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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/ 
Associated Topics:
High School Imaginary/Complex Numbers

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