# How to get principal argument of complex number from complex plane?

I am just starting to learn calculus and the concepts of radians. Something that is confusing me is how my textbook is getting the principal argument ($\arg z$) from the complex plane. i.e. for the complex number $-2 + 2i$, how does it get $\frac{3\pi}{4}$? (I get $\frac{\pi}{4}$).

The formula is $\tan^{-1}(\frac{b}{a})$, and I am getting $\frac{\pi}{4}$ when I calculate $\tan^{-1}(\frac{2}{-2})$. When I draw it I see that the point is in quadrant 2.

So how do you compute the correct value of the principal argument?

• But $\tan\frac{\pi}4=1$, not $-1$, so how are you getting $\tan^{-1}(-1)=\frac{\pi}4$? You should be getting $-\frac{\pi}4$. And since that’s in the fourth quadrant, and you know that you need an angle with the same tangent in the second quadrant, you add $\pi$ to get $\frac34\pi$. – Brian M. Scott Jul 23 '12 at 9:25
• Just so you know, it's the principal argument. – Potato Jul 23 '12 at 9:27
• Oh.. I didn't know you have to add pi. So if the angle is in the 3rd quadrant do I have to add 2pi? I can see these are silly questions, but small things like this are making it difficult for me to learn. I really want to understand this stuff. Thanks so much. – tb747 Jul 23 '12 at 9:36
• The period of $\tan$ is $\pi$ so that you must verify if $\pi$ has to be added (depending of the quadrant). That's why many programming languages include a 'atan2' function. – Raymond Manzoni Jul 23 '12 at 9:39

## 3 Answers

The principal value of $\tan^{-1}\theta$ is always between $-\frac{\pi}2$ and $\frac{\pi}2$. The principal value of $\arg z$, on the other hand, is always in the interval $(-\pi,\pi]$. Thus, for $z$ in the first quadrant it’s between $0$ and $\frac{\pi}2$; for $z$ in the second quadrant it’s between $\frac{\pi}2$ and $\pi$; for $z$ in the third quadrant it’s between $-\frac{\pi}2$ and $-\pi$; and for $z$ in the fourth quadrant it’s between $0$ and $-\frac{\pi}2$. This means that the $\tan^{-1}$ function gives you the correct angle only when $z$ is in the first and fourth quadrants.

When $z$ is in the second quadrant, you have to find an angle between $\frac{\pi}2$ and $\pi$ that has the same tangent as the angle $\theta$ returned by the $\tan^{-1}$ function, which satisfies $-\frac{\pi}2<\theta\le 0$. The tangent function is periodic with period $\pi$, so $\tan(\theta+\pi)=\tan\theta$, and $$\frac{\pi}2=-\frac{\pi}2+\pi<\theta+\pi\le0+\pi=\pi\;,$$ so $\theta+\pi$ is indeed in the second quadrant.

When $z$ is in the third quadrant, you have to find an angle between $-\pi$ and $-\frac{\pi}2$ that has the same tangent as the angle $\theta$ returned by the $\tan^{-1}$ function, which satisfies $0\le\theta<\frac{\pi}2$. This time subtracting $\pi$ does the trick: $\tan(\theta-\pi)=\tan\theta$, and

$$-\pi=0-\pi<\theta-\pi<\frac{\pi}2-\pi=-\frac{\pi}2\;.$$

There’s just one slightly tricky bit. If $z$ is a negative real number, should you consider it to be in the second or in the third quadrant? The tangent is $0$, so the $\tan^{-1}$ function will return $0$. If you treat $z$ as being in the second quadrant, you’ll add $\pi$ and get a principal argument of $\pi$. If instead you treat $z$ as being in the third quadrant, you’ll subtract $\pi$ and get a principal argument of $-\pi$. But by definition the principal argument is in the half-open interval $(-\pi,\pi]$, which does not include $-\pi$; thus, you must take $z$ to be in the second quadrant and assign it the principal argument $\pi$.

• I tried to correct the two (very) small typos in your solution above, but edits must be at least 6 characters, and I couldn't find anything else to edit (this solution is very informative). Last sentence (before the semicolon), " ...the principal argument is in the interval $(-\pi,\pi]$, which does not include $-\pi$; ... ," and, at the very end of the same sentence, " ...assign it the principal argument $\pi$ ." – Procore Jan 21 '17 at 21:08
• I realize this post is from 2012, yet it was immensely helpful to me as it was the clearest explanation I could find! Thank you! – C. Ekinci Sep 10 '18 at 3:13

One of the most important functions in analysis is the argument function $${\rm arg}:\quad \dot{\mathbb R}^2\to {\mathbb R}/(2\pi{\mathbb Z}),\qquad (x,y)\mapsto {\rm arg}(x,y)\ ,$$ resp. $${\rm arg}:\quad \dot{\mathbb C}\to {\mathbb R}/(2\pi{\mathbb Z}),\qquad z\mapsto {\rm arg}z\ ,$$ where the dot indicates that the origin is removed.

Intuitively ${\rm arg}(x,y)$ denotes the polar angle of $(x,y)$ "up to multiples of $2\pi$". For "local" considerations there are continuous real-valued ("numerical") representants of ${\rm arg}$; but these are defined only in a suitable part of $\dot{\mathbb R}^2$. In particular the principal value $${\rm Arg}:\quad {\mathbb R}^2\setminus\{(x,0)|x\leq0\}\ \to {\mathbb R},\qquad (x,y)\mapsto {\rm Arg}(x,y)$$ is defined on the $(x,y)$-plane slit up along the negative $x$-axis. It has the simple symmetry property ${\rm Arg}\bar z=-{\rm Arg}z$, and for $x>0$ it is given by $${\rm Arg}(x,y)=\arctan{y\over x}\qquad(x>0)\ .$$ Note that the ${\rm arg}$ function has a well defined gradient given by $$\nabla{\rm arg}(x,y)=\Bigl({-y\over x^2+y^2},{x\over x^2+y^2}\Bigr)\qquad\bigl((x,y)\ne(0,0)\bigr)\ .$$

I've come up with this recipe for principal argument. It saves messing around adding or subtracting $\pi$.

$$\text{Arg} (z) = n\ \text{cos}^{-1} \left(\frac{x}{z}\right)$$

in which n = 1 if y ≥ 0 but n = -1 if y < 0.

I've tried to 'automate' the value of n, but the best I can do is

$$\text{Arg} (z) = \frac{y}{|y|}\ \text{cos}^{-1} \left(\frac{x}{z}\right).$$

Unfortunately this fails for y = 0 (real z), so the y = 0 case would still have to be catered for separately.

Edit: A very ugly self-contained recipe would be

$$\text{Arg} (z) = \text{sgn}\left(\text{sgn}(y) + \frac{1}{2}\right)\ \text{cos}^{-1} \left(\frac{x}{z}\right).$$

## protected by Zev ChonolesDec 25 '16 at 5:57

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?