Proof for the irrationality of $eπ$ ('complex'ly?).* Some of you may have seen the video: 
https://youtu.be/DLWpj34UNRk
It is a proof of the irrationality of $eπ$ by Ron (14years) , presented by 'blackpenredpen' Youtuber.
The proof goes something like this:
We know that $e^{iπ}=-1$ ("Euler's famous identity").
$e^{iπ}=i^2$
So, $ π=\frac{2\ln{i}}{i}$
We must prove that $eπ$ is irrational.
Let us assume, on the contrary, that $eπ$ is rational.
$eπ=\frac{a}{b}$  where $a$ and $b$ are rational numbers.
$$eπ=e\frac{2\ln{i}}{i}=\frac{a}{b}$$
$$2eb\ln{i}=ai$$
$$\ln{i^{2eb}}=ai$$
$$\ln{(-1)^{eb}}=ai$$
$$(-1)^{eb}=e^{ai}$$
Squaring on both sides,
$$1^{eb}=e^{2ai}$$
$$1=e^{2ai}$$
$$e^0=e^{2ai}$$
When the bases are identical, the powers are equal.
$$0=2ai$$
$$a=0$$
Hence,
$$eπ=\frac{0}{b}$$
$$eπ=0$$
This is not possible and therefore gives us a contradiction.
This contradiction has arisen because of our incorrect assumption the $eπ$ is rational.
We conclude that $eπ$ is irrational.
(*) Is this proof legitimate?  Or is there something dubious about it? If so, please point it out.
 A: Look at the proof starting here:
$e \pi =\frac{a}{b}$ where $a$ and $b$ are relatively prime integers. 
The rest of the proof never uses that $a,b$ are integers. If you replace $a,b$ by real numbers, the rest of the "proof" still holds.
Actually, if you write at this point in the proof $e \pi =\frac{a}{b}$ where $a=e \pi $ and $b=1$, the rest of the "proof" still derives a contradiction.
A: The first mistake I see is the claim that $2eb\ln{i}\stackrel{?}{=}\ln{i^{2eb}}$. If we're taking the principal branch of the logarithm and power functions, there's no reason for that to be true! Consider the simplest nontrivial example: $b=1$. The left-hand side is just:
$$2e\ln i=e\pi i$$
But the right-hand side is:
$$\ln{i^{2e}}=\ln e^{2e\ln i}=\ln{e^{e\pi i}}=e\pi i+2n\pi i=(e+2n)\pi i$$
...where $n$ is the integer that makes the whole expression as close as possible to $0$. Since $2<e<3$, this will be $n=-1$, so:
$$\ln{i^{2e}}=(e-2)\pi i\neq e\pi i.$$
As Eric Wofsey points out, there are other errors after that, which only compound the incorrectness of the argument.
A: This proof reminds me of a very famous mathematical paradox concerning complex logarithm: $$e^0=e^{2\pi i}$$
$$\ln(e^0)=\ln(e^{2\pi i})$$
$$0=2\pi i$$
In fact, we shall use only one branch of the complex logarithm, and stay with it always. There is no single branch of logarithm defined for both input of argument $0$ and $2\pi$.
