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Post by Admin on Sept 26, 2023 6:07:17 GMT
Mondo
What do you mean by "these vectors"? Which vectors?
Vasco
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Post by Admin on Sept 26, 2023 6:13:31 GMT
Well it must be this! Imagine moving both of them to the origin, then we cast $z+a$ onto $z-a$ (or the other way around) but in book we do it by multiplying the first one by $\cos(\gamma)$. So now we have them collinear. Next imagine we rotate them clockwise until they are aligned with real axis, now a big question - what is theirs modulus/length? It is a real part $Re()$. Done. Mondo What do you mean by "cast"? Vasco |
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Post by mondo on Sept 26, 2023 6:21:08 GMT
Mondo What do you mean by "these vectors"? Which vectors? Vasco Vectors $|z+a|$ and $|z-a|$ aka $\sigma, p$, which were created by addition and subtraction of $z$ and $a$ respectively. Well it must be this! Imagine moving both of them to the origin, then we cast $z+a$ onto $z-a$ (or the other way around) but in book we do it by multiplying the first one by $\cos(\gamma)$. So now we have them collinear. Next imagine we rotate them clockwise until they are aligned with real axis, now a big question - what is theirs modulus/length? It is a real part $Re()$. Done. Mondo What do you mean by "cast"? Vasco By "cast" I mean an operation where you "project" one vector on the other. |
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Post by mondo on Sept 26, 2023 6:23:02 GMT
So I say, having two vectors say, $c$ and $d$ their division $\frac{c}{d}$ is always $Re(\frac{c}{d})$ IF they are collinear.
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Post by Admin on Sept 26, 2023 8:54:35 GMT
Mondo
I agree with that, but I find it much easier to see it geometrically from figure 35, the way I did it. That's why Needham drew diagram 35.
It's a situation which often occurs.
Mondo
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Post by Admin on Sept 26, 2023 9:01:55 GMT
Mondo
This is a situation where it is much better to think of a complex number like $A$ as $|A|e^{i\theta}$ where $\theta$ is some angle. Then division and multiplication is so much easier to see geometrically.
Multiplication is adding angles and division is subtracting angles
Vasco
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Post by mondo on Sept 26, 2023 20:15:28 GMT
Vasco yes and no - yes because I agree that it is easier to perform both multiplication and division in polar form. Although the last equation on p.559 is not about division per se but rather on noting that the ration between them is a real part of their quotient. It seems to be important later one in calculations of duals and components of complex potential in general. In other words, we are not interested here on angles or results of division.
I also think I am right in my reply #8 - the equations (30) and (32) assume that we know the mapping functions, i.e in (30) we need to know a conformal mapping for $\theta^*$ otherwise we are unable to perform this integration WRT to plain $\theta$.
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Post by mondo on Sept 26, 2023 20:32:30 GMT
Vasco, or do you mean that $\frac{|z|e^{i\theta}}{|r|e^{i\theta}} = \frac{|z|}{|r|}e^{i(\theta-\theta)} = \frac{|z|}{|r|}$ and this is a real number. But still, I think we need to be careful - just because we got a real number it doesn't mean it is a real part of this quotient. For instance we can't say that $\frac{|z|}{|r|} = Re(\frac{z}{r})$ - it only is true when they are collinear.
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Post by Admin on Sept 26, 2023 20:53:29 GMT
Mondo
In your post 17 you call $|z+a|$ and $|z-a|$ vectors, but they are not, they are real numbers. Also $\sigma=|z+a|\cos\gamma$ and $\rho=|z-a|$, all real numbers. $\sigma\neq |z+a|$.
Vasco
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Post by Admin on Sept 26, 2023 21:01:04 GMT
Vasco, or do you mean that $\frac{|z|e^{i\theta}}{|r|e^{i\theta}} = \frac{|z|}{|r|}e^{i(\theta-\theta)} = \frac{|z|}{|r|}$ and this is a real number. But still, I think we need to be careful - just because we got a real number it doesn't mean it is a real part of this quotient. For instance we can't say that $\frac{|z|}{|r|} = Re(\frac{z}{r})$ - it only is true when they are collinear. Mondo Which of my posts are you referring to here? Vasco |
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Post by mondo on Sept 26, 2023 21:17:54 GMT
Mondo In your post 17 you call $|z+a|$ and $|z-a|$ vectors, but they are not, they are real numbers. Also $\sigma=|z+a|\cos\gamma$ and $\rho=|z-a|$, all real numbers. $\sigma\neq |z+a|$. Vasco Correct, I made a mistake. Vasco, or do you mean that $\frac{|z|e^{i\theta}}{|r|e^{i\theta}} = \frac{|z|}{|r|}e^{i(\theta-\theta)} = \frac{|z|}{|r|}$ and this is a real number. But still, I think we need to be careful - just because we got a real number it doesn't mean it is a real part of this quotient. For instance we can't say that $\frac{|z|}{|r|} = Re(\frac{z}{r})$ - it only is true when they are collinear. Mondo Which of my posts are you referring to here? Vasco Your reply #20. Which I agree with. However the subtraction of angles that I did above is not applicable to [35] as there we deal with vector lengths/modulus. |
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Post by Admin on Sept 26, 2023 21:31:15 GMT
Mondo
I disagree with your last sentence in your post 25. In figure 35 we are dealing with the vectors (the arrows) which are $z+a$ and $z-a$.
$\displaystyle\frac{z+a}{z-a}=\frac{|z+a|}{|z-a|}e^{i\gamma}=\frac{|z+a|}{|z-a|}(\cos\gamma+i\sin\gamma)$
So $\displaystyle\text{Re}\bigg[\frac{z+a}{z-a}\bigg]=\frac{|z+a|}{|z-a|}\cos\gamma$
Vasco
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Post by Admin on Sept 26, 2023 21:47:22 GMT
Mondo
I have updated my post 26
Vasco
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Post by mondo on Sept 26, 2023 21:59:42 GMT
Mondo I disagree with your last sentence in your post 25. In figure 35 we are dealing with the vectors (the arrows) which are $z+a$ and $z-a$. $\displaystyle\frac{z+a}{z-a}=\frac{|z+a|}{|z-a|}e^{i\gamma}=\frac{|z+a|}{|z-a|}(\cos\gamma+i\sin\gamma)$ So $\displaystyle\text{Re}\bigg[\frac{z+a}{z-a}\bigg]=\frac{|z+a|}{|z-a|}\cos\gamma$ Vasco Ok yes, this way it is really easy to see this relation. Thank you. |
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Post by mondo on Sept 27, 2023 6:49:58 GMT
Vasco, another negative sign surprised me - right below figure [35] p.560 it reads $f = T + iS = -\Omega$ the whole equation makes sense to me, likewise the integral below it but why $-\Omega$?
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