Why does i^i = 0.207879576?

Question

I was entering random expressions in my calculator and came across the result of ( i^i ), which is approximately 0.207879576. This result confuses me conceptually. How is it possible that the square root of negative one, raised to the power of the square root of negative one, yields a positive decimal? Could someone help clarify the mathematics behind this?

Anonymous · Accepted Answer

a^x = e ^(x log a)so i^i = e^(i log i)log(z) is by definition log (|z|) + i argument(z)where z is a complex number so log(i) = log(|i|) + i  argument (i) = 0 + ipi/2 then ilog(i) = i  i * pi/2 = -pi/2and thus i^i = e^(i ln i) = e^pi/2 =

Taurean Bahringer · Answer

you have to be introduced to the subject of complex variables and the complex logarithm to understand it.
let z = i^i
now take the complex log
log z = i*log(i)
(same rule as real logs applies here log a^b = b*log a)
now the complex log is defined as:
Log[w] = ln[|w|]+i(Arg[w]+2Pi*k), for any integer k
the capital L version means the general case for all k,
with lower case l the log uses the principle value
let's look at this piece by piece
ln[|w|] means the natural log of the magnitude of the complex number w. when w=i, |w| = 1, and ln(1)=0
Arg(w) means the angle with respect to the positive real axis the complex number w makes. i is 90 degrees from it, but it must be in radians: Pi/2
thus we have
log i = i*Pi/2
and ilog i = ii*Pi/2 = -Pi/2
but we want to know what i^i is and we so far calculated the log of i^i, so all we have to do is take the antilog of this (i.e. raise e to the power of log(i^i) since e^(log w) = w),
so i^i = e^(log(i)) = e^(-Pi/2)
and you can evaluate this on your calculator and get:
approximately .2079

Anonymous · Answer

Complex numbers do strange things.  The formula e^ix = cos x + i sin x (if I have remembered it right) obviously leads to a real result for appropriate values of x.  Also, we have e^i pi = -1. and e^pi is a real number.

Anonymous · Answer

We have the identity:e^(ix) = cosx + isinxe^(i&Iuml;&#128;/2) = cos(&Iuml;&#128;/2) + isin(&Iuml;&#128;/2) = 0 + i1 = iTake the natural log of both sides.i*&Iuml;&#128;/2 = ln iMultiply both sides by i.i&Acirc;&sup2;(&Iuml;&#128;/2) = i ln i = ln(i^i)-1(&Iuml;&#128;/2) = ln(i^i)-&Iuml;&#128;/2 = ln(i^i)Exponentiate both sides.e^(-&Iuml;&#128;/2) = i^ii^i = e^(-&Iuml;&#128;/2) &acirc;&#137;&#136; 0.2078795

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Why does i^i = 0.207879576?

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