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Why is the principal square root of a complex number defined as $\sqrt z = \sqrt r e^{-i \varphi / 2}$ for $\varphi \in (-\pi, \pi]$ ?

Wouldn't it be more natural to let $\varphi \in [0, 2\pi)$ as it is usual for polar coordinates?

If you think that you are free to choose whatever variant you like, then it's not quite the case. Software libraries definitely prefer the first one so if you are to take some assistance from a computer you are forced to use $\varphi \in (-\pi, \pi]$.

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I imagine that it is $[-\pi,\pi)$ because it is nice to have continuity in a neighborhood of positive real numbers, which the $[0,2\pi)$ version does not have. I imagine the software version was originally chosen to avoid a few extra operations. Going from the software version to the other is straightforward, if the argument of the result (call it $\eta$) satisfies $\eta \geq \frac{\pi}{2}$, then replace $\eta$ by $\eta-\pi$. –  copper.hat Dec 25 '12 at 19:17

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Not to leave the question unanswered, per @copper.hat 's comment.

The choice is driven by desirability to have continuous square root in a neighborhood of positive real numbers.

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