Results 1 to 7 of 7

Math Help - Complex function : composition of a translation, rotation, etc

  1. #1
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1

    Complex function : composition of a translation, rotation, etc

    I don't know how to approach the following problem : Write the transformation T(z)=\frac{i-z}{i+z} as a composition of translations, rotations, dilatations and inversions. Use the result to find the image of the semi-plane y>0.

    My attempt : My idea was to write T(z) in the form re^{i \pi \theta} but I didn't reach anything important I believe.
    So I guess there's a better way to approach the problem. I'd like to know it/them.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by arbolis View Post
    I don't know how to approach the following problem : Write the transformation T(z)=\frac{i-z}{i+z} as a composition of translations, rotations, dilatations and inversions. Use the result to find the image of the semi-plane y>0.

    My attempt : My idea was to write T(z) in the form re^{i \pi \theta} but I didn't reach anything important I believe.
    So I guess there's a better way to approach the problem. I'd like to know it/them.
    If you are interested in conformal mapping you can read the entire detailed thread here.

    \frac{i-z}{i+z} = -\frac{z-i}{z+i} = - \frac{z+i-2i}{z+i} = -1 + \frac{2i}{z+i}

    Let f: z\mapsto z+i, let g:z\mapsto \tfrac{1}{z}, let h: z \mapsto i, let F: z\mapsto 2z, let G:z\mapsto z-1. Thus, f is translation, g is inversion, h is rotation (by \tfrac{\pi}{2}), F is dilation, and G is translation.

    However, G\circ H \circ h\circ g \circ f: z\mapsto \tfrac{i-z}{i+z}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Thanks a lot TPH! (I'll take some time to read through the link... seems really interesting. I don't know if my course will introduce me conformal mapping, it is made for physicists.)
    Just a little question... why did you bother to rewrite \frac{i-z}{i+z} in the form -1 + \frac{2i}{z+i}?
    I see that you got rid of a z over z form... is it a general way to solve problem like these?


    Quote Originally Posted by ThePerfectHacker View Post
    If you are interested in conformal mapping you can read the entire detailed thread here.

    \frac{i-z}{i+z} = -\frac{z-i}{z+i} = - \frac{z+i-2i}{z+i} = -1 + \frac{2i}{z+i}

    Let f: z\mapsto z+i, let g:z\mapsto \tfrac{1}{z}, let h: z \mapsto i, let F: z\mapsto 2z, let G:z\mapsto z-1. Thus, f is translation, g is inversion, h is rotation (by \tfrac{\pi}{2}), F is dilation, and G is translation.

    However, G\circ H \circ h\circ g \circ f: z\mapsto \tfrac{i-z}{i+z}.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by arbolis View Post
    I see that you got rid of a z over z form... is it a general way to solve problem like these?
    Yes, we have to clear the z in the numerator. Because the basic inversion is 1/z, it has no z in numerator. Thus, we have to bring it into that form.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Quote Originally Posted by ThePerfectHacker View Post
    Yes, we have to clear the z in the numerator. Because the basic inversion is 1/z, it has no z in numerator. Thus, we have to bring it into that form.
    Thanks. I'm infinitely thankful to you.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by arbolis View Post
    Thanks. I'm infinitely thankful to you.
    You can bow before my feet and say, "Oh Master how much I love thee".
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Quote Originally Posted by ThePerfectHacker View Post
    You can bow before my feet and say, "Oh Master how much I love thee".

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Translation/Rotation help =(
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: April 22nd 2010, 04:02 AM
  2. Translation and then a rotation of a triangle
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: April 14th 2010, 07:29 AM
  3. Replies: 3
    Last Post: August 5th 2009, 09:03 AM
  4. Translation function
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: March 10th 2009, 10:51 PM
  5. [SOLVED] Complex rotation
    Posted in the Advanced Math Topics Forum
    Replies: 1
    Last Post: August 27th 2005, 11:42 PM

Search Tags


/mathhelpforum @mathhelpforum