• #71
Frabjous said:
No. You want to view a+ib as a vector and re as a scalar which makes “=“ ambiguous. They are both complex numbers with a clean definition of “=“.
I still don't get it. When I say ##y = 2 + 3## I know that ##y = 5## and that ##5 = 2 + 3##. I can, looks like, unpack the ##5## into ##3 + 2##. Can I do something to ##e^{i \varphi}## (is there an algebraic algorithm?) to get to ##a + ib##?

I do know that ##e^{i \varphi} = \cos \varphi + i \sin \varphi##, but it's the same issue here too. Is there summation of a series or a limit of a function or something else involved in proving this equality? 🤔
 
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  • #73
Agent Smith said:
Gracias, but do I have a case if I say that the equality ##re^{i\varphi} = a + ib## is not as evident/obvious as the one you see in ##5 = 5##? 🤔
Good point. In fact, the definition of what a complex exponent means is not so simple. The correct meaning of ##e^{i\theta}## can only be accepted after some work. The work is not elementary. I think that the definition that stays within the more basic facts is the limit definition. The associated proofs are still not elementary. That is why it took a genius like Euler to discover Euler's formula.

PS. You should get comfortable in knowing that there are solid proofs that Euler's formula, ##e^{i\theta}=\cos(\theta)+i \sin(\theta)##. Then you are free to use it and forget the details of the proof. It is very useful.
 
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  • #74
Agent Smith said:
I still don't get it. When I say ##y = 2 + 3## I know that ##y = 5## and that ##5 = 2 + 3##. I can, looks like, unpack the ##5## into ##3 + 2##. Can I do something to ##e^{i \varphi}## (is there an algebraic algorithm?) to get to ##a + ib##?

I do know that ##e^{i \varphi} = \cos \varphi + i \sin \varphi##, but it's the same issue here too. Is there summation of a series or a limit of a function or something else involved in proving this equality? 🤔
Imagine any point in a planar, two-dimensional coordinate system, except the origin that requires particular handling since the distance between the origin and the origin is zero and we cannot measure an angle between two lines of length zero.

So given any point ##z## in that grid, the coordinate system of the plane. How would you describe it to me if I asked you to show me a way from the origin to the point ##z##?

As I see it, you could either tell me to go ##a## units to the right and ##b## units to the top, which means in equations that ##z=(a,b)=a+ib## if we identify the plane with complex numbers, or you could tell me turn myself and look horizontally to the right, then turn myself by an angle of ##\varphi ## counterclockwise and walk ##r## units into the new direction, which means ##z=(a,b)= r\cdot e^{i \varphi }.##
1721578600367.png

##r=|z|=\sqrt{a^2+b^2}## (Pythagoras) and times ##e^{i \varphi }## is the turning from looking horizontally to looking towards ##z.##

There are several ways to see why times ##e^{i \varphi }## is a counterclockwise rotation of the point ##(r,0)## to the point ##(a,b).## One is to look at Euler's formula (as linked to twice in previous posts) or by the study of complex multiplication (see picture in post #59). The only ambiguity here is, that the vertical coordinate is once the second in ##(a,b),## the ##b## and at the same time identified with the complex number ##i b##. You can imagine this as units. We have the unit ##[1]## horizontally and the unit ## [\mathrm{i}]## vertically. We do not write the units in ##(a,b)## if we refer to a point on the grid, and do write the units if we identify such a point with a complex number: ##(a,b)=a+ i b.##
 
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  • #75
martinbn said:
Some people call certain objects "comex curves", other people call them "Riemann surfaces".
Ah, the commercial exchange. They got great curves ;).
 
  • #76
fresh_42 said:
Separating complex numbers into real and imaginary parts is exactly what this article wants to put into the second row of consideration, and viewing complex numbers as elements of one field in the first place rather than reducing them to a simple, real vector space.

