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
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xaxis (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
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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.
 Fall '03
 StefanWaner
 Math, Algebra, Geometry, Complex Numbers

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