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Post by mondo on Sept 10, 2022 18:39:42 GMT
There is a lot of things I can't get in this subchapter but this is the first one:
1. On the page 91 author refers to Figure [29] saying that if $z$ travels around an arbitrary loop that does not encircle the origin then its image $\sqrt[3]{z}$ does the same and both return to theirs initial positions. However if $z$ rounds around the origin then $\sqrt[3]{z}$ does only one third of each round of $z$. Why is that? I don't see any proof in the text. What is so magical about the origin in this case?
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Post by Admin on Sept 10, 2022 19:10:09 GMT
There is a lot of things I can't get in this subchapter but this is the first one: 1. On the page 91 author refers to Figure [29] saying that if $z$ travels around an arbitrary loop that does not encircle the origin then its image $\sqrt[3]{z}$ does the same end both return to theirs initial positions. However if $z$ rounds around the origin then $\sqrt[3]{z}$ does only one third of each round of $z$. Why is that? I don't see any proof in the text. What is so magical about the origin in this case? Mondo You must think of $z$ as a vector from the origin to the point $z$. As $z$ moves around, the vector rotates about the origin sometimes in one direction sometimes in the other direction. Unless the vector circles the origin before returning to its original position, its angle remains the same. Vasco |
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Post by Admin on Sept 10, 2022 19:20:01 GMT
Mondo
if $z=re^{i\theta}$ then $\sqrt[3]{z}=r^{1/3}e^{i\theta/3}$ and so this answers your last question above.
Vasco
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Post by mondo on Sept 10, 2022 22:37:14 GMT
You must think of $z$ as a vector from the origin to the point $z$. As $z$ moves around, the vector rotates about the origin sometimes in one direction sometimes in the other direction. Unless the vector circles the origin before returning to its original position, its angle remains the same. if $z=re^{i\theta}$ then $\sqrt[3]{z}=r^{1/3}e^{i\theta/3}$ and so this answers your last question above. Vasco Yes I see it fully algebraically, the above equations make perfect sense to me. I only don't get this "winding" idea. Let me ask this question, when you say $\sqrt[3]{z}=r^{1/3}e^{i\theta/3}$ what was the path of $z$? Was it moving around the origin? Apparently yes because otherwise $\sqrt[3]{z}=r^{1/3}e^{i\theta}$ with no angle change right? |
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Post by Admin on Sept 11, 2022 9:39:28 GMT
Mondo
It doesn't have to move. If $z=re^{i\theta}$ where $\theta$ is any angle from $-\infty$ to $+\infty$, then $\sqrt[3]{z}=r^{(1/3)}e^{i\theta/3}$.
For example, if we choose the point $z$ with a specific angle $\theta=\phi$ then $\sqrt[3]{z}=r^{(1/3)}e^{i\phi/3}$.
If we then move $z$ around the plane then this relationship continues to hold, and if we move $z$ back to its original position without encircling the origin, then the angle $\theta$ of $z$ will still be equal to $\phi$ and so $\sqrt[3]{z}=r^{(1/3)}e^{i\theta/3}$ will still be equal to $r^{(1/3)}e^{i\phi/3}$.
If instead $z$ goes back to its original position in the plane after having gone round the origin once anticlockwise then its angle will now be $\theta=\phi+2\pi$, and so now $\sqrt[3]{z}=r^{(1/3)}e^{i(\phi+2\pi)/3}=r^{(1/3)}e^{i\phi/3}e^{i2\pi/3}$, and so its value has changed.
It is equal to the original value multiplied by $e^{i2\pi/3}$ (which means it has rotated through an angle $2\pi/3$).
So although $z$ is in the same position in the plane, its angle has changed and so the angle of its cube root has changed.
If $z$ were to go round the origin twice then the value of the cube root would again be rotated through a further angle of $2\pi/3$ and we would obtain a third cube root of $z$ while $z$ is still in the same position in the plane.
If we continue doing this then the 3 cube roots remain in the same 3 positions in the plane while the angle of $z$ keeps increasing.
