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Math Help - complex analysis mapping question

  1. #1
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    complex analysis mapping question

    Hi, can anyone please help me with this question? I have no idea how to start.

    (a) Consider the map f(z)=z^2. Prove that under f, lines parallel to the real axis are mapped to parabolas.
    (b) Consider the map g(z)=\sqrt{z}, for some branch of the square root. Prove that under g lines parallel to the real axis are mapped to hyperbolas.

    I did try to do the questions, so I substitute z=x+iy, then get f(x,y)=x^2+y^2-2ixy, but I really don't know how to show this mapping thing. Please help me. Thanks a lot.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    For all t\in \mathbb{R} we have f(t+bi)^2=t^2+b^2-2bti=u+iv then, the transformed curve of the line L:\; x=t,y=b i s

    f (L)\equiv\begin{Bmatrix}u=t^2+b^2\\v=-2bt\end{matrix}

    ( parabola if b\neq 0) .
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  3. #3
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    Quote Originally Posted by FernandoRevilla View Post
    For all t\in \mathbb{R} we have f(t+bi)^2=t^2+b^2-2bti=u+iv then, the transformed curve of the line L:\; x=t,y=b i s

    f (L)\equiv\begin{Bmatrix}u=t^2+b^2\\v=-2bt\end{matrix}

    ( parabola if b\neq 0) .
    Thank you very much. I did get this, but I just can't continue, because I don't understand why this tells f is parabola? I thought need to show something like v can be written as u^2, then that means parabola. Could you please explain a bit more to me? Thanks a lot for your time.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by tsang View Post
    Thank you very much. I did get this, but I just can't continue, because I don't understand why this tells f is parabola? I thought need to show something like v can be written as u^2, then that means parabola. Could you please explain a bit more to me? Thanks a lot for your time.

    Substitute t=-v/2b in the first equation. What do you obtain?.
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    Part (b) is still quite hard, how do I get it to hyperbolas?
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Hint

    If \sqrt{t+bi}=x+iy then, 2xy=b .
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    Quote Originally Posted by FernandoRevilla View Post
    Hint

    If \sqrt{t+bi}=x+iy then, 2xy=b .

    Thanks for your hint, Professor, I'm do like analysis, but not that good with abstract maths, as I'm only first year doing third year Complex Analysis. I need to keep trying and see if I can get it.
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  8. #8
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by tsang View Post
    I need to keep trying and see if I can get it.

    Patience.
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  9. #9
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    In attempting to do a nearly identical question, I end up with a function:

    t=u^2-v^2

    or

    b=2(t-u^2)(t+v^2)

    I can't see any way of eliminating t from this (the non-constant part of the line being mapped)?
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  10. #10
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    Quote Originally Posted by Aylix View Post
    In attempting to do a nearly identical question, I end up with a function:

    t=u^2-v^2

    or

    b=2(t-u^2)(t+v^2)

    I can't see any way of eliminating t from this (the non-constant part of the line being mapped)?
    That's true, I get the idea, but can't finish that bit of algebra manipulation.
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