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Math Help - Riemann Mapping (Complex)

  1. #1
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    Riemann Mapping (Complex)

    Use conformal maps or combinations of conformal maps such as linear fractional transformations, powers, roots, sin z, log z, etc., to find a one-to-one analytic function mapping the given region D onto the upper half-plane U.

    (1) D = \{z: |Arg z| < \alpha\}, \alpha \leq \pi

    (2) D = \{z= x + iy: x,y > 0\}

    (3) D = \{z= x+ iy: |y-1| < 2\}

    (4) D = \{z: |z-z_0| < r_0\}

    I'll try to do (1)

    -\alpha < Arg z < \alpha

    z=re^{i\gamma}
    z^{\alpha}=r^{\alpha}e^{i\theta\alpha}

    -\alpha\gamma < \theta\gamma <\alpha\gamma

    2\alpha\gamma = \pi

    \gamma = \frac{\pi}{2\alpha}

    Answer: iz^{\frac{\pi}{2\alpha}}

    If anyone could tell me if I'm on the right track and provide me with some help on the other problems, I would really appreciate it, thanks!
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  2. #2
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    Quote Originally Posted by shadow_2145 View Post
    Use conformal maps or combinations of conformal maps such as linear fractional transformations, powers, roots, sin z, log z, etc., to find a one-to-one analytic function mapping the given region D onto the upper half-plane U.

    (1) D = \{z: |Arg z| < \alpha\}, 0< \alpha \leq \pi
    Try to visual this set. Use several values to \alpha to motive you. If \alpha = \pi then |\arg z| is the cut-plane at the non-positive axis (cut along (-\infty,0]). The square root map z\to z^{1/2} (where the branch of the square root is choose along the same cut i.e. principle branch) will map this set to |\arg z| < \tfrac{\pi}{2} i.e. the right half-plane. But you want the upper-half plane to rotate your picture one-quater turn i.e. z\to iz will do the trick. This gives the conformal mapping z\to iz^{1/2}. Thus, in general if we want to shrink (or expand) |\arg z| < \alpha to |\arg z| < \tfrac{\pi}{2} we should use z\mapsto z^{\pi / 2\alpha} as you correctly said. But this only gives us the right half plane, therefore, we need z\mapsto iz^{\pi /2\alpha}. Good job .
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  3. #3
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    Quote Originally Posted by shadow_2145 View Post
    (2) D = \{z= x + iy: x,y > 0\}
    Again visualize this set. This is the first quadrant. You can also think of this set as \{ r e^{i\theta} | r>0, 0<\theta<\tfrac{\pi}{2} \}. Consider the mapping z\mapsto z^2. Under this mapping re^{i\theta} \mapsto r^2 e^{2i\theta}. Therefore, 0 < 2\theta < \pi while r^2 > 0. This is the upper half-plane. That is therefore the conformal mapping that will do the trick.

    <br />
D = \{z= x+ iy: |y-1| < 2\}<br />
    Again visualize this set. This is the horizontal infinite strip between -1 and 3. Let us raise this stip up to the positive side by the mapping z\mapsto z+1. Now we will compress this strip to 0 < y < \pi the reason for this will soon be apparent. We do this by a dilation mapping in this case z\mapsto \tfrac{\pi}{4}z. Note the following mapping z\mapsto e^z and look what happens with the points. The point x+iy, -\infty < x < \infty, 0 < y < \pi becomes e^{x+iy} = e^x e^{iy}. And what is that? That is the upper half plane!
    Therefore, the required mapping is z\mapsto e^{ \pi (z+1)/4}.

    Can you the last one?
    You can find a detail discussion on conformal mapping here. I hope it helps you.
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