This narrowed view as ##a+ib## is in my opinion what hides the beauty of complex analysis, or the algebraic background of complex numbers. It is a widespread disease and not really helpful. This article was all about
$$
\mathbb{C} \neq \mathbb{R}^2.
$$
At what level of structure? Clearly one is an Algebra, while Euclidean space isn't. Though they are Topologically but not Geometrically identical.
 
  • #77
martinbn said:
Some people call certain objects "comex curves", other people call them "Riemann surfaces".
Before getting into more advanced subjects like Riemann surfaces, there are conformal mappings between simply connected open sets in the complex plane.
 
  • #78
WWGD said:
At what level of structure? Clearly one is an Algebra, while Euclidean space isn't. Though they are Topologically but not Geometrically identical.
The complex numbers and the real numbers carry many structures. I wanted to point out the field structure because that is what is necessary for doing complex calculus. The vector space structure is insufficient. Two-dimensional real analysis and complex analysis are very different. Yet, students are introduced to complex numbers via the Gaußian plane. They start with a handicap for no reason and everything appears to be miraculous when complex analysis starts - only because some teachers prefer to draw little arrows instead of teaching the algebra as a field behind it. And the algebraic part of an algebraic closure and the field axioms aren't even difficult. But no, little arrows rule. This is in my opinion a bad approach. The little arrows are necessary, but after the algebraic subject is settled, not before. There is a reason people come up with equations like
$$
-1=\sqrt{-1}\cdot \sqrt{-1} =\sqrt{(-1)\cdot(-1)}= \sqrt{1}=1 \quad (*)
$$
The reason is, that in the world of little arrows, everything looks real, just two-dimensional. I know, that my opinion is inconvenient since it is against the inertia of the old argument: We always did it this way.

I need the two-dimensional view, too, by writing ##\mathbb{C}\cong \mathbb{R}[T]/ \bigl\langle T^2+1 \bigr\rangle ,## of course not without mentioning that ## \bigl\langle T^2+1 \bigr\rangle## is a maximal ideal. My imaginary unit would be the holomorphic image ##t## of the variable ##T## and ##(*)## would be impossible to set equal:
$$
-1 \equiv t^2 \equiv t\cdot t \not\equiv 1\cdot 1 =1.
$$
The algebraic view is also the key to understanding that complex numbers don't have an Archimedean order and that squares are not automatically positive anymore. If these concepts are settled, let's speak about our handicapped possibilities to visualize complex numbers despite the fact that
$$
\mathbb{R}[T]/ \bigl\langle T^2+1 \bigr\rangle \neq \mathbb{R}[T]/ \bigl\langle T^2 \bigr\rangle
$$
 
  • #79
fresh_42 said:
The complex numbers and the real numbers carry many structures. I wanted to point out the field structure because that is what is necessary for doing complex calculus. The vector space structure is insufficient. Two-dimensional real analysis and complex analysis are very different. Yet, students are introduced to complex numbers via the Gaußian plane. They start with a handicap for no we everything appears to be miraculous when complex analysis starts - only because some teachers prefer to draw little arrows instead of teaching the algebra as a field behind it. And the algebraic part of an algebraic closure and the field axioms aren't even difficult. But no, little arrows rule. This is in my opinion a bad approach. The little arrows are necessary, but after the algebraic subject is settled, not before. There is a reason people come up with equations like
$$
-1=\sqrt{-1}\cdot \sqrt{-1} =\sqrt{(-1)\cdot(-1)}= \sqrt{1}=1 \quad (*)
$$
The reason is, that in the world of little arrows, everything looks real, just two-dimensional. I know, that my opinion is inconvenient since it is against the inertia of the old argument: We always did it this way.

