# Ln (-1)

1. Jan 18, 2012

### Benn

I saw a post discussing (-1)(1/3). After I saw it, I was a little confused about how the value of ln(-1) is derived. I could see from euler's identity that it should equal i*π; however, I'm not sure where the multiplicity (2k+1) comes from. I searched the forums (and google) but couldn't find anything. Could anyone clear this up for me?

2. Jan 18, 2012

### micromass

Well, saying that $ln(-1)=(2k+1)i\pi$ for $k\in \mathbb{Z}$ simply means (by definition) that

$$e^{(2k+1)i\pi}=-1$$

Why does this hold? Well, by definition of the complex exponential, we have

$$e^{(2k+1)i\pi}=\cos((2k+1)\pi)+i\sin((2k+1)\pi)$$

From elementary trigonometry, we know that $\cos((2k+1)\pi)=-1$ and $\sin((2k+1)\pi)=0$. So we indeed see that the above expression equals -1.

However: most authors don't like the logarithm to take on multiple values. They want one value for the complex logarithm. That's why you will always see $ln(-1)=i\pi$. We don't mention the other values because we only want one answer for ln(-1). We say that $i\pi$ is the principal value of the logarithm.