What is the argument of $0$? Very easy but bit more thinking is making it complicated. What is the argument of $0$? The number $0$ can be written as $0+0i$ thus its argument is $\tan^{-1}(0/0)$ which is undefined; so, is the argument of $0$ undefined?
 A: The number $0$ may have arbitrary argument, that's why we usually don't define one. For example take the exponential form of complex number (or trigonometric one), then:
$$0=0\cdot e^{i\theta}\quad\text{or}\quad 0=0\cdot(\cos(\theta)+i\sin(\theta))$$
Any $\theta$ satisfies these equations.
Indeterminate form of type $0/0$
You have already found that we may write $$\arg(0)=\arctan(0/0)$$
Formally it is not true, but we can try to describe it using limits.
$$\arg(x+yi)=\arctan(y/x)$$
Take some path i.e. two sequences tending to $0$:
$$x_n\to0\quad\text{and}\quad y_n\to0$$
Value of $\arg(0)$ is described as
$$\lim_{n\to\infty} \arctan\left(\frac{y_n}{x_n}\right)$$
The problem is that for different sequences $x_n$ and $y_n$ the limit may give different results. We usually say that the limit
$$\lim_{(x,y)\to(0,0)}\arctan\left(\frac{y}{x}\right)$$
doesn't exist as it's path-dependent.
Transformation
Another way of thinking is the following. Take the Jacobian of transformation
$$x=r\cos(\theta),\quad y=r\sin(\theta)$$
It is equal to
$$\frac{\partial(x, y)}{\partial(r,\theta)}=r$$
The transformation is not invertible if the Jacobian is equal to $0$, therefore for $r=0$.
A: It is, indeed, undefined, since $0=0e^{i\theta}$ for any real $\theta.$ It is the only complex number with an undefined principal argument, though one could argue (pun intended) that it has any real argument we like.
A: It is undefined. An easy way to see that it would be difficult to define it uniquely is to consider that the argument of a product is the sum of arguments, i.e:
$$\arg(z_{1}z_{2})=\arg(z_{1})+\arg(z_{2})$$
If we consider $z_{1} = 0$, we find:
$$\arg(0)=\arg(0)+\arg(z_{2})$$
So therefore, $\arg(0)$ has an arbitrary value, we cannot uniquely define it.