I need the two-dimensional view, too, by writing ##\mathbb{C}\cong \mathbb{R}[T]/ \bigl\langle T^2+1 \bigr\rangle ,## of course not without mentioning that ## \bigl\langle T^2+1 \bigr\rangle## is a maximal ideal. My imaginary unit would be the holomorphic image ##t## of the variable ##T## and ##(*)## would be impossible to set equal:
$$
-1 \equiv t^2 \equiv t\cdot t \not\equiv 1\cdot 1 =1.
$$
The algebraic view is also the key to understanding that complex numbers don't have an Archimedean order and that squares are not automatically positive anymore. If these concepts are settled, let's speak about our handicapped possibilities to visualize complex numbers despite the fact that
$$
\mathbb{R}[T]/ \bigl\langle T^2+1 \bigr\rangle \neq \mathbb{R}[T]/ \bigl\langle T^2 \bigr\rangle
$$
An interesting question is whether we can define s multiplication in ## \mathbb R^2## to turn it into an Algebra. The standard inner product won't work, but why not other operation? Edit:nd ##<T^2>## is not maximal, so quotient wont be a field. But, good point, the two aren't order-isomorphic. Similar for ##\mathbb C^{n}## and ##\mathbb R^{2n}##.
 
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  • #80
WWGD said:
An interesting question is whether we can define s multiplication in ## \mathbb R^2## to turn it into an Algebra. The standard inner product won't work, but why not other operation? And ##<T^2>## is not maximal, so quotient wont be a field.
You could multiply it componentwise, or use a Lie multiplication of a two-dimensional real Lie algebra. However, you won't get a field, except if you use the graduation ##\mathbb{R}\oplus i \mathbb{R}## with ##(i\mathbb{R})\cdot (i\mathbb{R}) \subseteq \mathbb{R}.##
 
  • #81
Agent Smith said:
I still don't get it. When I say ##y = 2 + 3## I know that ##y = 5## and that ##5 = 2 + 3##. I can, looks like, unpack the ##5## into ##3 + 2##. Can I do something to ##e^{i \varphi}## (is there an algebraic algorithm?) to get to ##a + ib##?

I do know that ##e^{i \varphi} = \cos \varphi + i \sin \varphi##, but it's the same issue here too. Is there summation of a series or a limit of a function or something else involved in proving this equality? 🤔

It requires taylor series.
 
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  • #82
MidgetDwarf said:
It requires taylor series.
That is one good way. There are a couple of ways to define the complex exponent and each definition of ##e^{i\theta}## has its own proof of Euler's formula. Wikipedia has a few options here.
 
  • #83
FactChecker said:
Good point. In fact, the definition of what a complex exponent means is not so simple. The correct meaning of ##e^{i\theta}## can only be accepted after some work. The work is not elementary. I think that the definition that stays within the more basic facts is the limit definition. The associated proofs are still not elementary. That is why it took a genius like Euler to discover Euler's formula.

PS. You should get comfortable in knowing that there are solid proofs that Euler's formula, ##e^{i\theta}=\cos(\theta)+i \sin(\theta)##. Then you are free to use it and forget the details of the proof. It is very useful.
Asante sana. That's quality advice. So there are intriguing bridges (@MidgetDwarf ), like The Taylor Series.
It was like A = B and B = C, hence A = C than a more direct A = C (here B is the "intriguing bridge")
 
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  • #84
fresh_42 said:
Imagine any point in a planar, two-dimensional coordinate system, except the origin that requires particular handling since the distance between the origin and the origin is zero and we cannot measure an angle between two lines of length zero.

So given any point ##z## in that grid, the coordinate system of the plane. How would you describe it to me if I asked you to show me a way from the origin to the point ##z##?

As I see it, you could either tell me to go ##a## units to the right and ##b## units to the top, which means in equations that ##z=(a,b)=a+ib## if we identify the plane with complex numbers, or you could tell me turn myself and look horizontally to the right, then turn myself by an angle of ##\varphi ## counterclockwise and walk ##r## units into the new direction, which means ##z=(a,b)= r\cdot e^{i \varphi }.##

##r=|z|=\sqrt{a^2+b^2}## (Pythagoras) and times ##e^{i \varphi }## is the turning from looking horizontally to looking towards ##z.##

