What is #(1+i)^i# ?
1 Answer
Nov 19, 2017
for any integer
Explanation:
Note that:
#1+i = sqrt(2) e^((2npi + pi/4) i) = e^(1/2 ln(2) + (2npi + pi/4) i)" "# for any integer#n#
So:
#(1+i)^i = (e^(1/2 ln(2) + (2npi + pi/4) i ))^i#
#color(white)((1+i)^i) = e^(1/2 ln(2)i - (2npi + pi/4))#
#color(white)((1+i)^i) = e^(- (2npi + pi/4)) (cos(1/2 ln(2)) + i sin(1/2 ln(2)))#
for any integer