Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10 5/3/83; site houem.UUCP Path: utzoo!linus!decvax!mcnc!unc!ulysses!mhuxl!houxm!houem!agd From: agd@houem.UUCP (A.DEACON) Newsgroups: net.math Subject: More on i**i Message-ID: <229@houem.UUCP> Date: Mon, 13-Feb-84 10:32:50 EST Article-I.D.: houem.229 Posted: Mon Feb 13 10:32:50 1984 Date-Received: Tue, 14-Feb-84 01:53:02 EST Organization: Bell Labs, Holmdel NJ Lines: 39 In response to Seaman and Rentsch: The value of i^i is very well defined. I gave the values in a previous article and they are: i -(4n+1)*pi/2 i = e for all integer n. There is no problem in defining log(i) either: for any complex number z ln(z) = ln|z| + i(theta + 2n*pi) for all integer n where -pi < theta <= pi and |z| is the modulus of z. Theta is called the principle argument of the ln. As you can see there are an infinite number of values for ln(z). Of course the ln function cannot be extended continuously to the entire complex plane because of the pole. However, if you consider the Riemann surface, it can be. In the exp(x) sum, we need x=i*ln(i) to compute i^i, so it does require something "funny". Being clever has nothing to do with it. That's the way it is. For Seaman: i -(2n+1)*pi (-1) = e for all integer n and i -(4n-1)*pi/2 (-i) = e for all integer n. Art Deacon AT&T Bell Labs