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Question about i^i

  1. Dec 17, 2009 #1

    fedaykin

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    Why is i^i = 0.2078 ?
     
  2. jcsd
  3. Dec 17, 2009 #2
    you need to know about the complex log function. it's because [tex]i^i = e^{iLog(i)} = e^{i(ln|i| + iArg(i))} = e^{i(0 + i\pi/2)} = e^{-\pi/2}[/tex] (it's weird that [itex]i^i[/itex] is a real number)
     
    Last edited: Dec 17, 2009
  4. Dec 17, 2009 #3

    jgens

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    Adding to what fourier jr noted, [itex]e^{-\pi/2}[/itex] is one solution for [itex]i^i[/itex], but there actually exist infinitely many such solutions.

    Edit: To give a more thorough (and perhaps student friendly explanation) about why [itex]i^i[/itex] you'll need to know Euler's formula, which states that [itex]e^{ix} = \cos{(x)} + i\sin{(x)}[/itex]. If you're unfamiliar with the formula, you can derive it relatively easily using the power series for the sine, cosine, and exponential functions. Another approach (my personal favorite) uses an argument involving differential equations. If you happen to be interested in these proofs, you can find them here: http://en.wikipedia.org/wiki/Euler's_formula. Moving on to the problem in question, we'll begin with the equality

    [tex]i^i = e^{i\log{(i)}}[/tex]

    Thus, we've already reduced the problem to evaluating [itex]\log{(i)}[/itex]. Now, suppose that [itex]\log{(i)}[/itex] has a solution in the field complex numbers, then

    [tex]\log{(i)} = a + bi[/tex]

    for [itex]a,b \in \mathbb{R}[/itex], and consequently,

    [tex]i = e^{a+bi} = e^a e^{bi} = e^a[\cos{(b)} + i\sin{(b)}][/tex]

    Since the real part of [itex]\cos{(b)} + i\sin{(b)}[/itex] must necessarily be zero, this means that [itex]b = \pi/2 + k\pi[/itex] for [itex]k \in \mathbb{Z}[/itex]. We can further refine the appropriate values for [itex]b[/itex] by noting that [itex]\sin{(b)} > 0[/itex] and conclude that any [itex]b[/itex] satisfying [itex]b = \pi/2 + 2k\pi[/itex] should suffice.

    Alright, now for any choice of [itex]b[/itex] we have that [itex]\sin{(b)} = 1[/itex] and consequently [itex]i = ie^a[/itex]. This implies that [itex]e^a = 1[/itex] and similarly that [itex]a = 0[/itex]. Putting all this information together we find that,

    [tex]\log{(i)} = i[\frac{\pi}{2} + 2k\pi][/tex]

    Substituting this into our first equation we have

    [tex]i^i = e^{i\log{(i)}} = e^{-[\frac{\pi}{2} + 2k\pi]}[/tex]

    And finally, setting [itex]k = 0[/itex] gives us the principal solution

    [tex]i^i = e^{-\pi/2}[/tex]

    Hopefully you'll find this helpful (and hopefully it's not riddled with too many mistakes!).
     
    Last edited: Dec 17, 2009
  5. Dec 17, 2009 #4
    [tex]
    e^{i\pi} = -1
    [/tex]

    [tex]
    \sqrt{( e^{i\pi} )} = e^{i \pi/2} = i

    [/tex]
    Now raise to the i.

    [tex]
    (e^{i \pi/2})^i = e^{i^2 \pi/2} = e^{-\pi/2} = i^i
    [/tex]
     
  6. Dec 17, 2009 #5
    Nice, l'Hopital!
     
  7. Dec 17, 2009 #6

    jgens

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    Wow! That's a neat proof (and considerably more elegant than my brute force approach).
     
  8. Dec 17, 2009 #7

    Char. Limit

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    I must admit, that proof is one of the best I've ever seen of what [tex]i^i[/tex] equals. (And [tex]e^{i\pi}+1=0[/tex] is my personal favorite math equation, so I like that you used it.)
     
  9. Dec 18, 2009 #8

    arildno

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    Too bad..

    Here we go:
    [tex]i=e^{(2n+\frac{1}{2})\pi{i}),n\in\mathcal{Z}\to{i}^{i}=e^{-(2n+\frac{1}{2})\pi}[/tex]
    Infinite values, that is..
    It is not an equation, but an equality.
     
  10. Dec 18, 2009 #9

    Char. Limit

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    Too bad? What do you mean? It wasn't even my proof. Explain, please.

    Not so. For if Euler's Identity were not an equation, how could it also be known as Euler's Equation?

    Also, since Merriam-Webster states that:

    And since the equation (or equality, if you prefer) [tex]e^{i \pi }+1=0[/tex] is stating the equality of the left and right sides, and they all seem to be mathemetical expressions, I would say that Euler's Identity is an equation. It is also an equality since, I believe, all equalities are equations, once formally stated. It is also an identity. What it is not is a formula (and I've been berated before when I called it a formula, so I'm only agitated because my previous berator told me to call it an equation).
     
  11. Dec 18, 2009 #10
    I think that an equation is an equality involving expressions; however, an equality is a bit more involved. For example, the two sets {1,2,3} and {2,3,1} might be considered the same, or equal, if order does not matter.
     
  12. Dec 19, 2009 #11

    arildno

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    An equation is a sub-type of an equality, in which a set of argument values are implied (or best, SPECIFIED) (for example, the real numbers), for which only a sub-set of measure zero contains solutions to the equality.

    For example, in the equation x+y=0, our implied set of argument values is the (x,y)-PLANE, whereas the solution set is a line, i.e, a subset that has measure zero relative to the plane.

    An identity is an equality that holds for ALL values within the implied set.
    For example, x+x=2x is an identity, however we choose our set of argument values (for example, for the integers, the reals or the complex numbers).

    Any other types of equalities are..equalities (although on occasion, the above expression is called Euler's identity, but NEVER Euler's equation).
     
  13. Dec 21, 2009 #12

    Char. Limit

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    I really must remember not to state anything on this forum unless I know it is true and always true, apparently...

    If it's only occasionally referred to as Euler's Identity, why is that the article's TITLE? (I can use all caps too)

    And if it's never called Euler's Equation, why don't you prove that, with cited sources, like that Wikipedia article, which you definitely, deliberately, and undeniably contradicted, uses?

    I prefer to believe an encyclopedia that cites the sources it uses to a single person who doesn't, no matter how many "awards" he or she has.

    However, since you've been here for far longer than I have, and have connections that I don't, I'm probably just going to be banned instead. (Cynicism is strong in me...)
     
    Last edited: Dec 21, 2009
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