Write $w = \sqrt{3} - i$ in polar form. enter image description here

How is $\theta = \frac{-1}{\sqrt3} \textrm{ converted into } \frac{-\pi}{6}$? I understand that $w$ lies in the fourth quadrant of the unit circle, but that's about all I understand. I tried to convert degrees to radians, but I don't think that was the smart decision (I failed).

  • 1
    $\begingroup$ Perhaps you mean $w=3-i$? And $\tan \theta = \frac{-1}{\sqrt{3}}$ implies $\theta = -\frac{\pi}{6}.$ $\endgroup$
    – Dzoooks
    Feb 6, 2019 at 16:50
  • $\begingroup$ ah, i meant $w = \sqrt 3 - i$. And how does tan$\theta$ imply that? Where does the assumption come from? $\endgroup$
    – Evan Kim
    Feb 6, 2019 at 16:53
  • $\begingroup$ That's just how functions work. For example, if $\sin(\theta)= \frac{\sqrt{2}}{2}$ and $\theta$ is in the first quadrant, then the unique solution is $\theta= \frac{\pi}{4}$. Even simpler: let $f(x):=x^2$. Then if $x$ is positive, the unique solution to $f(x)=9$ is $x=3.$ $\endgroup$
    – Dzoooks
    Feb 6, 2019 at 16:55
  • $\begingroup$ $\theta$ isn't equal to $\frac{-1}{\sqrt{3}}$, $\tan \theta$ is. You have to find $\tan^{-1} (\frac{-1}{\sqrt{3}})$, which is equal to $\frac{-\pi}{6}$. This is basic trignometry. $\endgroup$ Feb 6, 2019 at 16:57
  • $\begingroup$ It should be pointed out that tan($5 \pi /6$) is also $-1 \over {\sqrt 3}$, but $5 \pi /6$ is not in the desired quadrant $\endgroup$ Feb 6, 2019 at 16:59

4 Answers 4


You should always draw the picture when you try to find the modulus and argument.

In an Argand Diagram, the complex number $x+\mathrm i y$ corresponds to the point $(x,y)$ on the $xy$-plane.

Put the point $(\sqrt 3,-1)$ on the $xy$-plane, and join it to the origin by a chord (part of a line).

enter image description here

You can see the chord joining the origin to the complex number as the hypotenuse of a right-angled triangle. The length of that chord is the modulus $|z|$, which can be found using Pythagoras:

$$|z|^2 = \left(\sqrt 3\right)^2 + 1^2 = 4$$

This gives $|z| = 2$, since $|z| \ge 0$.

The argument is the angle of the chord measures from "3 o'clock", i.e. the positive real axis. Anti-clockwise rotations are measured as positive, while clockwise rotations are measure as negative.

This may seem odd, but right is positive and left is negative on the number line.

We can use basic SohCahToh Trigonometry to find the argument. First, find the angle between the chord and the $y$-axis, then subtract from $90^{\circ}$ or $\frac{1}{2}\pi$ rads to get the angle between the chord and the positive $x$-axis.

$$\tan \theta = \frac{\mbox{opp}}{\mbox{adj}} = \frac{\sqrt 3}{1} = \sqrt 3$$

The angle between the chord and the $y$-axis is then $\arctan \sqrt 3 = 60^{\circ}$ or $\frac{1}{3}\pi$ rads. That means the argument, i.e. the angle between the chord and the positive $x$-axis (remembering that clockwise is negative) is $-\frac{1}{6}\pi$ or $-30^{\circ}$.

Finally, $|z|=2$ and $\arg z = -\frac{1}{6}\pi$, so $$z = r(\cos \theta + \mathrm i \sin \theta) = 2\left(\cos\left(-\frac{1}{6}\pi\right)+\mathrm i \sin\left(-\frac{1}{6}\pi\right)\right)$$

  • $\begingroup$ the visual helped a lot. How did you draw a graph like that? $\endgroup$
    – Evan Kim
    Feb 6, 2019 at 19:17
  • $\begingroup$ @EvanKim I used the website Desmos. Here's a link to what I did: desmos.com/calculator/4nvyw6xnav $\endgroup$ Feb 6, 2019 at 19:58

The solution to $\tan \theta = \frac {-1} {\sqrt 3}$ is $\theta = \frac {-\pi} 6 + n \pi$.

