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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:

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


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

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?


marked as duplicate by Patrick Stevens, Asaf Karagila, Community Apr 2 '17 at 9:57

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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 $\ln(ab)=\ln a+\ln b$.

  • $\begingroup$ So this is the same as how $\sqrt{4} = 2 \text{ and } -2$, but $2 \neq -2$ $\endgroup$ – Beta Decay Aug 22 '17 at 13:06
  • 2
    $\begingroup$ More or less, yes. On the positive reals, with logarithms as well as square roots, we have decided that we shall use the one value which is real and positive. For all other inputs, there is no single choice which sticks out as nicer than the others. $\endgroup$ – Arthur Aug 22 '17 at 13:17

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