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Math Help - sin(z) = cos(z)

  1. #1
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    sin(z) = cos(z)

    z\in\mathbb{C}

    sin(z)=cos(z)\Rightarrow sin(z)-cos(z)=0

    \displaystyle \frac{e^{\mathbf{i}z}-e^{-\mathbf{i}z}}{2\mathbf{i}}-\frac{e^{\mathbf{i}z}+e^{-\mathbf{i}z}}{2}=0

    \displaystyle e^{\mathbf{i}z}-e^{-\mathbf{i}z}-\mathbf{i}e^{\mathbf{i}z}-\mathbf{i}e^{-\mathbf{i}z}=0

    \displaystyle e^{\mathbf{i}z}(1-\mathbf{i})=e^{-\mathbf{i}z}(1+\mathbf{i})

    \displaystyle e^{2\mathbf{i}z}\frac{1-\mathbf{i}}{1+\mathbf{i}}=1

    \displaystyle -\mathbf{i}e^{2\mathbf{i}z}=1\Rightarrow e^{2\mathbf{i}z}=\mathbf{i}

    Book's answer \displaystyle z=\frac{\pi}{4}(4k+1) \ k\in\mathbb{Z}.


    Thanks.
    Last edited by dwsmith; December 19th 2010 at 12:58 PM. Reason: Error fixed
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  2. #2
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    Quote Originally Posted by dwsmith View Post
    z\in\mathbb{C}

    sin(z)=cos(z)\Rightarrow sin(z)-cos(z)=0

    \displaystyle \frac{e^{\mathbf{i}z}-e^{-\mathbf{i}z}}{2\mathbf{i}}-\frac{e^{\mathbf{i}z}+e^{-\mathbf{i}z}}{2}=0

    \displaystyle e^{\mathbf{i}z}-e^{-\mathbf{i}z}-\mathbf{i}e^{\mathbf{i}z}-\mathbf{i}e^{-\mathbf{i}z}=0

    \displaystyle e^{\mathbf{i}z}(1-\mathbf{i})=e^{-\mathbf{i}z}(1+\mathbf{i})

    \displaystyle e^{2\mathbf{i}z}\frac{1-\mathbf{i}}{1+\mathbf{i}}=0 Mr F says: This line should be \displaystyle e^{2\mathbf{i}z}\frac{1-\mathbf{i}}{1+\mathbf{i}}=1.

    [snip]
    ..
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  3. #3
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    Picking up from the correction:

    \displaystyle e^{\mathbf{i}z}=\pm\sqrt{\mathbf{i}}\Rightarrow \mathbf{i}z=ln(\pm\sqrt{\mathbf{i}})

    \displaystyle \Rightarrow z=-\mathbf{i}\left(ln|\sqrt{\mathbf{i}}|+\mathbf{i}\l  eft(\frac{\pm\pi}{2}+2\pi k\right)\right) \ k\in\mathbb{Z}

    \displaystyle z=\frac{\pm\pi}{2}+2\pi k=\pi \left(\frac{\pm 1}{2}+k\right) \ \ k\in\mathbb{Z}

    Something is still wrong though....

    The arg of z is negative and positive \displaystyle \frac{\pi}{2}+2\pi k and modulus of z is positive and negative i.
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  4. #4
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    You are solving \exp(2zi)=i.
    That means e^{-2y}\cos(2x)=0~\&~ e^{-2y}\sin(2x)=1.
    That gives 2y=0~&~2x=\frac{\pi}{2}+2k\pi or x=\frac{\pi}{4}+k\pi.
    That is the answer given the by the text.
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  5. #5
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    Hello, dwsmith!

    \sin z\,=\,\cos z, \;\;z\in \mathbb{C}

    How about this?

    Since neither \dfrac{\pi}{2} nor \dfrac{3\pi}{2} are roots of the equation, \cos z \ne 0

    Divide by \cos z\!:\;\;\dfrac{\sin z}{\cos z} \:=\:1 \quad\Rightarrow\quad \tan z \:=\:1


    Therefore: . z \;=\;\frac{\pi}{4} + \pi k

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  6. #6
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    That would have been a tad bit easier.
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