Why does $arg(z^{2})\neq 2arg(z)$? I was reading a text that i found about the argument of a complex number (http://scipp.ucsc.edu/~haber/ph116A/arg_11.pdf) but I dont truly understand the proof given in that text about why is it that $arg(z^{2})\neq 2arg(z)$, so if it is true, can you give me an idea of why does it happen, because most of the complex anaylisis textbooks say that this property is always true,$arg(z_{1}z_{2})=arg(z_{1})+arg(z_{2})$ if $z_{1},z_{2}$ are not zero, but if what I asked is true it means that it is false for $z_{1}=z_{2}$. I am really confused about this.

You are thinking of $\arg (z)$ as a real number modulo $2\pi$ or, equivalently, a point on the unit circle. The source you quote is thinking of $\arg (z)$ as the set of all possible angles that work. So for you
$$
\arg{i}= \pi/2 
$$
while for your source
$$
\arg{i}= \{ \ldots, \pi/2 - 2\pi  ,\pi/2, \pi/2 + 2\pi , \ldots \}
$$
If you double all those values you don't get all the members of the set representing your value of
$$
\arg(-1)= \pi .
$$
