Graduate How the value of ln(-1) is derived

[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?

 

[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.