I am beginning my study of Complex Analysis, and I stumbled on the following definition on how to compute the argument:

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

I am striving to understand why in the second and third quadrant we have $\arctan(\frac{y}{x}) + \pi$ and $\arctan(\frac{y}{x}) - \pi$ respectively. I have been searching for the reason why $\pi$ is summed and subtracted however this is presented as a formula and I have nor found any intuition that would explain those operations.


What is the reasoning behind summing and subtracting $\pi$?

Thanks in advance!

  • $\begingroup$ en.wikipedia.org/wiki/Atan2 $\endgroup$ – J.G. Feb 14 at 12:19
  • $\begingroup$ Answers below, but I can't help thinking that this is a horrible description of the argument. Simply draw a picture of z, find the angle in any convenient triangle with the real axis and identify arg(z) from there. $\endgroup$ – Paul Feb 14 at 14:07

The maximum and minimum values of arctan are π/2 and -π/2, so all the values will be between these.

You must be already knowing that arg(z) is the angle made by line joining the point representing our complex number on Cartesian plane with the origin, from the x axis (where x represents real component and y represents imaginary component). Lets see the cases of 2nd and 3rd quadrant.

Case-1 second quadrant: Your x part will be negative and y is positive hence arctan(y/x) will give you a value between -π/2 and 0. If you will directly plot this angle on the Cartesian plane it will be somewhere below x axis hence will be wrong as it will be representing negative y and positive x. So we add π to solve this problem.

Similar logic is applicable for the third quadrant where both x and y are negative hence arctan(y/x) will give you an angle somewhere in the 1st quadrant i.e between 0 and π/2. To get the correct angle you might add or subtract π.

This whole adjustment is done due to the limits of arctan function which does not cover whole 0 to 360 degrees instead just covers -90 to 90 degres i.e (-π/2 to π/2).

Hope this solves your doubt.


First of all observe that $\arctan(y/x) \in ( - \frac{\pi}{2}, \frac{\pi}{2}).$

Example: let $z=-1+i$, hence $x=-1,y=1$ . Draw a picture and you will see that $\arg(z)=\frac{3}{4} \pi.$

We have $\arctan(y/x)= \arctan(-1)=-\frac{1}{4} \pi$ and $\arctan(y/x)+ \pi=\frac{3}{4} \pi= \arg(z).$

Can you take it from here ?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.