Natural Log of negative number

1. Dec 15, 2007

BackEMF

I've got a simple question that's been bugging me for a while. I think I know where the problem is, I'd just like a formal mathematical reason why I can't say this:

$$\ln{(-1)}^2 = \ln(1) = 0$$

That part is fine...but then:

$$\ln{(-1)}^2 = 2 \ln(-1) = 2 (i \pi)$$

when they should obviously be equal.

It presumably displays the fact that you can't take the square and put it in front of the log as a "2" when you're dealing with logs of negative numbers. I'd like to know if there's any formal theory behind why this can't be done.

Thanks!

2. Dec 15, 2007

Ben Niehoff

What you're running into is the fact that $\ln z$ is multi-valued. Define $\ln z$ as the contour integral

$$\ln z = \int_{\gamma} \frac{d\zeta}{\zeta}$$

where $\gamma$ is some contour running from 1 to z.

Now, note that for each time the contour $\gamma$ winds around the origin in a positive sense, you get $2\pi i$ added to the integral (you can show this by the method of residues). Therefore, $\ln z$ is multivalued:

$$\ln z = \ln_p z + 2n\pi i$$

where $\ln_p z$ is the principle value, and $n$ is any integer.

3. Dec 15, 2007

mathman

A simple way of seeing this is noting that exp(2n[pi]i)=1. Therefore ln(1)=2n[pi]i where n is any integer.

4. Dec 15, 2007

BackEMF

Ah great! That makes perfect sense.

So the natural log of 1 is, infact, multivalued if you allow a mapping to the complex plane?

Thanks for cleaning that up anyway Ben & mathman.