### 3 Answers

- 6 years agoFavorite Answer
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)

- PhilipLv 66 years ago
The way I went about this problem many years ago was to develop the general formula for Z1 to the power Z2

where Z1 = r1exp(itheta1), Z2 = r2exp(itheta2). The question I posed was obviously just a special case of the

more general problem expressed here. As a student I generally preferred to develop the most general formulae

possible. I intuitively feel that you, Gavin, would share that preference. If so, I'm sure you'll have fun with the

general problem above.

- Anonymous5 years ago
i dont ******* know