It is possible for $z = x+iy$ to have more than two Arg(z) values? I have:
$$z = -4 - 4i$$
Then I know that $\theta = \arctan(y/x)$ which gives me:
$$\arctan(1) = \pi/4$$
But, if we draw the angle for $z = -4 - 4i$, we can observe that we are in the 3thd quadrant.
So, one can just takes:
$$\frac{\pi}{4} - \pi = -\frac{3\pi}{4}$$
and I have my $Arg(z)$, which is $\in [-\pi, \pi]$.
My result, which is in polar form is:
$$4\sqrt{2} e^{-\frac{3\pi}{4}i}$$
What I did (and I don't know if its correct) is:
$$z = -4 - 4i = -4(1 + i) = -4\sqrt{2}(\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}}i)$$
and my final result, which is in polar form is:
$$-4\sqrt{2} e^{\frac{\pi}{4}i}$$
My question: It is possible to have more than two Arg(z) values?
Note that $\frac{\pi}{4}$ is also $\in [-\pi, \pi]$.
Thank you very much for your help!
 A: Both your restults are correct and reprensent the same number : 
$$4\sqrt{2}e^{-\frac{3\pi}4 i} = 4\sqrt{2} e^{-\pi i + \frac{\pi}4i}=4\sqrt{2}\times(-1) \times e^{-\frac\pi4 i}$$
In general, the "polar form" is the first you wrote (since we want the radius to be positive).
Use the atan function is a "bad good idea": you encounters problems because the results of this function is always in $(-\pi/2,\pi/2)$. You should use remarkable values of both sine and cosine function to determine the argument: for instance
$$z=4\sqrt{2}(-\frac{1}{\sqrt{2}} - \frac 1{\sqrt{2}}i)$$
we know that $cos(\theta)=-\frac1{\sqrt{2}}=\sin(\theta)$. Hence (draw a circle if needed) $\theta=-3\frac{\pi}4$.

EDIT : 
Hum well... the matter is not really in the atan function it self. When you calculate $y/x$, if $x$ and $y$ both are negative then $y/x$ is positive and y've lost the information on the sign of $x$ and $y$. 
We can use the atan function but you have to check the sign of $y$ before. For instance an "good algorithm" would be: $$\arg(x+yi) =\left\{\begin{matrix}atan(y/x) \quad\text{ if }y\ge0 \\ atan(y/x)+\pi \quad\text{if } y<0\end{matrix}\right.$$
