Results 1 to 7 of 7

Math Help - i^i

  1. #1
    Junior Member hoeltgman's Avatar
    Joined
    Aug 2005
    Posts
    43

    i^i

    I read somewhere that:
     i^i = e^{-\frac{\pi}{2}}
    Where i is the complex unit number
    There wasn't any further explanation on this. Only a short developpment.
    I don't have any problem with the developpment. But what puzzles me is the fact the the complex power of a complex number is in fact a real number.
    Does anybody have a logical explanation for this. I think it's quite a unique phenomenon.
    If you take a number out of R and take its power, it will be a number in R again and not Q.
    \pi^\pi \text{\; is in R for example}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Check out this mathworld site: http://mathworld.wolfram.com/i.html

    Basically you start out with the Euler formula e^{ix}=\cos(x)+i\sin(x). When you test the value \frac{\pi}{2}, you simply get i, so now we can say that i=e^{i\frac{\pi}{2}}. So, i^i=(e^{i\frac{\pi}{2}})^i. And simple power rules show from here that it reduces to e^{-\frac{\pi}{2}}

    It makes sense after I go through all of the steps, but I agree I don't see a logical explanation of why this works the way it does. The mysteries of math.

    Jameson
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Does anybody have a logical explanation for this. I think it's quite a unique phenomenon.
    If you take a number out of R and take its power, it will be a number in R again and not Q.
    How about (-1/2)^(-1/2)?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Quote Originally Posted by CaptainBlack
    How about (-1/2)^(-1/2)?
    \left(-\frac{1}{2}\right)^{-\frac{1}{2}}... touché!

    Jameson
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member Rebesques's Avatar
    Joined
    Jul 2005
    From
    At my house.
    Posts
    534
    Thanks
    8

    Unhappy

    If you take a number out of R and take its power, it will be a number in R again and not Q.
    You mean C, and you are altogether right.

    Consider the function f(z)=z^i. Its graph lies in C ^2. And, it happens to cross the real axis for z=i. Nothing extraordinary here, just a complex stranger, coming from C to the house of the reals for tea. and returning to his imaginary field.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    i^i=e(-pi/2) derived from Euler's Formula:
    e^(ni)=cos(n)+isin(n).
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Nov 2005
    Posts
    21
    Here's another way to think of this:

    For all complex a and b, a^b:=e^(b*Ln(a)), where ln(a)=ln|a|+i*Arg(a)+2n*i, n is an integer, and the principal value, Ln(a), is the value for which n=0.

    Anyway, Ln(i)=ln(1)+i*Pi/2=i*Pi/2, so i^i=e^(-Pi/2)


    As for an example of irrational a and b such that a^b is rational, just consider (2^Sqrt(2))^Sqrt(2)=4.
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum