Why does i^i = 0.207879576?

Question

I was entering random things in my calculator and came across this. It makes no sense to me conceptually. How can the square root of negative one raised to the power of the square root of negative one be a positive decimal?

Viola Towne · 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 =

Viola Towne · Answer

you have to be introduced to the subject of complex variables and the complex logarithm to understand it.let z = i^inow take the complex loglog 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 kthe capital L version means the general case for all k,with lower case l the log uses the principle valuelet's look at this piece by pieceln[|w|] means the natural log of the magnitude of the complex number w. when w=i, |w| = 1, and ln(1)=0Arg(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/2thus we havelog i = i*Pi/2and ilog i = ii*Pi/2 = -Pi/2but 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

Viola Towne · 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.

Viola Towne · Answer

We have the identity:e^(ix) = cosx + isinxe^(iÏ/2) = cos(Ï/2) + isin(Ï/2) = 0 + i1 = iTake the natural log of both sides. i*Ï/2 = ln iMultiply both sides by i.iÂ²(Ï/2) = i ln i = ln(i^i)-1(Ï/2) = ln(i^i)-Ï/2 = ln(i^i)Exponentiate both sides.e^(-Ï/2) = i^ii^i = e^(-Ï/2) â 0.2078795

Dr. Summer Schneider PhD

Why does i^i = 0.207879576?

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