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Raising things to i power, I came across 4.8104773809655

  1. Jun 10, 2005 #1
    I was working on the general meaning of taking things to the i power -- I was pondering the meaning of

    e^pi*i = -1

    The proof for this thing is well established, I was musing about meaning. An obvious question was:

    What to the i power goes to zero?

    As I was hunting for that number (on my TI-83 while driving to work -- and it is very hard to press [2nd] i on that thing with one hand), I came across this number:



    4.8104773809655^i => i

    ...at least on my TI-83. 4.81047738096535^i on the google calculator.

    What is this number, and who has done work on it?


    Steve Rives
    Last edited: Jun 10, 2005
  2. jcsd
  3. Jun 10, 2005 #2


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    [tex]x^i = i[/tex]

    The solutions for x are:

    [tex]e^{\frac{\pi}{2} + n\pi} \quad \forall n \in \mathbb{Z}[/tex]

    So if you hadn't already guessed:

    [tex]e^{\frac{\pi}{2}} \approx 4.8104773809653516554730356667 \ldots[/tex]

    As for the equation:

    [tex]i^x = 0[/tex]

    Well there exists no complex solution for x. In fact the only time:

    [tex]a^b = 0[/tex]

    For a and b in complex numbers is when [itex]a=0[/itex] and [itex]b \neq 0[/itex].
    Last edited: Jun 10, 2005
  4. Jun 10, 2005 #3
    Of coruse! Only, I found it the hard way just now:

    ln(4.81047738...) is 1.5707963

    And that's pi/2
  5. Jun 10, 2005 #4


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    he did not ask when i^x = 0, he asked (what)^i -->0?

    so we might look at a^i = e^(i ln(a)) = cos(ln(a)) + i sin(ln(a)).

    but again it is clear that this number has absolute value 1, so cannot approach zero, at least not for real a.

    now for complex a, we just need to solve for when e^z goes to 0.

    but e^(-n)-->0 for example, so therefore e^i(in) -->0 too. so x = in satisfies

    e^(ix) goes to zero as n goes to infinity, i.e. as x goes to i.infinity.
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