Evaluating e^(i*pi) and i^i
Date: 07/25/99 at 08:16:52 From: julie balen Subject: Investigating hyperbolic functions Hi, I need to evaluate e^ipi and i^i. I know that e^ipi - 1 = 0, and that this uses the 5 most important constants in mathematics, but I do not really understand how or why it is used, or what i^i is. I read somewhere that i^i has an infinite number of solutions, but how can that be? And is that true? Sorry to keep asking all these things, but they really interest me, and I need to know them for a project I am doing in school. Thank you so much! Julie
Date: 07/25/99 at 08:47:15 From: Doctor Jerry Subject: Re: Investigating hyperbolic functions Hi Julie, Do you mean e^(i*pi)? I'm easily confused, without parentheses. This can be done with the formula you gave, which is a special case of Euler's identity: e^(i*t) = cos(t) + i*sin(t). If t = pi, then e^(i*pi) = cos(pi) + i*sin(pi) = -1. Okay, that leaves the i^i question. The way to do these is to convert to polar coordinates and use logs. Let w be any complex number for which w = i^i. Writing w in polar form, which is r*e^(i*t), then ln(w) = ln(r) + i*t*ln(e) = ln(r) + i*t = i*ln(i) Writing i in polar form, i = 1*e^(i*pi/2) and so ln(i) = i*pi/2. So, ln(r) + i*t = i(i*pi/2) = -pi/2. So, t = 0 and ln(r) = -pi/2. So, w = r*e^(i*t) = e^(-pi/2) = 0.20787... There are other answers. You can get them by doing the above, but start with w = r*e^(i*(t+k*2pi)) where k is any integer. I took k = 0. - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/
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