The Natural Log of -1
Date: 12/13/2004 at 13:53:05 From: Derek Subject: The natural log of -1 I was playing around with my graphing calculator, and I found out that the natural log of -1 is equal to pi*i. Can you explain why ln(-1) = pi*i?
Date: 12/13/2004 at 14:57:07 From: Doctor Schwa Subject: Re: The natural log of -1 Hi Derek, Here are two answers from our archives that might help: Log of a Negative Number http://mathforum.org/library/drmath/view/61830.html Proof of DeMoivre's Theorem http://mathforum.org/library/drmath/view/53836.html The basic point is that (cos theta + i * sin theta)^n = cos (n*theta) + i*sin(n*theta) which is rather like an exponential: (something^theta)^n = something^(n*theta). It turns out that the something is exactly e^i ... wow! So e^(i theta) = cos theta + i sin theta, and from that, since cos (pi) = -1 and sin (pi) = 0, e^(i pi) = -1. I hope that helps clear things up! - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/
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