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Complex analysis

  1. Oct 6, 2011 #1
    i am trying to solve below problem but not getting start; so plz help

    The function f(z) = e^(z+i*pi) has infinitely many points in the fiber of each point in its range.

    (A) Find four points that map to 1
    (B) The natural inverse of f(z), say g(z) maps each point in its domain to infinitely points, as with log(z). For example, g(1) is infinitely many points, including the four you provided in the first part of the problem.
    i. Choose a branch of g(z)
    ii. Sketch the domain of this branch
    iii. Sketch the range of this branch
    IV, Indicate which point in the set g(1) is the image of 1 under your branch.

    i have tried using following formula:
    f(z)= az +b/cz+d
    f1(z)= z+d/c
    f2(z)= 1/z
    f3(z)= (-(ad-bc)/c^2)*z
    f4(z)= z+a/c

    but getting nothing. can anyone help me with this?
     
  2. jcsd
  3. Oct 6, 2011 #2
    What's with that fiber thing and all those things you trying? This has to do with the logarithm multifunction. For example, if:

    [tex]u=e^{\pi iz}[/tex]

    then it's inverse is the log multifunction:

    [tex]z=\frac{1}{\pi i} \log(u)=\frac{1}{\pi i}(\ln|u|+i(\theta+2n \pi))[/tex]

    So just set u=1 and use the first four n's like 0, 1, 2, 3 to get your four values. Each n corresponds to a particular single-valued branch of the log function. Choose one. It's domain is the complex plane minus it's branch-cut which you have to define. Usually, the branch cut is along the negative real axis. Try and answer your question from the perspective of this multi-valued log function and a particular "branch" you choose by letting the argument run through say [itex]n\pi \leq \theta<(n+2)\pi[/itex]
     
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