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 $z$?


  • $\begingroup$ this might help you en.wikipedia.org/wiki/Argument_%28complex_analysis%29 $\endgroup$ – jim Dec 11 '12 at 11:44
  • $\begingroup$ Just from knowing $r$ you can't get argument. If you also know $x,y$ you can get the angle. But there are more than one definition of argument, e.g. some say $-\pi < \arg z \le \pi$. $\endgroup$ – coffeemath Dec 11 '12 at 11:47

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.

  • $\begingroup$ It's not true that if $z=a+ib$ then $\operatorname{Arg}(z)= \arctan(b/a)$, for then $1+i$ would have the same argument as $-1-i$. Also the $\arcsin$ and $\arccos$ thing won't always work, because you need a value of $\theta$ satisfying both $|z|\sin \theta =b$ and $|z|\cos \theta = a$. $\endgroup$ – Matthew Towers Dec 11 '12 at 16:17
  • $\begingroup$ @mt_ I am sorry, I don't know how I forgot to mention that. I have edited my answer. I hope it is better than the previous version. $\endgroup$ – funktor Dec 12 '12 at 6:51
  • $\begingroup$ looks right to me! $\endgroup$ – Matthew Towers Dec 12 '12 at 8:40
  • $\begingroup$ Great, thanks.. $\endgroup$ – Billie Jan 8 '13 at 7:25
  • $\begingroup$ I know this is an old question, but I would just like to expand a bit. What about for when complex number a+bi while a=0 and b<0. Would the argument always be 3pi/2? (New to Math SE, haven't got all the markdown yet. Sorry) $\endgroup$ – user277943 Oct 7 '15 at 17:32

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 $x+iy$.


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.

  • $\begingroup$ It's not true that if $z=x+iy$ then $\operatorname{Arg}(z) = \arctan(b/a)$, for then $1+i$ would have the same argument as $-1-i$. $\endgroup$ – Matthew Towers Dec 11 '12 at 17:31

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