1. e^{-pi/2}

Using the identity a^{b} = e^{b ln(a}), we see that i^{i} = e^{i ln(i}).

ln(i) is i pi/2, because e^{i pi/2} = cos(pi/2) + i sin(pi/2) = i

So e^{i ln(i}) = e^{i2 pi/2} = e^{-pi/2}

2. Approximate value: z = 0.4382829367270321 + 0.3605924718713855i

If z=i^{i}^{i}^{.}.. then i^{z}=z (remember, exponentiation is right-associative)

In general, the solution to a^{z}=z, where a is any complex constant, is z=W(-ln(a))/(-ln(a)), were W is the Lambert W-function. Look up the Wikipedia article on the Lambert W function for details, or just memorize the key fact that W(a)=b means a=be^{b}, and use that fact to solve, as follows:

a^{z} = z

a^{-}z = 1/z

-z a^{-}z = -1

-z e^{-z ln(a}) = -1

-z ln(a) e^{-z ln(a}) = -ln(a)

-z ln(a) = W(-ln(a))

z = W(-ln(a))/(-ln(a))

So, for our problem, in which -ln(i) = -i pi/2,

z = W(-i pi/2)/(-i pi/2)

z = (i 2/pi) W(-i pi/2)

Unfortunately, that leaves you to evaluate the W function, which Mathematica calls "ProductLog". If your calculator does this, great, you will have the approximate value: z = 0.4382829367270321+ 0.3605924718713855i.

But if your calculator doesn't go there, then there's a way to zero in on a solution.

Using the fact that a^{b} = e^{b ln(a}), we can start with z1 = i, z2=i^{z}1 = i^{i} = e^{-}pi/2, and then find z3=i^{z}2, etc.

z1=0+1i

z2=0.207879576350761+0i

z3=0.947158998072379+0.320764449979308i

z4=0.0500922361093191+0.602116527036004i

z5=0.387166181086114+0.0305270816054832i

z6=0.782275682433955+0.54460655765799i

z7=0.142561823163664+0.400466525337087i

z8=0.519786396407854+0.118384196415812i

z9=0.568588617271898+0.605078406797807i

z10=0.242365246825208+0.301150592071316i

z11=0.578488683377103+0.231529735306831i

z12=0.427340132691626+0.548231217343565i

z13=0.330967104357644+0.262891842794588i

z14=0.574271015390431+0.328715623630499i

...

z37=0.438703698350747+0.370465460404472i, the first approximation of z within .01 of the true value.

...

z = 0.4382829367270321+0.3605924718713855i