$x=y\Rightarrow \arg x=\arg y$ I was wondering if the implication in the title holds.
Obviously it would depend on the definition of $\arg$, so my question is the following: is there any known definition that makes the implication false?
My take is that there isn't one, because such definition is ill-defined. (I was once told that a function is well-defined if $x=y$ implies $f(x)=f(y)$.)
However I was told that there can be a difference of $2\pi$ between the left hand side and the right hand side, which I think is absurd according to the reason above.
I know that the argument function is multivalued in general, but I think the implication should hold no matter how we define a equality on a multivalued function.
Who is correct and what's wrong?
 A: You are actually correct, and unfortunately the other answers fail to make that clear. The point is, regardless of how you define $\arg$, if $x = y$ then by meaning of equality $x,y$ are literally identical and hence $\arg(x) = \arg(y)$.
There is also no such thing as a function that is not well-defined. People may make a false claim that some function exists satisfying some property, but doing so does not mean that they have any function that is not well-defined! Rather, it means that they have failed to define a function in the first place! For example, if they claim that there is a function $f$ on $ℚ$ such that $f(k/m) = k$ for any $k,m∈ℤ$, it does not mean that $f$ is not well-defined, but rather they simply made a false claim!
Even if you define $\arg(z) = \{ \ t : r∈ℝ^+ ∧ t∈ℝ ∧ r·\exp(i·t) = z \ \}$, everything I said above holds. Whoever told you otherwise is downright wrong and got completely confused with the fact that $\exp(i·t) = \exp(i·u)$ may not imply $t = u$, which has nothing to do with the fact that $x = y$ implies $\arg(x) = \arg(y)$.
A: There are three ways of looking at this. But first bear in mind that argument is not well defined for zero, so you need to be clear about that in any definition.
First we can say that the argument is multi-valued - here if $x=y$ then $x$ and $y$ have the same set of possible values for the argument, but some calculation may lead to values which differ when you expected them to be the same.
Second we can define a principal value for the argument giving each number a definite value and making the equation true. However, step changes may be needed in the argument mid calculation.
Third we can sat that the argument function is from complex numbers to the unit circle - each complex number other than zero maps to the point on the unit circle which is met by the line joining that point to the origin (doing geometry in the complex plane). This also leads to equality. This has technical advantages in that the circle is a smooth curve with no jumps or breaks.
I would further note that the multiple possible values for the argument do lead to some issues where the logarithm of a complex number is taken and used in a calculation - the logarithm becomes a "multi-valued function"
A: "Is there any ...?"
If $\phi : [a,b] \to \mathbb C\setminus \{0\}$ is continuous, then there is a continuous choice of $\arg(\phi(t))$.  If we want such a continuous choice (it is useful in some situations, for example contour integration), then we must allow that it could happen that $\phi(t_1) = \phi(t_2)$ but $\arg(\phi(t_1)) \ne \arg(\phi(t_t))$.
[Image from Wikipedia]

Here, we integrate twice along a segment of the negative real axis, where the argument in the $M$ direction is $\pi$ and the argument in the $N$ direction is $-\pi$.
