For the convergence of $|z|$ , notice that if the sequence ${z_{n}}$ converges, the sequence ${|z_{n}|}$ also converges.
if $$ lim_{n->\infty}z_{n} = lim_{n->\infty}(x+iy)= w = (a+bi) $$ it means: $lim_{n->\infty}x_{n}=a$ and $lim_{n->\infty}y_{n}=b$.
Then $$lim_{n->\infty}|z_{n}|=lim_{n->\infty}\sqrt{x_{n}^{2}+y_{n}^{2}}=\sqrt{a^{2}+b^{2}}$$
The sequence of $arg(z_{n})$ need not to converge. To notice this look at this example:
$$z_{n}=-1+(-1)^{n}\frac{i}{n}$$
However it is possible to find a convergent sequence of $Arg(z_{n})$