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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

$$\varphi=\arctan\left|\frac{y}{x}\right|=\arctan\left|\frac{-\sqrt{3}}{1}\right|=\arctan\sqrt{3}=\frac{\pi}{3}.$$

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?

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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, $\,\dfrac{5\pi}{3}\,$

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  • $\begingroup$ Actually I don't know, but now it seems comprehensible what you said. $\endgroup$ – Christian Ivicevic Nov 17 '12 at 12:30
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The arguement of an imaginary number, Z is the angle, $-\pi<\phi<\pi$, given by

$\phi(Z)=arctan(\frac{Im(Z)}{Re(Z)})$

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

enter image description here

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