#i ^i = # ?

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Answer 1 · 2016-06-25T16:01:53

#i^i=0.20788#

Any complex number can be written as

#x +i y = (sqrt(x^2+y^2))e^{i phi}# with #{x,y}in RR^2#

where #phi = arctan(y/x)#

then

#(x +i y)^{x+iy} equiv ((sqrt(x^2+y^2))e^{i phi})^{(sqrt(x^2+y^2))e^{i phi}}#

making #x = 0, y = 1#

#i^i = (e^{i pi/2})^{e^{i pi/2}} = (e^{i pi/2})^i = e^{-pi/2} = 0.20788#