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    I'm trying to show that \frac{d}{dz}cos(z)=-sin(z) by expressing them in terms of the exponential function.

    I have that cos(z)=\frac{e^{iz}+e^{-iz}}{2} and sin(z)=\frac{e^{iz}-e^{-iz}}{2i}

    But I got that \frac{d}{dz}cos(z)=\frac{d}{dz}(  \frac{e^{iz}-e^{-iz}}{2i})=sin(z) which is not what I was looking for, -sin(z)

    :hmmmm2:
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    (Original post by RamocitoMorales)
    :hmmmm2:
    Yes, you made a calculation error. Post details if it's still not coming out, though you'll probably see it if you check through yourself.
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    (Original post by ghostwalker)
    Yes, you made a calculation error. Post details if it's still not coming out, though you'll probably see it if you check through yourself.
    \frac{d}{dz}\frac{(e^{iz}+e^{-iz}}{2})=\frac{(ie^{iz}-ie^{-iz})\cdot 2-0\cdot (e^{iz}+e^{-iz})}{2^{2}}=\frac{2ie^{iz}-2ie^{-iz}}{4}=\frac{e^{iz}-e^{-iz}}{2i}=sin(z)
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    (Original post by RamocitoMorales)
    \frac{d}{dz}\frac{(e^{iz}+e^{-iz}}{2})=\frac{(ie^{iz}-ie^{-iz})\cdot 2-0\cdot (e^{iz}+e^{-iz})}{2^{2}}=\frac{2ie^{iz}-2ie^{-iz}}{4}=\frac{e^{iz}-e^{-iz}}{2i}=sin(z)
    Not sure of the methodology, as my complex analysis is non-existent, however:

    On your last step, depending on how you did it, when you multiply top and bottom by i, you get i^2=-1, or if you divided by i, then you get 1/i = -i on the bottom.
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    (Original post by ghostwalker)
    Not sure of the methodology, as my complex analysis is non-existent, however:

    On your last step, depending on how you did it, when you multiply top and bottom by i, you get i^2=-1, or if you divided by i, then you get 1/i = -i on the bottom.
    Ah, thanks, I've got it. I made the mistake of treating i as a real number. :facepalm:
 
 
 
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