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Computing i^i

  1. Jul 27, 2011 #1
    To compute this, we’ll make use of Euler’s formula cis(x) = eix = cos(x) + i·sin(x):

    ei(π/2) = cos(π/2) + i·sin(π/2) = i, and exponentiating by i we get
    ii = (ei(π/2))i = ei·i(π/2) = e-π/2 ∈ ℝ.

    But we also have cis(2πk + π/2) = i, k ∈ ℤ. Thus by the same logic, we get ii = e-(2πk + π/2) for k ∈ ℤ.

    Infinitely many evaluations, so ii doesn't have a distinct/unique value? This seems like a discrepancy. I don't have any strong arguments, but I have this W|A computation telling me that ii is indeed distinct, since none of e-5π/2, e-9π/2, e-13π/2, etc are equal.

    Also, since the exponential function from ℝ onto ℝ+ is certainly injective, each of these numbers e-(2πk + π/2) must be distinct. If ii is distinct, why is its value e-π/2 = 0.2078795... ?

    Can anyone explain this?
     
    Last edited: Jul 27, 2011
  2. jcsd
  3. Jul 27, 2011 #2
    Well, there's nothing surprising about that. xy is by definition ey log(x), and log is a multivalued function. Except for integer y, this always leads to multiple possible values of xy. One usually resolves this by the somewhat artificial dodge of using the principal branch of log y, defined by convention, (just as [itex]\sqrt{y}[/itex] is taken to mean the positive square root of y, even though there is also a negative one).
     
  4. Jul 27, 2011 #3
    That's true, in that case I'll accept that ii cannot be uniquely evaluated. But I've seen in other places (including on this forum) that ii has a distinct value. This is clearly untrue by the logic in the OP post. What's the discrepancy?
     
  5. Jul 27, 2011 #4
    It does have a unique value -- if you choose the principal branch. (Which is to say, it has a unique value if you define it to.)
     
  6. Jul 27, 2011 #5

    micromass

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    Indeed, the complex exponential is defined as

    [tex]x^y=e^{yLog(x)}[/tex]

    The log is multivalued, so there will be an infinite number of possible [itex]x^y[/itex]. But in many applications, we want a principal value of [itex]x^y[/itex]. This is done by the formula

    [tex]x^y=e^{yLog(x)}[/tex]

    But where the Log is now the principal branch and is single-valued. That is, we restricted the range of the Log such that it becomes single-valued.

    This principal value of [itex]x^y[/itex] pops up in many places, for example in the definition of the Riemann-zeta function:

    [tex]\sum{\frac{1}{n^s}}[/tex]

    where the s can be complex.
     
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