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Math Help - Complex Trig Function

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    Complex Trig Function

    Show that \overline{sin(iz)}=sin(i\bar{z}) if and only if z=n \pi i, (n=0, \pm 1, \pm 2, ...).

    Anyways, I got from the LHS to the RHS but didn't find the restriction. Where should that pop up?
    Last edited by davesface; April 8th 2010 at 10:24 PM. Reason: fixing latex
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    Quote Originally Posted by davesface View Post
    Show that \bar{sin(iz)}=sin(i\bar{z}) if and only if z=n \pi i, (n=0, \pm 1, \pm 2, ...). (There should be a complex conjugate bar over the whole first expression, but I'm not sure how to do that in Latex.)

    Anyways, I got from the LHS to the RHS but didn't find the restriction. Where should that pop up?
    \sin(z)=\cos(x)\sinh(y)+i\sin(y)\cosh(x)
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    Quote Originally Posted by davesface View Post
    Show that \bar{sin(iz)}=sin(i\bar{z}) if and only if z=n \pi i, (n=0, \pm 1, \pm 2, ...). (There should be a complex conjugate bar over the whole first expression, but I'm not sure how to do that in Latex.)

    Anyways, I got from the LHS to the RHS but didn't find the restriction. Where should that pop up?

    You must have made the same mistake I did the first part:

    \overline{\frac{e^{i\cdot iz}-e^{-i\cdot iz}}{2i}}=\overline{\sin iz} =\sin i\overline{z}=\frac{e^{i\cdot i\overline{z}}-e^{-i\cdot i\overline{z}}}{2i}\Longleftrightarrow \overline{e^{-z}-e^{z}}=e^{-\overline{z}}-e^{\overline{z}} ...and the mistake's already here! It must be

    On the left hand: \overline{\frac{e^{-z}-e^z}{2i}}=\overline{-i\,\frac{e^{-z}-e^z}{2}}=i\,\frac{e^{-z}-e^z}{2}

    On the right hand: \frac{e^{-\overline{z}}-e^{\overline{z}}}{2i}=-i\,\frac{e^{-\overline{z}}-e^{\overline{z}}}{2} ,and mulitiplying by \frac{2}{i} and assuming the equality (as in the beginning) we continue:

    e^{-z}-e^z=e^{\overline{z}}-e^{-\overline{z}}\Longleftrightarrow putting z=x+iy\,,\,\,x,y\in\mathbb{R} and comparing real and imaginary parts I get x=0 as unique condition...

    Tonio
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