Vasco
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Post by mondo on Sept 18, 2022 6:01:55 GMT
Thank you Vasco for this explanation. I got it but have some more questions that arose after reading further:
1. Below the first equation on page 96 there is a description with an example of three cubic roots for $f(0)$. The first one equal $1$ is obvious as when $z = 0$ we have $1^{\frac{1}{3}}$. The second one $e^{i\frac{2\pi}{3}}$ I believe comes from writing $1 = e^{i0}$ and after rotation of $\frac{2\pi}{3}$ that become $e^{i\frac{2\pi}{3}}$. The third one must be a result of further rotation by $\frac{2\pi}{3}$ hence we have $e^{i\frac{4\pi}{3}}$ but how is it $e^{-i\frac{2\pi}{3}}$?
2. For [37] - rays are mapped to horizontal lines because their angle is content? Similarly circles are mapped to vertical lines because of what is shown on figure [36]? That the mapping goes "almost" entirely vertically upward? That can not be the case because here the author talks about straight vertical lines.
3. In the second paragraph on page 99 author says $|log(z)|$ tends to infinity as $z -> 0$ but in fact it is $- \infty$ right?
4. How did we get from the last equation on page 100 to the first on 101? The last one on page 100, let's call it $A$ is in actually $e^L(z)$ so in order to get $L(z)$ we would need to apply $log(e^{L(z)})$ but log is what we try to get. Hence I don't get the last operation.
Thank you
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Post by Admin on Sept 19, 2022 10:59:50 GMT
Mondo
1. $e^{i4\pi/3}$ and $e^{-i2\pi/3}$ are the same point in the complex plane.
Did you delete your question about $\theta_2$ oscillating? Presumably that means that you understand it now, is that right?
2. I don't understand what you mean here. But notice that figure 36 is using $\log z$ and figure 37 is using Log $z$, two different things. Also it does different things to circles whose centre is not the origin.
3. No it's $+\infty$ don't forget what |...| means.
4. Since at the bottom of page 100 we have on the LHS that $1+\frac{L}{n}=1+....$ then we can cancel the 1 on each side and multiply the RHS by $n$ and this gives us $L$ as the RHS at the top of page 101.
Vasco
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Post by mondo on Sept 19, 2022 20:10:14 GMT
1. $e^{i4\pi/3}$ and $e^{-i2\pi/3}$ are the same point in the complex plane. I can see that from the position in the complex plane of both of them but I wonder how to show algebraically they are the same points? If I were to calculate three roots of $f(z)$ I would write $1$, $e^{i2\pi/3}$ and $e^{i4\pi/3}$. Did you delete your question about $\theta_2$ oscillating? Presumably that means that you understand it now, is that right? Yes I got it. The key to understand this for me was to visualize how $(z+i)$ and $(z-i)$ were drawn on [34b]. Meaning the addition is a vector from $-i$ to $z$. Subtraction is a vector from $i$ to $z$. 2. I don't understand what you mean here. But notice that figure 36 is using $\log z$ and figure 37 is using Log $z$, two different things. Also it does different things to circles whose centre is not the origin. Vasco, but the only difference between $log(z)$ and $Log(z)$ is the angle which in later case is limited in the range $-\pi < Arg(z) \leq \pi$. Hence the first problem I have is how can we talk about circles mapping if this angle range doesn't allow for it? I mean we need to have full $2\pi$ range to represent a circle while we are only allowed for half of it here. Next, the mapping - they say rays are mapped to horizontal lines. My understanding is this is because rays have constant angle since as we move on a ray its image will only change the real part - hence horizontal line. Similarly for circles, as we travel it, its image has constant real (module) part but angle varies hence we grow only on imaginary axis. Is that right? 3. No it's $+\infty$ don't forget what |...| means. Ok, I overlooked. Thanks 4. Since at the bottom of page 100 we have on the LHS that $1+\frac{L}{n}=1+....$ then we can cancel the 1 on each side and multiply the RHS by $n$ and this gives us $L$ as the RHS at the top of page 101. Ahh sure, that was easy. Thanks. |
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Post by Admin on Sept 20, 2022 7:21:59 GMT
Mondo
1. Since $1=e^{2\pi n}$ where $n=0,\pm 1,\pm 2,...$ then $1^{1/3}=e^{i2\pi n/3}$ where $n=0,\pm 1,\pm 2,...$ then we can see that if we choose $n=0,\pm 1$ then we get $e^{-2\pi/3},e^{0},e^{2\pi/3}$
2. $\pi<\text{Arg}(z)\leq\pi$ is the full range. $\pi-(-\pi)=2\pi$
You need to read the paragraph on page 98 that refers to figure 36 and study 36 until you understand it. Notice that only one of the circles on the left is centred on the origin and so is mapped to a straight vertical line. The other two circles are not centred on the origin and so are not mapped to vertical straight lines.
