Likely fake proof of the irrationality of a rational multiple of pi/pi I would like to know what has gone wrong in this 'proof'. 
Suppose that $$k\frac 1\pi\pi=\frac ab\operatorname{,where}a,b\in \mathbb Z\operatorname {and}k\in \mathbb Q.$$
Then, we multiply both sides by $i$ and raise  $e$ to the power of both sides. We find that $$e^{i(k\frac 1\pi)\pi}=e^{i\frac{2a}{2b}}$$ This implies that $$e^{i(2bk\frac 1\pi)\pi}=e^{i2a}.$$ Using laws of expontents, we rewrite this as $$\begin{pmatrix}e^{i2b\pi}\end{pmatrix}^{\frac k\pi}=e^{i2a}.$$ Since b is an integer, the left hand side is equal to $1$. We take the natural log of both sides using the definition of the natural log in the complex domain.Thus,$$0=i(2a+2n\pi)\operatorname{for some integer }n.$$ Since the imaginary part of $0$ is $0$, $$a+n\pi=0.$$ Rearranging the equation for $\pi,$ we find that $\pi=-\frac an$, which is a contradiction as $\pi$ is irrational.
This is very odd because $k\frac 1\pi=k$ and $k$ is assumed to be a rational number. I have no idea what I have proven despite trying being careful, particularly when using complex numbers. Could someone point out an error?   

There is no error there, except the error of change of definition
You know that $e^{\pi*i} = -1$
Which implies that $e^{2*\pi*i} = 1$
But we know also know that $e^0 = 1$
Does this mean that $2*\pi*i = 0$?
We know that $\sqrt(4) = ±2$
Does this mean than $+2 = -2$?
Mathematics is biased, so when twisting definition we should be careful
