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Why is i^i a real number?

  1. Nov 2, 2005 #1
    This is something that I've pondered for a while and I can't see a logical explanation for. I'll go ahead and demonstrate that it is in fact a real number before you guys think I haven't done my homework. :tongue2:
    Starting with the Euler formula:

    (1) [tex]e^{ix}=\cos(x)+i\sin(x)[/tex],

    and using [itex]\frac{\pi}{2}[/itex] for x, it follows that

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

    Now using that identity, [itex]i^i[/itex] can be expressed as:

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

    which is equivalent to:

    (4) [tex]e^{i^2\frac{\pi}{2}}=e^{-\frac{\pi}{2}}[/tex].

    This was all borrowed from Mathworld's site, thanks to them. Mathworld - i

    Now, can anyone provide an analytic explanation for this? I admit I haven't taken a course on complex analysis, so my skills with complex numbers are limited, but any insight would be appreciated.

    Many thanks,
    Last edited: Nov 2, 2005
  2. jcsd
  3. Nov 2, 2005 #2


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    eix = cos(x) + isin(x), not sin(x) + icos(x). I assume that's a typo because your equation (2) is correct. In general, zz' takes on many values, but you can choose a principal branch. By definition:

    zz' = ez'log(z)

    where log(z) = log|z| + iarg(z) by definition. The expression above has multiple values because log(z) does, and log(z) has multiple values because arg(z) does (arg(z) is b + 2k(pi) for every integer k). Choose arg(z) between 0 and 2(pi), then you fix a value for log, and hence for zz'. Working through the definitions, we get:

    ii = eilog(i) = ei(log|i| + iarg(i)) = ei(i(pi)/2) = e-pi/2

    which is what you have. Now what sort of analytic explanation are you looking for other than the fact that it follows from definition? The exponential function is defined by:

    [tex]e^z = \sum _{n=0} ^{\infty} \frac{z^n}{n!}[/tex]

    very similar to what it is for real numbers. Also, defining the derivative of a complex function in a way identical to how you define it for real functions:

    [tex]f'(z) = \lim _{h \to 0} \frac{f(z+h) - f(z)}{h}[/tex]

    the exponential function is also the solution, f, to the IVP:

    [tex]f'(z) = f(z),\ f(0) = 1[/tex]

    like it is for reals. This should give you a good reason as to why the complex exponential is defined as it is. log is defined such that elog(z) = z, as you'd want, and zz' is defined such that zz' = ez'log(z) as you'd want. So all these definitions are very natural. And a consequence of these definitions is that ii is real. Does that make sense?
  4. Nov 2, 2005 #3


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    Also, maybe you can graph f(z) = zi and f(z) = iz for z purely imaginary, z purely real, |z| = 1, Im(z) = k for some fixed real k, Re(z) = k for some fixed real k. Maybe this will give you a sense of why ii fits in where it does.
  5. Nov 6, 2005 #4


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    geometrically, e^iz squashes the top half of the complex plane down into the infinite horizontal strip between the x axis and y = -1, and extends the lower half of the complex plane down below y = -1, so that the whole former plane lies in the lower half plane.

    then it rotates the lower half plane 90degrees cc, so that it becomes the right half plane.

    then it roitates the right half plane around and around the origin, so that the vertical line x = 1 goesa orund the unit circle, and lines between x = 0 and x=1 go around concentric smaller circles, and vertical olines to the right of x = 1 go around larger concentric circles.

    thus you see that all orginally horizontal lines in the upper half complex plane, such as y = pi/2, where i pi/2 lies, wind around the origin on circles.

    thus some points on these lines such as i pi/2, muct cross the x axis, as these circles go around and around the origin.

    i pi/2 is one of these points, so e^i[i pi/2] = [e^{i pi/2}]^i = i^i winds up on the real axis.
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