There are several ways to see why times ##e^{i \varphi }## is a counterclockwise rotation of the point ##(r,0)## to the point ##(a,b).## One is to look at Euler's formula (as linked to twice in previous posts) or by the study of complex multiplication (see picture in post #59). The only ambiguity here is, that the vertical coordinate is once the second in ##(a,b),## the ##b## and at the same time identified with the complex number ##i b##. You can imagine this as units. We have the unit ##[1]## horizontally and the unit ## [\mathrm{i}]## vertically. We do not write the units in ##(a,b)## if we refer to a point on the grid, and do write the units if we identify such a point with a complex number: ##(a,b)=a+ i b.##
Arigato gozaimus for the explanation. It's clearer now. We have 2 ways of representing the same object, one with polar coordinates (length, angle) and the other with normal Cartesian coordinates (with an x and a y). I did a bit of informal history of mathematics years ago, but seem to have forgotten which of the two ideas, vectors/complex numbers, came first. Either way, it's nice to see them complement each other like this.
 
  • #85
fresh_42 said:
The complex numbers and the real numbers carry many structures. I wanted to point out the field structure because that is what is necessary for doing complex calculus.
The field structure is not enough, you need the absolute value. Otherwise you could do calculus in any field, which is not the case.
fresh_42 said:
The vector space structure is insufficient. Two-dimensional real analysis and complex analysis are very different.
Yes, but the two dimensional real space has additional structure, the complex structure. It is possible, and often done, to view any complex manifold as a real manifold with a complex structure.
fresh_42 said:
Yet, students are introduced to complex numbers via the Gaußian plane. They start with a handicap for no reason and everything appears to be miraculous when complex analysis starts - only because some teachers prefer to draw little arrows instead of teaching the algebra as a field behind it.
I don't think the first introduction of complex numbers is always done with the idea that they are needed for complex analysis. Some think of them from a more algebraic point of view. I think the choice to emphasize the complex plane is purely pedagogical. Also, I think that if you stress that the complex numbers are a field, complex analysis will still look like a miracle.
fresh_42 said:
And the algebraic part of an algebraic closure and the field axioms aren't even difficult. But no, little arrows rule. This is in my opinion a bad approach. The little arrows are necessary, but after the algebraic subject is settled, not before. There is a reason people come up with equations like
$$
-1=\sqrt{-1}\cdot \sqrt{-1} =\sqrt{(-1)\cdot(-1)}= \sqrt{1}=1 \quad (*)
$$
The reason is, that in the world of little arrows, everything looks real, just two-dimensional. I know, that my opinion is inconvenient since it is against the inertia of the old argument: We always did it this way.
No, I don't think your opinion is inconvenient, but I also don't see the problem with the standard aproach to complex numbers. The more ways to see something the better the understanding.
fresh_42 said:
I need the two-dimensional view, too, by writing ##\mathbb{C}\cong \mathbb{R}[T]/ \bigl\langle T^2+1 \bigr\rangle ,## of course not without mentioning that ## \bigl\langle T^2+1 \bigr\rangle## is a maximal ideal. My imaginary unit would be the holomorphic image ##t## of the variable ##T## and ##(*)## would be impossible to set equal:
$$
-1 \equiv t^2 \equiv t\cdot t \not\equiv 1\cdot 1 =1.
$$
The algebraic view is also the key to understanding that complex numbers don't have an Archimedean order and that squares are not automatically positive anymore. If these concepts are settled, let's speak about our handicapped possibilities to visualize complex numbers despite the fact that
$$
\mathbb{R}[T]/ \bigl\langle T^2+1 \bigr\rangle \neq \mathbb{R}[T]/ \bigl\langle T^2 \bigr\rangle
$$
It is funny you should say that, because the dual numbers ##\mathbb{F}[T]/ \bigl\langle T^2 \bigr\rangle## can be used to define the arrows that you objected to.
 
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  • #86
martinbn said:
I think that if you stress that the complex numbers are a field, complex analysis will still look like a miracle.
Very well said. I can't think of a good way to go from field properties to geometric properties like conformal mappings, although it does set the stage for inner product spaces.
 
