How to figure out the Argument of complex number?

Solution 1:

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.
$\varphi = \arg(z) = \begin{cases} \theta & \mbox{if } x > 0 \\ \theta + \pi & \mbox{if } x < 0 \mbox{ and } y \ge 0\\ \theta - \pi & \mbox{if } x < 0 \mbox{ and } y < 0\\ \frac{\pi}{2} & \mbox{if } x = 0 \mbox{ and } y > 0\\ -\frac{\pi}{2} & \mbox{if } x = 0 \mbox{ and } y < 0\\ \mbox{indeterminate } & \mbox{if } x = 0 \mbox{ and } y = 0. \end{cases}$

Solution 2:

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 $x+iy$.