Natural Log of negative number

[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!

 

[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.

 

[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.

 

[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.