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Thread: real number value for imaginary unit

  1. #1
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    real number value for imaginary unit

    Since $\displaystyle \cos(i)$ is a real number (1.54308...)

    Does this mean that I can substitute "i" as follows?

    $\displaystyle i=\arccos(1.54308...)$

    If not, why not?


    Also, since $\displaystyle i^i=\frac{1}{e^\frac{\pi}{2}}$ does this mean that

    $\displaystyle i=\arccos(1.54308...)=\left(\frac{1}{e^\frac{\pi}{ 2}}\right)^\frac{1}{i}$

    ?
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  2. #2
    Junior Member Dark Sun's Avatar
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    The first one follows. If $\displaystyle \mbox{cos }(i)=1.54308...$, then $\displaystyle i=\mbox{arccos}(1.54308...)$

    The second equation follows as well. Provided that $\displaystyle i^i=\frac{1}{e^{\pi /2}}$, then by taking the $\displaystyle \frac{1}{i}$ power of both sides of the equation, we have:

    $\displaystyle (\frac{1}{e^{\pi /2}})^{\frac{1}{i}}=i=\mbox{arccos}(1.54308...)$.

    I'm not sure what you are going to use these facts to solve, but each of these steps are correct. Are you certain that $\displaystyle \mbox{cos }(i)=1.54308...$? Are we speaking in terms of hypothetical? Imaginary numbers are tricky, it would help if I knew where you were coming from.

    I suspect that each of these steps are correct, but that the premise may be in error.
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  3. #3
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    According to wikipedia's page on the imaginary unit:

    $\displaystyle \cos(i) = \cosh(1)= (1.54308...)$

    The page doesn't show how this is derived.

    I don't have a particular use in mind for this info. I was just cruising wikipedia and this jumped out at me.

    Even if cos(i) is a real number, I guess this does not give us a real number value for "i" itself. Acos(cos(i)) does not exist, or is imaginary.

    Same with $\displaystyle i^i$.
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