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Complex logs

  1. Oct 23, 2006 #1
    Hi, I'm kinda confused with the following, I must be looking at it wrong

    Determinie all values of log(z) and Log(z):

    a.) -9+2i

    im confused when it comes to the angle. It is going to be in the form [tex]re^{i \theta}[/tex]. where [tex]r= \sqrt{85}[/tex]

  2. jcsd
  3. Oct 24, 2006 #2


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    Know that [tex]\log (z)= \mbox{Log} (z) +2k\pi i[/tex] for [tex]k\in\mathbb{Z}[/tex] is the multivalued logarithm function, and that [tex]\mbox{Log} (z)= \mbox{Log} |z| i\Arg (z)+2k\pi i[/tex] is the principal branch of the multivalued function [tex]\log (z)[/tex].

    Now if [tex]z=re^{i\theta}[/tex] then [tex]\log (z) =\log (re^{i\theta}= \mbox{Log} (r) + i\theta + 2k\pi i[/tex] and hence for [tex]z=-9+2i[/tex] we have [tex]r=\sqrt{85}[/tex] and [tex]\theta = \tan ^{-1} \left( -\frac{9}{2}\right)[/tex] so that

    [tex]\log (-9+2i) = \mbox{Log} \sqrt{85} + i \tan ^{-1} \left( -\frac{9}{2}\right) + 2k\pi i, \mbox{ for } k\in\mathbb{Z}[/tex]

    is all values of [tex]\log (-9+2i)[/tex] and [tex]\mbox{Log} (-9+2i)[/tex] is the principal value of [tex]\log (-9+2i)[/tex] (so put [tex]k=0[/tex] in the above to get

    [tex]\mbox{Log}(-9+2i) = \mbox{Log} \sqrt{85} + i \tan ^{-1} \left( -\frac{9}{2}\right)[/tex].
    Last edited: Oct 24, 2006
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