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It's all fun and games till someone shoots their i out. This turns out to be the best one: the square root is taken to be in -pi/2 to 3pi/2 instead of -pi to pi. This is done by hacking the 2-argument arctangent to first rotate the points clockwise by 90° and then adding the 90° back directly. This corresponds to taking branch cut along the imaginary axis from -infinity to -i, and -infinity to -i. We will see that the overlapping part *cancels out*, meaning that this amounts to a cut along the segment i[-1,1].
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Now here's the complex square root in trigonometric form (this is the only form that generalizes to higher order roots -- see a separate Desmos notebook for this)
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Now use sqrt(z-i)sqrt(z+i) with the branch for the sqrt being this choice.
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Now this actually gives complete coverage of the complex plane when defined on a disk (formally, of course, we need to omit the interval [-1, 1] which is the image of the cut on i[-1,1]
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