1. e^-pi/2

Using the identity a^b = e^{b ln(a}), we see that iⁱ = 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^ii... 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^z1 = iⁱ = e^-pi/2, and then find z3=i^z2, 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