and take Ln: {z | z ≠ 0 and arg(z) ≠ π} Æ C I 4. ln 0 is still undefined, as there is no complex number w such that ew = 0. Examples 3.7 (a) ln1 = 0 + 2nπi = 2nπi; (b) ln4 = 1.386... + 2nπi; (c) If r is real, then ln r = the usual value of ln r + 2nπi; (d) lni = πi/2 + 2nπi...
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This document was uploaded on 03/20/2014 for the course MATH 144 at Hofstra University.

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