✔ 最佳答案
i^i=e^(ilni) from e^ix=cosx+isinx then lni=pi*i/2+2kpi(so there are technically infinite answers to this problem) since sinx=1 and cosx=0 like how the sqrt(4) could be -2 or 2 but be take 2 as the value we generally take i*pi/2 as the principal value in accordance with the ln(z) formula=lnmod(z)+iargz here=ln(1)+ipi/2 you can get that formula through expressing z=Re^ix(where R is the modulus and x the argument) and natural logging both sides and using the rules of logs so i^i is generally defined to be e^(ipii/2)=e^(-pi/2)=0.208(3dp)