How to figure out the Argument of complex number? I have the absolute value of complex number ,
$$ r = |z| = \sqrt{x^2 + y^2}$$
when $z = x + iy$ is a complex number.
How can I calculate the Argument of $z$? 
Thanks.
 A: Note that the "answer" $arctan(y/x)$ is just wrong. To see it check out the example $-1-i$: $\arctan(-1/-1) = 45°$ but correct would be $225°$. And this is not just a problem with the definition of the range of the argument.
The correct answer is given by Wikipedia:
$\varphi = \arg(z) =
\begin{cases}
\arctan(\frac{y}{x}) & \mbox{if } x > 0 \\
\arctan(\frac{y}{x}) + \pi & \mbox{if } x < 0  \mbox{ and } y \ge 0\\
\arctan(\frac{y}{x}) - \pi & \mbox{if } x < 0 \mbox{ and } y < 0\\
\frac{\pi}{2} & \mbox{if } x = 0 \mbox{ and } y > 0\\
-\frac{\pi}{2} & \mbox{if } x = 0 \mbox{ and } y < 0\\
\mbox{indeterminate } & \mbox{if } x = 0 \mbox{ and } y = 0.
\end{cases}$
Because of that many programming languages have the function $\operatorname{atan2}(y,x)$ which gives the above correct argument for $x+iy$.
A: You should know that any complex number can be represented as a point in the Cartesian ($x$-$y$) plane. That is to say that a complex number $z=a+b\text i$ is associated with some point (say $A$) having co-ordinates $(a,b)$ in the Cartesian plane.
You might have heard this as the Argand Diagram. Let $\tan \theta$ be the direction ratio of the vector $\vec{OA}$ (Assume the line joining the origin, $O$ and point $A$ to be a vector)

Then, $$\tan\theta =\frac ba \implies \theta= \arctan 
\Big (\frac ba\Big )$$

However, we can't go about claiming $\theta$ to be $\operatorname {Arg}(z)$ just yet. There is a small detail that we need to keep in mind (Thank you to a user for pointing that out!). We need to watch out for the quadrant on which our complex number lies and work accordingly. 
Example Say there are 2 complex numbers $z=a+b\text i$ and $w=-a-b\text i$. Then, $$\operatorname{Arg}(w)=\arctan\Big( \frac {-b}{-a} \Big )= \arctan\Big( \frac {b}{a} \Big )= \operatorname{Arg}(z)$$
which is just preposterous. It suggests that $w$, which lies on the third quadrant on the Argand Diagram,  has the same argument as a complex number ($z$) which in the first quadrant. To correct this issue, we'll have to put forth some simple conditions. As we just saw, one of them could go something like: $\text{ if } a,b<0 \text{ then } \operatorname{Arg}(z)=\theta -\pi$
Here is a list of conditions for computing the Argument (This has already been mentioned in one of the answers above and I am just re-posting it here). Once you get a intuitive feel for this, it should come to you naturally.
$\varphi = \arg(z) =
\begin{cases}
\theta & \mbox{if } x > 0 \\
\theta + \pi & \mbox{if } x < 0  \mbox{ and } y \ge 0\\
\theta - \pi & \mbox{if } x < 0 \mbox{ and } y < 0\\
\frac{\pi}{2} & \mbox{if } x = 0 \mbox{ and } y > 0\\
-\frac{\pi}{2} & \mbox{if } x = 0 \mbox{ and } y < 0\\
\mbox{indeterminate } & \mbox{if } x = 0 \mbox{ and } y = 0.
\end{cases}$

Alternatively, you can use the fact that $|z| \sin \theta= b$ (or, $|z| \cos \theta= a$) and then solve for $\theta$. However, you might still have to make last minute amendments (like we did earlier) to come up with the correct answer. So this is not the shorter of the two methods.
A: First, remember that $\sin(\theta)$ is a quotient: Take any point $(x,y)$ such that the line through $(x,y)$ and the origin makes an angle of $\theta$ with the positive $x$ axis.  Then $\sin(\theta) = \frac{y}{\sqrt{x^2 + y^2}}$.
In our specific case, $z$ may be thought of as our point (in the complex plane) .  Take the $\sin^{-1}$ of this value, and voila, you're almost there.  Just make sure you're living in the right quadrant. 
Through a similar argument, if $z = x + iy$, then $Arg(z) = \tan^{-1}(\frac{b}{a})$, if you are more comfortable with tagent.