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  • #87
I believe ##e^z## is the only Analytic extension to the Complex Plane satisfying ##\frac {de^z}{dz}=e^z##.
But you have a valid point, imo. If ##z## is a Complex variable, why is it broken into Real and Complex parts in so many settings?

Are there intermediate extensions between ## \mathbb Q \subset \mathbb R \subset \mathbb C ##
?
 
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  • #88
Never mind. Tower theorem makes the intermediate extensions impossible.
 
  • #89
martinbn said:
It is funny you should say that, because the dual numbers ##\mathbb{F}[T]/ \bigl\langle T^2 \bigr\rangle## can be used to define the arrows that you objected to.
Exactly, that's my point. This is the arrow world and ##\mathbb{F}[T]/ \bigl\langle T^2+1 \bigr\rangle## something completely different.

I conclude that you are so hardened in your personal opinions that you aren't willing to even consider my point of view. There is no way to reason against "We always did it this way!".

What's next? "If everybody would do this?" or "I have my rules."?

You could at least accept that not everybody thinks that "We always did it this way!" is a satisfactory argument.
 
  • #90
fresh_42 said:
Exactly, that's my point. This is the arrow world and ##\mathbb{F}[T]/ \bigl\langle T^2+1 \bigr\rangle## something completely different.
What exactly do you mean by an arrow?
fresh_42 said:
I conclude that you are so hardened in your personal opinions that you aren't willing to even consider my point of view. There is no way to reason against "We always did it this way!".
How? How do you know my opinion? I haven't expressed it!
fresh_42 said:
What's next? "If everybody would do this?" or "I have my rules."?

You could at least accept that not everybody thinks that "We always did it this way!" is a satisfactory argument.
 
  • #91
I have no problem with the Gaußian plane. It is useful and necessary. I have a problem with the fact that its inadequacy is lost. It cannot represent ##(\mathrm{i}\mathbb{R})\cdot (\mathrm{i}\mathbb{R})\subseteq \mathbb{R}## in a sufficient way. It is information that got lost by representing a one-dimensional field in a two-dimensional Euclidean plane. But it is a very important information. Starting with the plane does not adequately take this into account and must be painfully corrected later on.
 
  • #92
fresh_42 said:
I have no problem with the Gaußian plane. It is useful and necessary. I have a problem with the fact that its inadequacy is lost. It cannot represent ##(\mathrm{i}\mathbb{R})\cdot (\mathrm{i}\mathbb{R})\subseteq \mathbb{R}## in a sufficient way. It is information that got lost by representing a one-dimensional field in a two-dimensional Euclidean plane. But it is a very important information. Starting with the plane does not adequately take this into account and must be painfully corrected later on.
I am not sure what you mean. The information is not lost, noone defines the complex numbers without the multiplication. So surely the product of imaginary numbers is real is clear no matter what the definition is. By the way what was your preferred way of defining the complex numbers?
 
  • #93
Agent Smith said:
What is its geometric meaning, if I may ask?
I'm afraid that I did you a disservice by concentrating on the geometry of Euler's formula. The geometric properties of general analytic and harmonic functions are beautiful.
 
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  • #94
FactChecker said:
I'm afraid that I did you a disservice by concentrating on the geometry of Euler's formula. The geometric properties of general analytic and harmonic functions are beautiful.
As long as ##e^{i \varphi}## is a rotation of ##1## by an angle ##\varphi##, it's ok. Is it?

The general formula being ##re^{i \varphi}##, where the length ##r## is being rotated countreclockwise by ##\varphi##.
 
  • #95
Agent Smith said:
As long as ##e^{i \varphi}## is a rotation of ##1## by an angle ##\varphi##, it's ok. Is it?

The general formula being ##re^{i \varphi}##, where the length ##r## is being rotated countreclockwise by ##\varphi##.
You mean scaling by a factor of ##1##? But, yes, Complex multiplication is equivalent to scaling and rotation.
 
