I am just learning about Euler's identity. When messing around with it, I am getting some unsettling identities, which, I believe, is probably due to my lack of application of certain rules.
Starting with:$$e^{\pi i}=-1$$ Is this statement true: $$\pi i = \ln(-1)$$ $$\to \ln(-1*n)= \pi i+\ln(n)$$ Also this one clearly does not hold up: $$\ln(1)=0$$ $$\ln(1)=\ln(-1*-1)= \ln(-1)+\ln(-1)=\pi i + \pi i$$ $$\to 2\pi i=0 \to2=0$$ Please advise me on what I am missing.
Like square root, log is treated as a "multi-valued function." It's just that when we restrict to the positive real line, we can make it single-valued. (Just as we do with square root in algebra 1.)
So the first few things you derived are right: one of the logarithms of $-1$ is $\pi i$. The others all differ from it by multiples of $2\pi i$. So when you find that $\log(1) = 0$ and $\log (1) = 2\pi i$, that's fine: it just says that there are many values of the log. (Sorta like writing $1^\frac{1}{2} = 1$ and $1^{\frac{1}{2}} = -1$, which each mean that the thing on the right, squared, gives you $1$.)
To say this a bit more elegantly: there is no single log function defined on the whole complex plane (except the origin) that's continuous on the whole complex plane (except the origin), just as there's no continuous square root function defined on the whole complex plane.
By long tradition, we agree to not try to make log continuous on the negative real axis; we can then define a single consistent logarithm that's continuous everywhere except the negative real axis (and 0, where it's undefined). This single consistent log, for which $\log(1) = 0$, is called the "principal branch" of the logarithm. It's sometimes denoted $\mathrm{Log}$, I believe.
There's another perfectly good logarithm-like function, which I'll call $\widehat{\log}$: it's defined by
$$ \widehat{\log}(x) = \log x + 2 \pi i $$ It has all the properties of $\log$ except that $\widehat{\log}(1) = 2\pi i$ instead of $0$.
You can also define other log-like things, by adding $4 \pi i$, $-6 \pi i$, etc. to $\log$.
Taken together, we get something for which $$ \log(ab) = \log(a) + \log(b) $$ where here "log" denotes any of the functions I mentioned above, holds in a kind of limited sense: if you add $\log(a)$ and $\log(b)$, you'll get $\log(ab)$...but only up to multiples of $2 \pi i$. That is to say: $$ \log(ab) - \log(a) - \log(b) = 2\pi i ~n $$ for some integer $n$.
The first identity is true. Pretty neat, right?
However, in general for the complex extension of the logarithm $$ \log(ab) \ne \log(a) + \log(b) $$ which clears up the other troubling identities you arrive at.
The log function is constructed from the exponential function as follows: You want a function $\log(z)=u(x,y) + iv(x,y)$ such that $z = e^{u+iv} = e^u e^{iv}$ so that $e^u = \lvert z \rvert \implies u = \ln{\lvert z \rvert}$ and $v=\arg(z)$ so that we have $$ \log(z) = \ln{\lvert z \rvert} + i \arg({z}) $$ Note that we do not have a restriction on the range of $\arg(z)$ so that we have a multivalued function. This implies that $\arg(ab) \neq \arg(a) + \arg(b)$ in general so that in general $$\log(ab) = \ln{\lvert ab \rvert} + i \arg(ab) \neq \ln{\lvert a \rvert \lvert b \rvert} + i(\arg(a)+\arg(b)) = (ln {\lvert a \rvert} + i \arg(a)) + (\ln(\lvert b \rvert) + i \arg(b) ) = \log(a) + \log(b).$$
