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Homework Help: Complex logaritm question

  1. Sep 25, 2007 #1
    1. The problem statement, all variables and given/known data
    Find where the function: [tex]f(z)=Log(z-2i+1)[/tex] is analytic and where it is differentiable.


    2. Relevant equations

    Cauchy-Riemann equations?

    3. The attempt at a solution

    Here's where I am so far:
    [tex]Log(z-2i+1)=Log((x+1)+i(y-2))=ln(\sqrt{(x+1)^{2}+(y-2)^{2}})+iArg(z)[/tex]

    since I'm only looking at the principal value of the logarithm, [tex]0<\theta\leq2\pi[/tex] (this is the text book's choice of principal arguement), then
    [tex]ln(\sqrt{(x+1)^{2}+(y-2)^{2}})[/tex] will be discontinuous at [tex]x=-1[/tex] and [tex]y=2[/tex]... and the function is undefined everywhere on the positive real axis (because of the choice of argument). So, [tex]f(z)[/tex] is non-differentiable at [tex]z=-1+2i[/tex] and everywhere in the positive direction on the real axis extending from the point [tex]z=-1[/tex], because that is the "center" of my mapping.

    Now, I'm not sure how to show that it is differentiable everywhere else... Am I supposed to apply the cauchy-riemann equations to [tex]u=ln(\sqrt{(x+1)^{2}+(y-2)^{2}})[/tex] and [tex]v=Arg(z)[/tex]? If so, how do I take a partial derivative of [tex]Arg(z)[/tex]?
     
  2. jcsd
  3. Sep 25, 2007 #2

    Dick

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    Homework Helper

    Do you know where Log(z) is analytic? I think it's everywhere except along the branch cut you use to make it well defined, right? Then just translate by 2i-1. I don't think you need to get carried away with details.
     
    Last edited: Sep 25, 2007
  4. Sep 25, 2007 #3
    i was thinking something along the lines of using [tex]Arg(z)=arctan(\frac{y-2}{x+1})[/tex] but I'm unsure how to define the inverse tangent... but if I let [tex]Arg(z)=2arctan(\frac{y-2}{x+1})+C[/tex], where C is an arbitrary constant, then the cauchy-riemann equations will be satisfied... not sure why yet... my trig is not good....

    but anyway, yeah, i like your way better.... you're right in that it's probably sufficient for this excercise...

    thanks!
     
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