$\theta = \frac {-\pi} 6$ is a solution in the fourth quadrant, as desired.


The easiest way to do this kind of conversion is often just to treat it as geometry: plot $x+iy$ as a point $(x, y)$, draw a line joining it to the origin, draw another meeting the $x$-axis at a right angle, and look at the resulting triangle.

In this case it'll be half an equilateral triangle, so you could even do it without trig functions if you wanted.

And if you do use trig functions, the triangle tells you which to use.


The first thing to realize is that if you ever have $x^2 + y^2 = 1$ then there is a $\theta$ so that $x = \cos \theta; y = \sin \theta$ and $\tan \theta = \frac yx$.

And second thing to realise is that if you have any $a,b$ where they don't both equal $0$, you can rewrite $a = (\sqrt{a^2+b^2})a'$ and $b = (\sqrt{a^2 + b^2})b'$ where $a' = \frac a{\sqrt{a^2 + b^2}}$ and $b' = \frac b{\sqrt{a^2 + b^2}}$ and you will then have $a'^2 + b'^2 = 1$.

Which means there is $\theta$ so that $\tan \theta = \frac {b'}{a'} = \frac ba$, $a' = \cos \theta$ and $a = (\sqrt{a^2 + b^2})\cos \theta$ and $b' = \sin \theta$ and $b = (\sqrt{a^2 + b^2})\sin \theta$.

That means for any $a + bi \ne 0$ so that $a+bi$ is a complex number and $a$ is its "real" part and $bi$ is its "imaginary part", we can write:

$a+bi = (\sqrt{a^2 + b^2})(\frac a{\sqrt{a^2 + b^2}} + \frac b{\sqrt{a^2+b^2}}i)$

and that means there exist a $\theta$ so that

$a + bi = (\sqrt{a^2 + b^2})(\cos \theta + i\sin \theta)$.

If we imagine $a+bi$ as the point $(a,b)$ on a complex plane, this should not be surpising. $\sqrt{a^2 + b^2}$ is simply the distance from $(a,b)$ to $(0,0)$ and $\theta$ is simply the angle between $(a,b)$ and $(0,0)$ and the real axis.

And we can always find this $\theta$ by solving $\tan \theta = \frac ba$ and determining the proper quadrant by the positive and negative values of $a,b$.

So if $w = \sqrt 3 - i$ then if $r = \sqrt {\sqrt 3^2 + 1} = 2$ we have

$w = 2(\frac {\sqrt 3}2 - \frac 12 i)$.

And we can solve $\cos \theta = \frac{\sqrt{3}}2$ and $\sin \theta = -\frac 12$ and $\tan \theta = \frac 1 {\sqrt{3}}$. That theta is $-\frac {\pi}6$.


$w = \sqrt 3 - i$

$=2(\frac {\sqrt 3}2 - \frac 12 i)$

$= 2(\cos (-\frac \pi 6) + \sin (-\frac \pi 6) i)$.

And .... that's that.


Oh! Important post-script!

If you have $w = a + bi$ and you want to find the right $\theta$ and you think, "Okay, I've got to figure out $\sqrt {a^2 + b^2}$ and I have to divide $a$ and $b$ by it..."

You don't. Note: $\tan \theta = \frac {\frac a{\sqrt{a^2+b^2}}}{\frac b{\sqrt{a^2 + b^2}}} = \frac ab$. So $\theta = \arctan \frac ab$ or $\theta = \arctan \frac ab +\pi$.

So given $w = \sqrt 3 - i$ we need to simply solve $\tan \theta = \frac {-1}{\sqrt 3}$ with $\theta$ in the fourth quadrant.

As to WHY that works is because $w = 2(\frac {\sqrt 3}2 - \frac 12 i)$ but we don't have to do the (somewhat) tedious calculation $a' = \frac {\sqrt 3}{\sqrt{(\sqrt 3)^2 + (-1)^2}}$ and $ b' = \frac {-1} {\sqrt{(\sqrt 3)^2 + (-1)^2}}$


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