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Math Help - Mapping involving hyperbolic functions

  1. #1
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    Mapping involving hyperbolic functions

    The question:
    For the mapping f(z) = sinh(z), find and sketch the image of Re(z) = c

    My attempt:
    Unless I'm mistaken, Re(z) = C is just a line.

    I'm having problems with the f(z) = sinh(z) part. I set f(z) = w, and used the definition of sinh(z) as follows:

    w=sinh(z)
    w = \frac{e^z - e^{-z}}{2}
    w = \frac{e^{x+yi} - e^{-(x+yi)}}{2}
    w = e^x(cos(y) + isin(y)) - e^{-x}(cos(y) - isin(y))
    w = cos(y)(e^x - e^{-x}) + isin(y)(e^x+e^{-x})

    Now I'm stuck. :/ Have I attempted this correctly, and if so, how do I proceed? Any help would be greatly appreciated!
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  2. #2
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    Yes, it's correct.

    So now that you know the real part is equal to a constant C

    cos(y)(e^x - e^{-x}) = C

    cos(y)(e^x - 1/e^x) = C

    cos(y)(e^{2x} - 1) = Ce^x

    cos(y) = Ce^x/(e^{2x} - 1)

    y = arccos[Ce^x/(e^{2x} - 1)].

    I'm sure you can sketch this.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Glitch View Post
    w = cos(y)(e^x - e^{-x}) + isin(y)(e^x+e^{-x})

    That is right.

    \textrm{Re\;}w:\mathbb{C}\to \mathbb{R}

    (\textrm{Re\;}w)(x+iy)=2\cos y\sinh x

    Now, for y=0 and x\in\mathbb{R} , we have that 2\sinh x varies in \mathbb{R}. So, the image of \textrm{Re\;}w is \mathbb{R}.

    For \textrm{Im}\;z use that 2\cosh x varies in [2,+\infty) .

    Edited: I misread the question. I thought the problem only was to find the image of:

    \textrm{Re}\;w\;,\;\textrm{Im}\;w
    Last edited by FernandoRevilla; April 30th 2011 at 04:41 AM.
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    @Prove It

    Are you sure? The solution is:

    (\frac{u}{sinh(c)})^2 + (\frac{v}{cosh(c)})^2 = 1; c != 0
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  5. #5
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    Where did u and v come from?
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  6. #6
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    Quote Originally Posted by Prove It View Post
    Where did u and v come from?
    Judging by past questions, the convention is:

    w = u + iv
    z = x + iy
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  7. #7
    MHF Contributor FernandoRevilla's Avatar
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    We have

    u+iv=f(c+iy)\Leftrightarrow \begin{Bmatrix} u=2\cos y\sinh c\\v=2\sin y\cosh c\end{matrix}

    (i)\quad c\neq 0

    \dfrac{u^2}{(2\sinh c)^2}+\dfrac{v^2}{(2\cosh c)^2}=1

    and we obtain an ellipse.

    (ii)\quad c=0

    u+iv=f(0+iy)\Leftrightarrow \begin{Bmatrix} u=0\\v=2\sin y\end{matrix}

    and we obtain the set

    \{0\}\times [-2,2]
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