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Complex Analysis-Isomorphisms

  1. Jun 22, 2010 #1
    1. The problem statement, all variables and given/known data
    1. Find and isomorphism of the region [tex] |arg z|< \frac{\pi}{8} [/tex] on the region:
    [tex] |z|<1,z Not In [0,1) [/tex].

    2. Find and isomorphism of the region [tex]D= (z=i+re^{it}, \frac{5 \pi}{4} < t<\frac{7 \pi}{4}, r \geq 0 ) [/tex] on the region [tex] R=(z: 0<Rez< \pi , Imz>0) [/tex].


    2. Relevant equations
    Mobiuos Transformations...Elementry Functions

    3. The attempt at a solution
    I have no idea about 2...
    About 1- I thought about first taking the function [tex] z \to z^8 [/tex] which will copy the region to [tex] C-(- \infty, 0] [/tex] ... I was thinking about using LOG or something from that point, but I have no idea how to continue this composition of functions

    Hope you'll be able to help me

    Thanks
     
  2. jcsd
  3. Jun 22, 2010 #2
    I'll give you a hint to point you in the right
    direction for both of these. The second one is
    just one additional step from the first.

    A basic beginning for both of these problems is
    to begin by mapping the region in question to
    the region R = Re z > 0, Im z > 0.

    For 1, the function is f(z) = z^2*exp(i*pi/4).

    Now, consider applying g(z) = 1/(z+1), which maps
    the Im z = 0,Re z > 0 boundary of R to the line
    segment (0,1) and the Re z = 0, Im z > 0
    boundary of R to the upper half of the circle
    |z-1/2| = 1/2...do you see that? So, g maps R
    onto the upper half of the disk |z-1/2| < 1/2.

    Then, note that h(z) = (2z-1)^2 maps this upper
    half disk onto |z| < 1, z not in [0,1).

    Then, the whole function is the composition h(g(f(z))).

    For 2, recall that exp(z) is an isomorphism from the
    horizontal strip 0 < Im z < 2pi, Re z < 0 onto
    |z| < 1, z not in [0,1).

    Then, the so called principle branch of log(z) is an
    isomorphism in the opposite direction. Now, proceed
    as for 1 to find an isomorphism from the region onto
    |z| < 1, z not in [0,1), then compose with log(z).
    Finally, you have to rotate and compress (multiply
    by 1/2*exp(-i*pi/2)).

    Hopefully, that helps.
     
  4. Jun 23, 2010 #3
    Wow! Thanks a lot for the detailed guidance! It was awsome...
    Unfortunately, there is one thing I can't understand:
    1. As far as I know, when we take the map [tex] \frac{1}{z} [/tex] , it mappes Im z = 0,Re z > 0 to (0,1)... When we take the map [tex] \frac{1}{z+1} [/tex] , it mappes this region to (1,2)...Am I right?
    2. I can't figure out why this map copies Im z > 0,Re z = 0 to the upper half of the circle
    |z-1/2| = 1/2 ... This is the part I couldn't understand in your guidance...
    I thought the map [tex] \frac{1}{z} [/tex] takes the first quadrant to :
    [tex](z: |z|<1 , Imz>0,Rez>0)[/tex] ... So the map [tex] \frac{1}{z+1} [/tex] takes this region to the circle:[tex](z: |z-1|<1 , Imz>0,Rez>0)[/tex] .... I realy can't understand where is my mistake in understanding these mappings... Hope you'll be able to help


    Thanks a lot again
     
  5. Jun 23, 2010 #4
    First, sorry that I crapped out of texifying...my first time posting here, and I didn't know how to do it.

    Note that if [tex] z\in (0,1) [/tex] then [tex] \frac{1}{z} \in (1,\infty), [/tex] and [tex] z\in (1,\infty) [/tex] then [tex] \frac{1}{z} \in (0,1), [/tex] so you are not quite right.

    This is the same reason that [tex] \frac{1}{z+1} [/tex] maps [tex] (0,\infty) \longrightarrow (0,1) [/tex]

    Second, remember that Mobius transformations map lines to circles. The vertical line [tex] Re z = 0 [/tex] gets mapped to some kind of circle by [tex] \frac{1}{z+1} [/tex]. Note [tex] 0\longrightarrow 1 [/tex] and [tex]\infty\longrightarrow 0[/tex].

    Hopefully, it is clear why [tex] Re z = 0, Im z > 0 [/tex] maps to half of the circle. I actually mispoke, it's the bottom half - figure out why and go from there. You're welcome - it was a welcome diversion, I'm deployed in Iraq right now.
     
  6. Jun 25, 2010 #5
    I think it's completely understandble now...Thanks again!
     
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