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Complex Variables. Problem about complex sine.

  1. Feb 29, 2008 #1
    [SOLVED] Complex Variables. Problem about complex sine.

    1. The problem statement, all variables and given/known data

    Proof that the function
    [tex]\begin{displaymath}
    \begin{array}{cccc}
    f: &A=\left\{z\in\mathbb{C}\mid-\frac{\pi}{2}<\Re z<\frac{\pi}{2}\right\} &\longrightarrow &B=\mathbb{C}-\left\{z\in\mathbb{C}\mid \Im{z}=0,\,\left\vert\Re z\right\vert\geq 1\right\}\\
    &z &\mapsto &\sin(z)
    \end{array}
    \end{displaymath}[/tex]
    is a biyection.

    2. Relevant equations



    3. The attempt at a solution

    I have already proven (on the previous exercise of my book) that the complex sine is inyective on a vertical band of the complex plane of width [tex]\pi[/tex]. Then I have only to proof that the function is surjective or that there exists left inverse (more difficult since its multivaluated in general).

    My problem is that I have no idea how to proof that since I have saw that the complex sine maps vertical lines in the complex plane to hyperbolas and horizontal ones to ellipses. Then I think it's imposible that the sine could map the set [tex]A[/tex] to the set [tex]B[/tex] one-one because the hyperbolas and the ellipses I have mentioned get out from [tex]B[/tex].
     
  2. jcsd
  3. Feb 29, 2008 #2
    I did this very same problem recently. Fortunately, the book had already proven that sin(z) maps the strip [itex]\{x + iy : 0 \le x \le \pi/2[/itex] and [itex]y \ge 0\}[/itex] in a one-to-one fashion onto the first quadrant.

    Then, using the fact that [itex]\sin(-\bar{z}) = -\overline{\sin(z)}[/itex], one may show that sin(z) maps the strip [itex]\{x + iy : -\pi/2 \le x \le \pi/2[/itex] and [itex]y > 0\}[/itex] to the upper half-plane bijectively.

    Similarly, one may show that sin(z) maps the strip [itex]\{x + iy : -\pi/2 \le x \le \pi/2[/itex] and [itex]y < 0\}[/itex] to the lower half-plane bijectively.

    Now all that's left is to consider the strip [itex]\{x + iy : -\pi/2 \le x \le \pi/2[/itex] and [itex]y = 0\}[/itex] which I leave to you.
     
  4. Mar 1, 2008 #3
    Thanks for your help
     
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