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Taking a number to a complex number power

  1. Dec 16, 2011 #1
    How do you compute the following?

    2it where t is a real number

    while I am at it, how do you compute powers that are not integers

    ie: 23.14
  2. jcsd
  3. Dec 16, 2011 #2
    2^(it) = exp(it(log2)) = exp(i(tlog2)) = cos(tlog2) + isin(tlog2)

    I guess we would define

    2^(3.14) = exp(3.14(log2))

    I'm not sure if i could write it in any other way, and i doubt i could compute that in my head but they are equivalent.

    Sorry for the bad formatting if it isn't clear.

    Edit: These logs are base e.
    Last edited: Dec 16, 2011
  4. Dec 16, 2011 #3


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    "3.14", as opposed to [itex]\pi[/itex], is a rational number. It is, in fact, [itex]314/100= 157/50[/itex] so [itex]2^{3.14}= 2^{157/50}= \sqrt[50]{2^{157}}[/itex].

    [itex]\pi[/itex] is not rational but there exist a sequence of rational numbers that converge to it (the sequence 3, 3.1, 3.14, ..., for example). [itex]2^\pi[/itex] is equal to the limit of the sequence [itex]2^3[/itex], [itex]2^{3.1}= 2^{31/10}= \sqrt[10]{2^{31}}=[/itex], [itex]2^{3.14}= 2^{314/100}= \sqrt[50]{2^{157}}[/itex],...

    Of course, to actually calculate those things you would use [itex]2^a= e^{a ln(2)}[/itex] as The1337gamer said.
  5. Dec 17, 2011 #4
    Thank you all, both addressed all of my issues. Halls of Ivy: where can I find more about this sequence that converges to Pi? What is the formula for each term?
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