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Thread: Trig. hyp. complex no.s

  1. #1
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    Trig. hyp. complex no.s

    Can someone help explain why Modulus(sinZ)^2=sin^2(x)+sinh^2(x)

    When Z=x+iy

    many thanks
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  2. #2
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    It is well known that $\displaystyle \sin (z) = \sin (x)\cosh (y) + i\cos (x)\sinh (y)$.

    So
    $\displaystyle \begin{gathered}
    \left| {\sin (z)} \right|^2 = \sin ^2 (x)\cosh ^2 (y) + \cos ^2 (x)\sinh ^2 (y) \hfill \\
    \left| {\sin (z)} \right|^2 = \sin ^2 (x)\left( {1 + \sinh ^2 (y)} \right) + \left( {1 - \sin ^2 (x)} \right)\sinh ^2 (y) \hfill \\
    \left| {\sin (z)} \right|^2 = \sin ^2 (x) + \sinh ^2 (y) \hfill \\
    \end{gathered}$
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  3. #3
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    Quote Originally Posted by Bryn View Post
    Can someone help explain why Modulus(sinZ)^2=sin^2(x)+sinh^2(x)

    When Z=x+iy

    many thanks

    $\displaystyle \sin z=\frac{e^{iz}-e^{-iz}}{2i}=\frac{e^{
    xi-y}-e^{-xi+y}}{2i}$ $\displaystyle =\frac{e^{-y}(\cos x+i\sin x)-e^{y}(\cos x-i\sin x)}{2i}=\frac{-\cos x(e^{y}-e^{-y})}{2i}+\frac{\sin x(e^y+e^{-y})}{2i}i$ $\displaystyle = \sin x\frac{e^y+e^{-y}}{2}+\cos x\frac{e^{y}-e^{-y}}{2}i$ , so:

    $\displaystyle |\sin z|^2=Re(z)^2+Im(z)^2=\sin^2\!x\cosh^2\!y+\cos^2\!x \sinh^2\!y$ $\displaystyle =\sin^2\!x(1+\sinh^2\!y)+\cos^2\!x\sinh^2\!y=$ $\displaystyle (\sin^2\!x+\cos^2\!x)\sinh^2\!y+\sin^2\!x=\sinh^2\ !y+\sin^2\!x$

    Note that there's a mistake in your post: the hyperbolic sine is of y, not of x.

    Tonio
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