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Math Help - [SOLVED] solve cosh(x)= i

  1. #1
    crb
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    [SOLVED] solve cosh(x)= i

    any help will be appreciated....thank you

    best i got out of this was x=ln{1+ or - 2^(1/2)}i ...but unsure if that is correct or even a valid expression...thank you for any help ...peace
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  2. #2
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    \begin{array}{l}<br />
 \cosh (z) = \cosh (x)\cos (y) + i\sinh (x)\sin (y) = i \\ <br />
 z = \sinh ^{ - 1} (1) + \left( {\frac{{1}}{2}\pi } \right)i \\ <br />
 \end{array}
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  3. #3
    Super Member angel.white's Avatar
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    Quote Originally Posted by crb View Post
    any help will be appreciated....thank you

    best i got out of this was x=ln{1+ or - 2^(1/2)}i ...but unsure if that is correct or even a valid expression...thank you for any help ...peace
    Initial equation
    cosh(x)=i

    Take inverse
    x=arcosh(i)

    Identity
    x=ln(i+\sqrt{i-1}\sqrt{i+1})

    Multiply Square Roots
    x=ln(i+\sqrt{(i-1)(i+1)})

    FOIL
    x=ln(i+\sqrt{i^2-1})

    Simplify
    x=ln(i+\sqrt{-2})


    --------------------
    VERIFICATION:

    Initial Equation
    cosh(x)=i

    Identity
    \frac{e^x}{2} + \frac{e^{-x}}{2}=i

    Rewrite
    \frac{e^x}{2} + \frac{1}{2e^x}=i

    Substitute in our Answer
    \frac{e^{ln(i+\sqrt{-2})}}{2} + \frac{1}{2e^{ln(i+\sqrt{-2})}}=i

    Simplify
    \frac{i+\sqrt{-2}}{2} + \frac{1}{2(i+\sqrt{-2})}=i

    Multiply by 2 and subtract i from both sides
    \sqrt{-2} + \frac{1}{i+\sqrt{-2}}=i

    Common denominator
    \frac{i\sqrt{-2}+\sqrt{-2}\sqrt{-2}}{i+\sqrt{-2}} + \frac{1}{i+\sqrt{-2}}=i

    Simplify
    \frac{i^2\sqrt{2}-\sqrt{2}^2+1}{i+\sqrt{-2}}=i

    Simplify
    \frac{-\sqrt{2}-1}{i+\sqrt{-2}}=i

    Multiply by the Denominator
    -\sqrt{2}-1=i(i+\sqrt{-2})

    Distribute
    -\sqrt{2}-1=i^2+i\sqrt{-2}

    Simplify
    -\sqrt{2}-1=i^2+i^2\sqrt{2}

    Simplify
    -\sqrt{2}-1=-1-\sqrt{2}

    Both sides are equal, thus the answer is correct.
    Last edited by angel.white; January 25th 2008 at 01:47 PM.
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  4. #4
    crb
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    thanks for replies...appreciate it......thank you
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  5. #5
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    Quote Originally Posted by angel.white View Post
    Initial equation
    cosh(x)=i

    Take inverse
    x=arcosh(i)

    ...
    I did not check your answer. But this is a questionable step. The cosh function defined on \mathbb{C} does not have an inverse. The way Plato has it is how I would do it.
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