By expressing the domain of a function in polar coordinates it is possible to take the function beyond it’s principle value (ie beyond 2$\pi$) to reveal the full beauty & complexity of it’s Riemann surface. Am I right in saying that Cartesian co-ordinates are inherently limited in this regard in a similar way that real numbers hide the greater truths revealed by using complex numbers?
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How would you use polar coordinates to reveal the full beauty and complexity of, say, $(z^2 - 1)^{1/2}$? – Jan 03 '18 at 10:46
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@Rahul: By taking the argument beyond 2\pi. Am I right in thinking that Cartesian coordinates cannot do this? – Peter Smallwood Jan 03 '18 at 11:00
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OK, let $z=re^{i\theta}$ with $\theta>2\pi$, so we have $f(z) = (re^{2i\theta}-1)^{1/2}$. Now what? – Jan 03 '18 at 11:22
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@Raul. If I may take a simpler function: F(z)=z^0.5 then as we go beyond 2PI we see the second sheet of its Riemann surface (see link below). I don't think this can be done with Cartesian co-ordinates (unless you tell me otherwise). https://en.wikipedia.org/wiki/Riemann_surface – Peter Smallwood Jan 03 '18 at 12:27