Find i^i , where i = radical(-1)

Suprisingly i^i is real

From Euler's equation
e^(xi)=cos(x)+isin(x) with x in radians
it can be shown that when x=pi
e^(pi*i)=cos (pi)+isin(pi)=-1+0*i=-1
taking the square root of e^(pi*i)=-1
sqrt(e^(pi*i))=sqrt(-1)=i
and taking the natural log
ln(i)=ln(sqrt(e^(pi*i))=(1/2)ln(e^(pi*i))=(1/2)(pi*i)=(pi*i)/2

now if you take the logrithm of i^i you get
ln(i^i)=i*ln(i)
substituting in ln(i)=(pi*i)/3
i*ln(i)=i*(pi*i)/2=(pi*i^2)/2=-pi/2
take antilog of i*ln(i)=ln(i^i)=-pi/2 then
i^i=e(-pi/2)

Neat huh!

Rap c∫;?/

i^i = e^(i*log(i))
= e^(i* ((pi)/2)*i) (Principal value)
= e^(-(pi/2)).

Yeah, neat enough.

VERY NEAT!!! THX A LOT!!