Does $1 = 1 + 2i\pi$?

Question

I was playing around with a method of how to do negative logs and cane up with the following method using Euler's identity:

$$e^{i\pi} = -1$$ $$\therefore \ln{-1} = i\pi$$

So therefore the following can be used for negative numbers:

$$\ln{x} = \ln{-x} + i\pi$$

E.g.

$$\ln{-e} = \ln{e} + i\pi$$ $$\ln{-e} = 1 + i\pi$$

However, I decided to try to use this with $\ln{e}$:

$$\ln{e} = \ln{-e} + i\pi$$ $$\ln{e} = 1 + 2i\pi$$

Since $\ln{e} = 1$, does this mean that $1 = 1 + 2i\pi$? Or have I made a mistake at some point?

Answer 1

No, it means that the natural logarithm is multivalued. $\ln e$ is both $1$ and $1+2\pi i$ at the same time, along with a host of other values (specifically, $1+2\pi i n$ for all integers $n$). Similarly, $\ln(-1)$ is both $\pi i$ and $-\pi i$ at the same time.

That's why, in order to actually call it a function, you have to make a choice. Usually that the imaginary part should be in the interval $[0,2\pi)$ or that it should be in the interval $[-\pi,\pi)$. Once you've made that choice, $\ln$ is a function again, but you lose some properties like continuity and $\ln(ab)=\ln a+\ln b$.