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Post by Admin on Mar 13, 2016 7:16:25 GMT
Gary
As I said in my previuos post, once we know everything about $\mathcal{K}$ other than in which direction it points, then we know everything about the new rotated $S$ other than in which direction it points, because they are in fact coincident lines, since they both pass through $p$. Then we just need to decide in which same direction they both must point. and we are home and dry. If you find it confusing to move $p$ along $S$ before we have decided which way it points, then think of it as moving along $\mathcal{K}$ instead or just as moving along a line parallel to $\mathcal{K}$ which passes through $p$. It's all the same thing.
When I say we know everything about $S$ other than in which direction it points, I mean that we could draw it accurately on a piece of paper, but we wouldn't know at which end of it to place the arrowhead. We know the orientation of it $\pm\pi$.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Mar 13, 2016 16:22:05 GMT
Vasco,
I do see that S is rotated to plus or minus $i\hat{\xi}$ and parallel to $\mathcal{K}$. That is no problem. I also see that we can choose to move p in either direction along S or $\mathcal{K}$. It just seems to me that Needham is instructing us to first choose an angle of S, and then choose one of the two orthogonal directions for $\mathcal{K}$. And only then rotate S and move p to see the effect on f(p). All in that order. With $e^z$, if we choose to put S on the real line or any horizontal line, what criterion enables us to select $i\hat{\xi}$ rather than $-i\hat{\xi}$ as the direction of $\mathcal{K}$
Gary
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Post by Admin on Mar 14, 2016 16:55:36 GMT
Gary I follow what you are saying but I don't think the wording of the question stops us from considering a line segment to represent $\mathcal{K}$, orthogonal to the original direction, and then deciding which way it must point in order to answer the question "Which?" I am making a distinction here between $\mathcal{K}$, which can be regarded as a directed line segment, and the line segment $S$. When Needham writes "The complex curvature $\mathcal{K}$ must therefore point in one of the two orthogonal directions" it seems to me he is inviting us (giving us a hint) to think of $\mathcal{K}$ as a line segment (oriented correctly as orthogonal to the original $S$, but with no arrowhead to give it direction) and then deduce the direction and so answer the question: Which?. When he then says "By considering the image of $S$ when it points in this direction" he is returning to $S$ and we then have the two directed line segments, complex numbers, whatever, pointing in the same direction. I have produced a document with more detail hereVasco
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