$\pi = 0$! It can't be! Keep in mind: I'm still in high-school so forgive my poor maths. Also remember that I'm in HIGH-SCHOOL so nothing to complex
I like to mess around with equations and I find it quite fascinating the results I can somehow come up with. Recently, I found Euler's Identity and started to mess around with this. Somehow I can up with the following result: $π = 0.$ I thought to myself, as anyone would, how is this possible?
I then decided to show my proof to my maths teacher and my parents and neither one could show me where I went wrong so I was hoping someone here could.
This is my proof:
$$e^{iπ} + 1 = 0$$
$$e^{iπ} = -1$$
$$e^{2iπ} = 1$$
$$2iπ = \ln(1) = 0$$
$$4\times-1\timesπ = 0$$
$$-4π^2 = 0$$
$$0 = 4π^2$$
$$0 = 2π$$
$$0 = π$$
Edit
As I pointed out above, my maths knowledge isn't that advanced, compared to people on this site. Therefore the answer(s) I accept and/or upvote will be based on understandability as well as how well they answer the question.
 A: If $x$ and $y$ are real, then $e^x=e^y$ implies $x=y$. 
But if $x$ and $y$ are complex, $e^x=e^y$ only implies that
$$\frac{x-y}{2\pi i}\in\Bbb Z$$
Therefore, from $e^{2\pi i}=e^0$ we can't deduce that $2\pi i=0$.
A: $e^z = e^w$ does not imply that $z = w$; the exponential function is not one-to-one on ${\mathbb C}$. Instead, it implies that $z - w = 2 \pi i n$ for some $n \in {\mathbb Z}$. So, the fourth equality of your 'proof' does not follow from the third equality.
A: The problem is simple. Note that if $0=\log(1)=\log(e^{2\pi I})=2\pi i$ than the logarithm is no mapping. Canonically one give a solution to this type of equations by restricting to the main branch of the complex logarithm.  
A: The $log(z)$ or $ln(z)$ is a set not a value!
It denotes all $z' \in \mathbb{C}$ such that $e^{z'} = z$
What you want is the $Log(z)$ with $arg(Log(z)) \in [0,2\pi)$ or $(-\pi,\pi)$. Depends on how you define it.
As other said already it's a bit more difficult in $\mathbb{C}$.
A: "2iπ=ln(1)=0"
In complex numbers $\ln$ is not single value but a multi-valued "function".  So $\ln 1$ doesn't just equal $0$.  It equals the entire set of $\{2k\pi i\}$ of which $0$ is the only real value in the set.
This is because $e^{a + bi} = e^a(\cos b + i\sin b)$ and trig functions are periodic so $z= e^{a + bi} = e^{a +bi + i2k\pi}$ is not a 1-1 function any more.  So $\ln z = \{a + i(b + i2k\pi)| k \in \mathbb Z\}$ has multiple values.
So from there we can conclude only that $0 = i 2k\pi$ for some integer value of $k$.  Which it does.  ... id $k = 0$.
