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.
 A: They are distinguishing between $arg$ (lower case) and $Arg$ (uppercase).
$arg (z) = Arg(z) + 2k\pi$ for any $k$.
So $arg(z) + arg(z) = Arg(z) + 2k \pi + Arg(z) + 2m\pi = 2Arg(z) + 2n\pi$ for any possible integer value of $n=k+m$.
But $2arg(z) = 2(Arg(x) + 2k\pi) = 2Arg(x) + 4k\pi$ for any integer value of $k$.
So $arg(z) + arg(z) \ne 2arg(z)$.  
Which is probably what they should have said at the start.
....
To be honest.... I'm not loving the text.  
I prefer to think of arguments as modulo classes and when we say $arg(z) = blah$ we mean $arg(z) \equiv blah \mod 2\pi$.  (I hope you know what that notation means.  Formally it means $arg(z) = \{blah + 2k\pi|k \in \mathbb Z\}$.
Hence I'd say $2arg(z)= 2blah$ to mean $2arg(x) \equiv 2blah \mod 2\pi$.
i.e.  $2\arg(z) = \{2blah + 2k\pi|k \in \mathbb Z\}$.  And $2\arg(z) \ne \{2\theta |\theta \in arg(z)\}=\{2blah + 4k\pi|k \in \mathbb Z\}$.
So in MY book, I would sat that $arg(z^2)$ DOES equal $2arg(z)$.
I honestly don't see the point of confusing students this way. 
...
On the other hand it will be very important that $arg (z^{\frac 1n}) \ne \frac 1n arg(z)$.  It is $\frac 1n arg(z) + \frac {2k}n\pi$.
A: The property $\arg(z_1z_2)=\arg(z_1)+\arg(z_2)$ is not true in general.
Assuming $\arg$ takes its values on $[0,2\pi)$, then what happens when $z_1=z_2=e^{\frac{3}{2}\pi i}=-i$?

If by $\arg$ the multivalued function is meant, then it depends on how you define $2\arg(z)$.
More generally, one can define $A+B$ for two subsets of the complex numbers (or, more generally, subsets of an additive group) by
$$
A+B=\{a+b:a\in A,b\in B\}
$$
Now, generally,
$$
2A=\{2a:a\in A\}\ne A+A\tag{*}
$$
The inclusion $2A\subseteq A+A$ is easy to see. For the other, one needs that, for every $a,b\in A$, there exists $c\in A$ with $a+b=2c$. This can happen or not, depending on the set $A$. For instance, if $A=\mathbb{Q}$ is the set of rational numbers, then $2\mathbb{Q}=\mathbb{Q}+\mathbb{Q}$; but if $A=\mathbb{Z}$ is the set of integers, then $\mathbb{Z}+\mathbb{Z}=\mathbb{Z}$, but $2\mathbb{Z}$ is the set of even integers.
In case the set $A$ is described as
$$
A=\{a_0+nc:n\in\mathbb{Z}\}
$$
where $a_0$ and $c$ are fixed complex numbers (in your case $c=2\pi i$), the equality
$$
2A=A+A
$$
doesn't hold. Indeed
$$
2A=\{2a_0+2nc:n\in\mathbb{Z}\}
$$
whereas $(a_0+0c)+(a_0+1c)=2a_0+c\in A+A$ and
$$
2a_0+c=2a_0+2nc
$$
cannot be satisfied with integer $n$.
Of course this could be easily repaired by defining
$$
2A=A+A
$$
instead of the naïve definition (*).
Why your textbook is insisting on this is unknown to me, but as you see it has nothing to do with arg, and is rather a property of addition.
A: 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
$$
\arg(-1)= \pi .
$$
