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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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