Post by Admin on Mar 11, 2016 9:54:01 GMT
Gary
If you look at figure 1(a) of my published solution to exercise 17 of chapter 5, you will see the directed line segment $S$ through $p$ making an angle
$\phi$ with the positive direction of the real axis. Notice that the point $p$ could be drawn as a vector by drawing a line from the origin to $p$ making an angle $\theta$ with the positive direction of the real axis.
If we imagine the line segment being pinned to the complex plane at $p$ then we can rotate $S$ clockwise or anticlockwise about the point $p$ and when we stop we make a note of the angle $\phi$ and that knowledge together with the fact that it passes through $p$, completely defines $S$. For example if we rotate $S$ about $p$ in either direction through an angle $\pi$ then the line segment points in the opposite direction.
When for example we talk about a point moving in the positive or negative direction along $S$ we understand the positive direction to mean the direction $e^{i\phi}$ and the negative direction to mean $e^{i(\phi\pm\pi)}$.
When you talk about a ray I assume that this means a line through the origin, such as the one from the origin to $p$. Again the positive direction along $p$ is $e^{i\theta}$ and the negative direction is $e^{i(\theta\pm\pi)}$.
To answer your specific question: "Is the positive direction of movement of $p$ on rotated $S$ the movement outward from the origin of unrotated $S$?", I'm not sure what you mean by origin here - for me $S$ does not have an origin. The positive direction of movement of $p$ on $S$ is as described above, namely $e^{i\phi}$. In my figure 8 in the appendix, it is also possible to describe the direction of movement of $p$ along $S$ in terms of the angle $\theta$ of $p$, but only if we already know the direction of $S$. So if $p$ moves along $S$ in the direction of increasing $\theta$ then that may correspond to the positive or negative direction of $S$.
I notice that in your document in figure 2 you draw $S$ as being a segment of the real axis. It could be, but in general it is just parallel to the real axis as in my figure 2(a).
Vasco
If you look at figure 1(a) of my published solution to exercise 17 of chapter 5, you will see the directed line segment $S$ through $p$ making an angle
$\phi$ with the positive direction of the real axis. Notice that the point $p$ could be drawn as a vector by drawing a line from the origin to $p$ making an angle $\theta$ with the positive direction of the real axis.
If we imagine the line segment being pinned to the complex plane at $p$ then we can rotate $S$ clockwise or anticlockwise about the point $p$ and when we stop we make a note of the angle $\phi$ and that knowledge together with the fact that it passes through $p$, completely defines $S$. For example if we rotate $S$ about $p$ in either direction through an angle $\pi$ then the line segment points in the opposite direction.
When for example we talk about a point moving in the positive or negative direction along $S$ we understand the positive direction to mean the direction $e^{i\phi}$ and the negative direction to mean $e^{i(\phi\pm\pi)}$.
When you talk about a ray I assume that this means a line through the origin, such as the one from the origin to $p$. Again the positive direction along $p$ is $e^{i\theta}$ and the negative direction is $e^{i(\theta\pm\pi)}$.
To answer your specific question: "Is the positive direction of movement of $p$ on rotated $S$ the movement outward from the origin of unrotated $S$?", I'm not sure what you mean by origin here - for me $S$ does not have an origin. The positive direction of movement of $p$ on $S$ is as described above, namely $e^{i\phi}$. In my figure 8 in the appendix, it is also possible to describe the direction of movement of $p$ along $S$ in terms of the angle $\theta$ of $p$, but only if we already know the direction of $S$. So if $p$ moves along $S$ in the direction of increasing $\theta$ then that may correspond to the positive or negative direction of $S$.
I notice that in your document in figure 2 you draw $S$ as being a segment of the real axis. It could be, but in general it is just parallel to the real axis as in my figure 2(a).
Vasco
