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Math Help - cosh (z) = i

  1. #1
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    cosh (z) = i

    cosh(z)=\mathbf{i} \ \ z\in\mathbb{C}

    e^{2z}-2\mathbf{i}e^z+1=0

    \displaystyle e^z=\frac{2\mathbf{i}\pm\sqrt{-4-4}}{2}=\mathbf{i}(1\pm\sqrt{2})

    \displaystyle z=ln(\mathbf{i}(1+\sqrt{2}))=ln(\mathbf{i}(1+\sqrt  {2}))+\mathbf{i}\left(\frac{\pi}{2}+2\pi k}\right) \ \ k\in\mathbb{Z}

    \displaystyle z=ln(\mathbf{i})+ln(1+\sqrt{2})+\mathbf{i}\left(\f  rac{\pi}{2}+2\pi k}\right)=\mathbf{i}\left(\frac{\pi}{2}+2\pi k}\right)+ln(1+\sqrt{2})+\mathbf{i}\left(\frac{\pi  }{2}+2\pi k}\right)

    \displaystyle =ln(1+\sqrt{2})+2\mathbf{i}\left(\frac{\pi}{2}+2\p  i k}\right)

    \displaystyle z=ln(1+\sqrt{2})+\mathbf{i}({\pi}+4\pi k}) \ \ k\in\mathbb{Z}

    I know there is another solution but my question is about this solution. The book doesn't \mathbf{i}({\pi}+4\pi k}) but instead has \displaystyle \mathbf{i}\left(\frac{\pi}{2}+2\pi k}\right).

    The book, when taking the modulus of z, didn't have the \mathbf{i} which would cause the additionally argument. Whenever I take the modulus, I have the \mathbf{i}. How can I get rid of it?
    Last edited by dwsmith; December 20th 2010 at 05:55 PM.
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  2. #2
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    Quote Originally Posted by dwsmith View Post
    cosh(z)=\mathbf{i} \ \ z\in\mathbb{C}

    e^{2z}-2\mathbf{i}e^z+1=0

    \displaystyle e^z=\frac{2\mathbf{i}\pm\sqrt{-4-4}}{2}=\mathbf{i}(1\pm\sqrt{2})

    \displaystyle z=ln(\mathbf{i}(1+\sqrt{2}))

    =ln(\mathbf{i}(1+\sqrt{2}))+\mathbf{i}\left(\frac{  \pi}{2}+2\pi k}\right) \ \ k\in\mathbb{Z} Mr F says: This is wrong.

    \displaystyle z=ln(\mathbf{i})+ln(1+\sqrt{2})+\mathbf{i}\left(\f  rac{\pi}{2}+2\pi k}\right)=\mathbf{i}\left(\frac{\pi}{2}+2\pi k}\right)+ln(1+\sqrt{2})+\mathbf{i}\left(\frac{\pi  }{2}+2\pi k}\right)

    \displaystyle =ln(1+\sqrt{2})+2\mathbf{i}\left(\frac{\pi}{2}+2\p  i k}\right)

    \displaystyle z=ln(1+\sqrt{2})+\mathbf{i}({\pi}+4\pi k}) \ \ k\in\mathbb{Z}

    I know there is another solution but my question is about this solution. The book doesn't \mathbf{i}({\pi}+4\pi k}) but instead has \displaystyle \mathbf{i}\left(\frac{\pi}{2}+2\pi k}\right).

    The book, when taking the modulus of z, didn't have the \mathbf{i} which would cause the additionally argument. Whenever I take the modulus, I have the \mathbf{i}. How can I get rid of it?
    You need to review the complex logarithm.
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  3. #3
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    z=\mathbf{i}(1+\sqrt{2})

    ln(z)=ln|z|+\mathbf{i}arg(z)

    \displaystyle |z|=\mathbf{i}(1+\sqrt{2}) \ \ arg(z)=\frac{\pi}{2}+2\pi k

    I don't know what is wrong.

    Oh, |\mathbf{i}|=1
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