${{2\pi i=0}}$? I was trying to solve for $i^i$.  I got distracted and did the following:
$e^{i \theta} = \cos(\theta) + i\sin(\theta)$
$e^{i 2\pi} = cos(2\pi) + i\sin(2\pi) = 1$ ||| Take ln() of two sides of equation:
$\ln(e^{i2\pi}) = \ln(1)$
$i2\pi = 0$
Clearly this is false. What did I do wrong?
 A: You cannot conclude that $2i\pi=0$.
If $f$ is an injective function, you can indeed argue that $f(x)=f(y)$ implies $x=y$. However the exponential function is not injective. Indeed, it is periodic with period $2i\pi$:
$$
e^{z+2i\pi}=e^z
$$
for every complex $z$. The complex logarithm is a wild beast, compared to the standard functions: it cannot be the inverse of the exponential exactly because the exponential is not injective; it's an example of a multivalued function and cannot be used like standard functions.
A: You should ask yourself, if you have understood the definitions of all the relevant objects?
It turns out, that defining $\ln$ is not so easy to do for complex numbers, and that a definition of the $\ln$ does not have the nice properties that they have for real numbers.
For example
$$
 \ln(e^z) = z
$$
does not need to be true for a complex number $z$.
(this can be regarded as a mistake in your question)
Just because something is true for real numbers, it doesnt mean that the same thing is also true for complex numbers.
A: You went wrong in assuming $z\to e^z $ was one to one.  As $e^{z+2\pi i}=e^z*e^{2\pi i}=e^z (\cos 2\pi +i\sin 2\pi)=e^z*1=e^z$, it clearly is not one-to-one .
Thus we can not assume $e^z=e^w\implies z=w $.
And thus if we define $\ln w=v $ if $e^v=w$, it is not a single value unique function.
In fact $z=a+bi;a,b\implies \mathbb R $ implies $e^z=e^a (\cos b +i\sin b)=e^a (\cos (b+k2\pi)+i\sin (b+k2\pi))=e^{z+2k\pi i} $ for some integer $k $. Thus we can conclude $e^v=e^w $ implies $v=w+2k\pi i$.
So when we define $\ln v=w $ if $e^w=v $ it isn't a single value function.  It is a "multivalue function" where $\ln z $ does not just equal one value $u$.  It equals an infinite number of values of complex numbers, some multiple of $2i\pi $ apart, of which $u $ is just one example.
So $e^{2\pi i}=e^0=1\implies \ln e^{2\pi i}=\ln 1\implies 2\pi i=0 +2k\pi i$ for some integer $k$.  Which it does!
