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Post by mondo on Jun 19, 2023 4:50:47 GMT
I have one more question for figure [4b], I agree that $dz = izd\theta$ is the correct complex number to represent $dz$. However now I wonder about its location. I mean $dz = izd\theta$ as it stands represents a vector emanating from the origin not from $z$ as on the figure [4b] and not as this integral round unit circle assumes. How is that it doesn't make a difference?
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Post by Admin on Jun 19, 2023 14:21:24 GMT
Mondo
A complex number is exactly like a vector and so is defined by its length and direction and two vectors are equal if their length and direction are equal, and so are two complex numbers.
We can think of them as emanating from where we wish.
In the example we have the vector $z$ and the vector $dz$. If we add them together as $z+dz=w$ then $dz$ emanates from $z$ and $w$ is the vector emanating from the tail of $z$. If we write $dz+z=w$ then $z$ emanates from $dz$ and $w$ is the same vector emanating from the tail of $dz$. Draw yourself a picture of the parallelogram.
In figure 4 on page 437 this is the new radius of the circle rotated through $d\theta$.
Vasco
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Post by mondo on Jun 19, 2023 20:54:59 GMT
I know how to add them and that $z + dz = w$ which itself is presented on [4b] as a vector emanating from the origin to the end of the small vector $dz$. With my question I was only interested in the fact that for the calculation of integral we are in fact using $dz$ that emanates from the origin and not from $z$, contrary to what [4b] illustrates.
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Post by Admin on Jun 20, 2023 13:04:57 GMT
I know how to add them and that $z + dz = w$ which itself is presented on [4b] as a vector emanating from the origin to the end of the small vector $dz$. With my question I was only interested in the fact that for the calculation of integral we are in fact using $dz$ that emanates from the origin and not from $z$, contrary to what [4b] illustrates. Mondo $dz$ is $dz$. I don't understand your problem. Try solving the integral the way you understand it, and see if you get a different answer. Vasco |
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Post by mondo on Jun 22, 2023 5:34:49 GMT
I know how to add them and that $z + dz = w$ which itself is presented on [4b] as a vector emanating from the origin to the end of the small vector $dz$. With my question I was only interested in the fact that for the calculation of integral we are in fact using $dz$ that emanates from the origin and not from $z$, contrary to what [4b] illustrates. Mondo $dz$ is $dz$. I don't understand your problem. Try solving the integral the way you understand it, and see if you get a different answer. Vasco Just to be clear on that $dz = izd\theta$ - $z$ times $d\theta$ based on arc length (ray times angle) and then finally times $i$ to give it correct orientation? As to the location of $dz$, looks like we actually want to have it at the origin so we get the infinitesimal increment needed in the integral calculation. The figure [4b] in fact shows $dz$ at the different location than the integral uses (origin) but this is just for the sake of presenting the idea. |
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Post by Admin on Jun 22, 2023 8:31:07 GMT
Mondo
This is a contour integral round the unit circle and so as the point $z$ moves round the circle (or as the vector $z$ rotates about the origin) so does the vector $dz$. $dz$ remains attached to the point $z$ as $z$ moves round $C$.
We can think of the point $z$ as moving round the circle, or alternatively as the vector $z$ rotating about the origin.
We can obtain the expression in at least two ways:
1. $z$ can be rotated anticlockwise through a right angle to give $iz$ which is a complex number of length $1$ like $z$ but at right angles to $z$, and then multiply it by $d\theta$ to get the complex number $dz$ at right angles to $z$ and of length $d\theta$.
2. If $z=e^{i\theta}$ (as it does here), then differentiating we have $dz=ie^{i\theta}d\theta=izd\theta$ substituting $z$ for $e^{i\theta}$.
Method 1. is more geometric and visual than method 2.
Try and get used to the idea that a complex number like $2+3i$ is not necessarily anchored at any particular point.
When we think of a complex number as defining a point then it is anchored at the origin, but when we think of it as a vector it isn't anchored at any particular point and we can think of it as being anchored anywhere we wish.
This idea was introduced in chapter 10. Remember?
Vasco
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Post by mondo on Jun 22, 2023 17:27:46 GMT
This is a very good and clear explanation. I only question one point, and my doubt is about just this, everything else is clear: $dz$ remains attached to the point $z$ as $z$ moves round $C$. So what I say is that it is anchored only on the figure, for the sake of visualizing $dz$ purpose. And here we really think about it as a vector. While in fact it is a complex number originating from the origin. For instance, let's say $z = 2 + 3i$ then $iz = -3 + 2i$, is another complex number originating from the origin. And you say when considered a vector we can move it freely as it doesn't change it's properties like length, direction, I think I can agree with that. So it looks like there is no way of writing $dz$ mathematically so it must start from $z$ in a counterclockwise, perpendicular direction? |
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Post by Admin on Jun 23, 2023 8:24:38 GMT
This is a very good and clear explanation. I only question one point, and my doubt is about just this, everything else is clear: $dz$ remains attached to the point $z$ as $z$ moves round $C$. So what I say is that it is anchored only on the figure, for the sake of visualizing $dz$ purpose. And here we really think about it as a vector. While in fact it is a complex number originating from the origin. For instance, let's say $z = 2 + 3i$ then $iz = -3 + 2i$, is another complex number originating from the origin. And you say when considered a vector we can move it freely as it doesn't change it's properties like length, direction, I think I can agree with that. So it looks like there is no way of writing $dz$ mathematically so it must start from $z$ in a counterclockwise, perpendicular direction? Mondo Glad you found it useful. I think the best way to think about this situation is to imagine the vector $z$, anchored at the origin, rotating round the unit circle $C$. This is exactly like the situation illustrated in figure 7 on page 11 of the book where we add the vector $M$ to the vector $Z(t)$ to obtain the vector $Z(t+\delta)$. It is useful to think of figure 4b with an arrow drawn from the origin to the point $z$ and an arrow from the origin to the new position of the moving point at $z'=z+dz$ After time $dt$ the vector $z$ has rotated through an angle $d\theta$ to become the new vector $z'=z+dz$ This is vector addition: move along the vector $z$ to the point $z$ on the circle $C$ and then move along $dz$ to arrive at the point $z'$ on the circle $C$. The point $z$ has moved around the circle to $z'$. We repeat this idea as the point moves all the way round the circle $C$. $|dz|$ is ultimately equal to the arc length $d\theta$ and $dz$ is perpendicular to $z$, as we go round $C$. Vasco |
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