How to translate $1-i$ into polar coordinates

To calculate $r$ is easy: $r=\sqrt{1^2+(-1)^2}=\sqrt{2}$. Now $\tan\theta=-1$ so $\theta=-\frac{\pi}{4}+k\pi$.

Now $a=1, b=-1$ which means that we're in the 4th quarter of xy axis.

-$\frac{\pi}{4}$ is already in the 4th quarter so I would be tempted to say that this is the answer, yet the answer is $-\frac{\pi}{4}+2\pi$.

Why do we need to add $2\pi$ if $-\frac{\pi}{4}$ is already in the 4th quarter?

• To get the minimum positive angle, I believe. Also two angles which differ by multiple of $2\pi$ are equivalent. Commented May 5, 2017 at 10:26
• yes, I think you are right. The polar co-ordinate of $1-i$ is $=\sqrt2\angle{-45}^\circ$ Commented May 5, 2017 at 10:27

Both $\sqrt2 e^{-i\pi/4}$ and $\sqrt2 e^{7i\pi/4}$ are valid answers, as they are "off" by a factor of $e^{2\pi i}=1$. However, it is often convention to have your angles in the interval $[0,2\pi)$. In these cases, you should add the $2\pi$. But this isn't strictly necessary, usually.
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