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  • #96
Agent Smith said:
As long as ##e^{i \varphi}## is a rotation of ##1## by an angle ##\varphi##, it's ok. Is it?

The general formula being ##re^{i \varphi}##, where the length ##r## is being rotated countreclockwise by ##\varphi##.
Correct.
 
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  • #97
Agent Smith said:
As long as ##e^{i \varphi}## is a rotation of ##1## by an angle ##\varphi##, it's ok. Is it?

The general formula being ##re^{i \varphi}##, where the length ##r## is being rotated countreclockwise by ##\varphi##.
My concern is that you can not really say that a "length" is rotated. A "length" does not have a direction, so it can not be rotated. A vector has both a length and a direction. It is better to say that ##re^{i \varphi}## is a vector in the complex plane of length ##r## pointing in the direction ##\varphi##. The vector from the origin to the positive number ##r## on the real line is rotated. A "rotation" happens when you multiply another complex number, ##z##, by ##e^{i \varphi}## and there will be the usual change in length if you also multiply by a real number, ##r##.
 
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  • #98
FactChecker said:
My concern is that you can not really say that a "length" is rotated. A "length" does not have a direction, so it can not be rotated. A vector has both a length and a direction. It is better to say that ##re^{i \varphi}## is a vector in the complex plane of length ##r## pointing in the direction ##\varphi##. The vector from the origin to the positive number ##r## on the real line is rotated. A "rotation" happens when you multiply another complex number, ##z##, by ##e^{i \varphi}## and there will be the usual change in length if you also multiply by a real number, ##r##.
Not to contradict you, but you mean to say that by adding direction to a length it now can be transformed? 🤔 Transformations can be/done on line segments, no? How does adding a direction enable transformations? Do triangles have direction?
 
  • #99
Agent Smith said:
Not to contradict you, but you mean to say that by adding direction to a length it now can be transformed? 🤔 Transformations can be/done on line segments, no? How does adding a direction enable transformations? Do triangles have direction?
Maybe rather than direction, you may say they have ( or be given), an orientation.
 
  • #100
Agent Smith said:
Not to contradict you, but you mean to say that by adding direction to a length it now can be transformed? 🤔
It can be rotated if it has an initial direction. Vectors have length and direction. A length (eg length=5 feet) does not. I wouldn't say that all transformations require a direction.
Agent Smith said:
Transformations can be/done on line segments, no?
A straight line segment would need to be given an orientation before you could say that it was "rotated".
Agent Smith said:
How does adding a direction enable transformations? Do triangles have direction?
Every side of a given triangle has a length and two endpoints. We would have to do more. Each side would need a starting point and a endpoint so that they have directions. For instance, there would be clockwise and counterclockwise orientations of the sides. And some might have mixed combinations of side orientations that are neither all clockwise or all counterclockwise.
 
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  • #101
FactChecker said:
It can be rotated if it has an initial direction. Vectors have length and direction. A length (eg length=5 feet) does not. I wouldn't say that all transformations require a direction.

A straight line segment would need to be given an orientation before you could say that it was "rotated".

Every side of a given triangle has a length and two endpoints. We would have to do more. Each side would need a starting point and a endpoint so that they have directions. For instance, there would be clockwise and counterclockwise orientations of the sides. And some might have mixed combinations of side orientations that are neither all clockwise or all counterclockwise.
It's likely that I was taught only baby transformations where some details were left out for simplicity. Si, I do recall losing points when I failed to recognize ##\triangle ABC## was not the same as ##\triangle CBA##. When does orientation become important? The example above was a step in the proof of the Pythagorean Theorem, cogito (fairly certain).
 
  • #102
Agent Smith said:
When does orientation become important?
One example -- which may not be what you are thinking about at all...

Suppose that you are contemplating a three dimensional surface enclosing a radiant source. This three dimensional surface may not be convex. It may be pretty twisty.