Vasco
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Post by mondo on Sept 20, 2022 18:27:35 GMT
1. Since $1=e^{2\pi n}$ where $n=0,\pm 1,\pm 2,...$ then $1^{1/3}=e^{i2\pi n/3}$ where $n=0,\pm 1,\pm 2,...$ then we can see that if we choose $n=0,\pm 1$ then we get $e^{-2\pi/3},e^{0},e^{2\pi/3}$ Yes clear. The other way to look at it is to realize that all roots are separated by $120^{\circ}$ and because the first one sits at the real axis the other two are symmetric in regards to $X$ axis hence must be of equal and opposite value. I found this explanation easier and more "visual" (as this book motto is)  2. $\pi<\text{Arg}(z)\leq\pi$ is the full range. $\pi-(-\pi)=2\pi$ Ok, I confused the meaning of this inequality. What it really means is we can go up to $\pi$ in both clockwise and anticlockwise directions. The point is to not cross $\pi$ You need to read the paragraph on page 98 that refers to figure 36 and study 36 until you understand it. Notice that only one of the circles on the left is centred on the origin and so is mapped to a straight vertical line. The other two circles are not centred on the origin and so are not mapped to vertical straight lines. Ok I see it, the circle centered at the origin, in sashed line has a "constant radius" in reference to orgin and hence its image is a straight line going downwards. If it were circling in the opposite direction then its image would go vertically upwards. The middle circle is almost a vertical line because its real component changes as we go. The smallest circle is again a circle under mapping because there is no singularity inside of it. Thank you |
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Post by Admin on Sept 20, 2022 19:35:11 GMT
Mondo
1. In your post #7 above you specifically asked for an algebraic explanation!!
2. I wouldn't say "... not cross $\pi$". That sounds very imprecise. Much better to say "... not cross the negative real axis"
Vasco
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Post by Admin on Sept 21, 2022 9:17:10 GMT
Mondo
The smallest circle is not mapped to a circle but to an ellipse!
Vasco
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Post by mondo on Sept 21, 2022 16:43:27 GMT
1. In your post #7 above you specifically asked for an algebraic explanation!! Yes I know, I like your algebraic proof, just wanted to appreciate the "visual" way of looking at it. 2. I wouldn't say "... not cross $\pi$". That sounds very imprecise. Much better to say "... not cross the negative real axis" I think I need some more help on why do we selected this "branch point". The book says to have a single valued branches. But I don't see it fully. What would happen when we cross that negative real axis? The smallest circle is not mapped to a circle but to an ellipse! True, can you see it in your mind without looking at the image of the mapping? I have such confidence with most plots in this chapter but this one is harder to imagine. Thank you. |
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Post by mondo on Sept 23, 2022 6:39:04 GMT
2. I wouldn't say "... not cross $\pi$". That sounds very imprecise. Much better to say "... not cross the negative real axis" I think I need some more help on why do we selected this "branch point". The book says to have a single valued branches. But I don't see it fully. What would happen when we cross that negative real axis? Ok, I think I got it. A branch point is defined as a place which when encircled the function will not return to its starting point. So to avoid it we decided to make a cut from the origin to infinity. The discontinuity (a jump on the figure [37]) is because when we cross negative X axis the $Arg(z)$ from large positive value become large negative one (due to our principal angle restriction). |
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