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

Convert $1-\sqrt{3}i$ to polar coordinates $(r,\varphi)$.

I started by computing $r=|1-\sqrt{3}i|=\sqrt{1^2+\sqrt{3}^2}=\sqrt{4}=2$. When I tried to compute the angle I did something like


Although this answer seems plausible to me, I am unsure, because the angle should be $-\frac{\pi}{3}$ otherwise the resulting coordinates would be the first quadrant rather than in the fourth. How do I have to compute $\varphi$ to match the right quadrant?

share|cite|improve this question
up vote 1 down vote accepted

Why did you do

$$\arg z=\arctan\left|\frac{y}{z}\right|??$$

It should be

$$\arg z=\arctan\frac{y}{z}=\arctan-\sqrt 3=-\frac{\pi}{3}\,,\,\frac{2\pi}{3}$$

Since in $\,z=1-\sqrt 3i\,\;$ the real part is positive and the imaginary part is negative, the vector(=the complex number) is in the fourth quadrant, so the answer must be $\,-\dfrac{\pi}{3}\,$ , or if a positive number is wanted, $\,\dfrac{5\pi}{3}\,$

share|cite|improve this answer
Actually I don't know, but now it seems comprehensible what you said. – Christian Ivicevic Nov 17 '12 at 12:30

The arguement of an imaginary number, Z is the angle, $-\pi<\phi<\pi$, given by


$= arctan(\frac{-\sqrt{3}}{1})=-\frac{\pi}{3}$

enter image description here

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.