Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The fact that it is true, seems very obvious, if one draws the complex number $z = (-2 + 0i)$ on the complex plane. The angle is certainly 180 degrees, or pi radians.

But how can this be proven by calculation? Using $\arg(z)=\arctan(b/a)$ or even the "extended" version $\arg(z) = \arctan(b/a) + \text{sign}(b)(1-\text{sign}(a))$ gives $0$.

share|cite|improve this question
This depends on what your definition for the argument is. Some authors consider the argument to lie in the interval $(-\pi,\pi]$; others consider the argument to lie in the interval $[0,2\pi)$; others have other variations. Likely though, it would indeed be $\pi$. – Cameron Williams Jun 18 '14 at 2:01
up vote 1 down vote accepted

We can arrive at this conclusion by using the definition of $\arg(z)$. Recall that every complex number can be written in the form,

$$ z = r( \cos(\theta) + i \sin(\theta) ), $$

where $r=\left| z \right|$ and $\theta = \arg(z)$. If $z=-2$ then we have two equations,

$$ -2 = r \cos(\theta) \qquad 0 = r\sin(\theta)$$

The first equation tells us that $r\neq 0$ which then combined with the second equation tells us that $\sin(\theta) = 0$ which is only satisfied if theta is a multiple of $\pi$, i.e. $\theta = n\pi$. We now substitute $\theta = n\pi$ into the first equation and get,

$$ -2 = r \cos(n\pi), $$

$$ \Rightarrow -2 = r (-1)^n,$$

$$\Rightarrow r = 2 (-1)^{n+1},$$

but we know that $r>0$ so $n$ must be odd for the last equality to hold. Therefore we know know that $\theta$ is an odd multiple of $\pi$. In other words,

$$ \arg(-2) = \pi + 2\pi k $$

share|cite|improve this answer

You must use the function $\text{arctan2}(y,x)$, not $\arctan$. Simply using $\arctan y/x$ will not give you the angle that the point $(x,y)$ makes with the origin, since the $\tan$ function repeats itself every $\pi$, i.e. $\tan x = \tan (x+\pi)$, i.e. does not have an inverse on the entirety of it's domain. This means that the $\arctan$ function (as conventionally defined) can only return angles between $-\pi/2$ and $\pi/2$.

The function $\text{arctan2}(y,x)$, will however return the correct angle, as it takes into consideration what quadrant the point is in.

share|cite|improve this answer
Good answer. This gives a nice definition: – NotNotLogical Jun 18 '14 at 2:03
Unfortunately, we can only upvote once... I am in a personal crusade against the formula $\theta=\arctan(\frac{y}{x})$, which is, to my despair, seen everywhere. – Taladris Jun 18 '14 at 10:53

$\arg(-2)=\pi-\arctan\left( \frac{0}{-2} \right)=\pi-\arctan(0)=\pi-0=\boxed\pi$

share|cite|improve this answer

More complicated than $\arctan(y/x)$, yet valid for all $z\ne0$ is $$ \arg(x+iy)=\left\{\begin{array}{cl} 2\arctan\left(\frac{y}{x+\sqrt{x^2+y^2}}\right)&\text{if }y\ne0\text{ or }x\gt0\\[6pt] \pi&\text{otherwise} \end{array}\right. $$ which is based on the identity $$ \tan(\theta/2)=\frac{\sin(\theta)}{1+\cos(\theta)} $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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