You may want to think about how much total radiant flow goes through the surface. So you decide to integrate. You integrate incremental cross-sectional area ##a## times the radiating charge ##q## divided by ##r^2## for the inverse square law. And you get the wrong answer.

Because not all of the area elements are facing the radiating source. And the shape is folded enough that some area elements are facing the wrong way -- the radiant flux is actually flowing "in" rather than "out" through some elements.

So you invent the idea of an oriented area element. So that the area element is ##\vec{a}## rather than just ##a##. And stick a vector dot product into your integral so that you can correctly track whether the area element is pointed "in", "out" or "mostly sideways".

I've not been exposed to such an idea, but it seems to me that one could calculate the oriented area of triangle ##\bigtriangleup {ABC}## in terms of the vector cross product of two oriented sides: $$\vec{a} = \frac {(\vec{C} - \vec{B}) \times (\vec{B} - \vec{A})} {2} = \frac{(\vec{A} - \vec{C}) \times (\vec{C} - \vec{B})} {2} = \frac{(\vec{B} - \vec{A}) \times (\vec{A} - \vec{C})} {2}$$Meanwhile, for the mirror image triangle ##\bigtriangleup {CBA}## we have
$$-\vec{a} = \vec{a_\text{mirror}} = \frac {(\vec{A} - \vec{B}) \times (\vec{B} - \vec{C})} {2} = \frac{(\vec{C} - \vec{A}) \times (\vec{A} - \vec{B})} {2} = \frac{(\vec{B} - \vec{C}) \times (\vec{C} - \vec{A})} {2}$$
In two dimensions, the notion of a winding number depends on the orientation of a closed planar curve. If you follow the curve from one end to the other, it will have (at least locally) a right side and a left side. You trace a path from a chosen point to infinity and count the number of times the path goes through the curve from left to right minus the number of times it passes from right to left. The result is the winding number.
 
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  • #103
Agent Smith said:
When does orientation become important? The example above was a step in the proof of the Pythagorean Theorem, cogito (fairly certain).
Any motion, force, acceleration, etc in 3-space has a direction in 3-space.
 
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  • #104
@jbriggs444 . "one way" of solving the problem in your post was to use vectors (quantity + direction). There was a snippet on eigen values, which was stated succinctly as a situation where vector computation is reduced/equivalent to scalar computation, the scalar components being eigen values. Am I correct? How now this? Gracias for engaging with me.

@FactChecker , danke for the examples. I failed to ask the right question, which is why did orientation of the triangles used in the proof of The Pythagorean Theorem matter? I was penalized for marking (say) ##\triangle ABC \cong \triangle DEF## as a valid step in the proof; The correct answer was ##\triangle CBA \cong \triangle DEF##. From the bits and pieces I remember, the corresponding angles had to be equal and in the former statement ##\triangle ABC \cong \triangle DEF## this was not the case, while in the latter ##\triangle CBA \cong \triangle DEF## it was.
 
  • #105
Agent Smith said:
@jbriggs444 . "one way" of solving the problem in your post was to use vectors (quantity + direction). There was a snippet on eigen values, which was stated succinctly as a situation where vector computation is reduced/equivalent to scalar computation, the scalar components being eigen values. Am I correct? How now this? Gracias for engaging with me.

@FactChecker , danke for the examples. I failed to ask the right question, which is why did orientation of the triangles used in the proof of The Pythagorean Theorem matter? I was penalized for marking (say) ##\triangle ABC \cong \triangle DEF## as a valid step in the proof; The correct answer was ##\triangle CBA \cong \triangle DEF##. From the bits and pieces I remember, the corresponding angles had to be equal and in the former statement ##\triangle ABC \cong \triangle DEF## this was not the case, while in the latter ##\triangle CBA \cong \triangle DEF## it was.
In the bottom case, congruence doesn't depend on orientation, but by a combination of relations between sizes of sides, angles. I'm not sure the ancient Greeks who laid out such notions were even aware of general notions of orientation, orientability.
 
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