Insights Views On Complex Numbers

[Hans de Vries]

This is a step in the opposite direction of Geometric Algebra/Calculus, which I consider a very good approach to a large number of advanced physics and engineering subjects in higher dimensions. The main advantage of GA is that it starts with a few basic geometric observations and (with good book-keeping) can derive a great deal from those. The main disadvantage of GA is that it has a significant learning curve and requires a lot of practice and book-keeping.

I don't know if you are responding to my post? It was not meant to diminish Geometric Algebra which I have used a lot in all kinds of fields in Physics. Be aware though that high levels of abstraction can sometimes obscure the incompleteness of a mathematical toolbox.

 

[Agent Smith]

Many people consider Euler's formula, $e^{i\theta} = \cos(\theta)+i\sin(\theta)$, to be the most important equation in mathematics. I like its geometric meaning.

What is its geometric meaning, if I may ask?

 

[fresh_42]

What is its geometric meaning, if I may ask?

We have the complex number $z=x+iy = e^{i \theta}.$ Its real $x$-component is $x=\sin \theta$ and its imaginary $iy$-component is $y=\cos \theta$ by the definition of sine and cosine at the right triangle. Together, this becomes
$$
z=x+iy=\cos \theta +i \sin \theta
$$
That the RHS equals $e^{i \theta }$ is Euler's formula.

 

[FactChecker]

What is its geometric meaning, if I may ask?

I can only give some rather vague intuitive reactions.
1) I see it as relating the XY coordinate system to the polar coordinate system.
2) The way that complex multiplication relates to rotations around the origin, then leads to the rotations in the other direction relating to division. So there is always a complex multiplicative inverse (unlike general matrices).
3) Rotations are significant in cyclic functions.
4) The behavior of the exponential function under differentiation is very significant in differential equations.

I'm sure that others can give a more concrete and coherent answer.

 

[Agent Smith]

View attachment 348633
We have the complex number $z=x+iy = e^{i \theta}.$ Its real $x$-component is $x=\sin \theta$ and its imaginary $iy$-component is $y=\cos \theta$ by the definition of sine and cosine at the right triangle. Together, this becomes
$$
z=x+iy=\cos \theta +i \sin \theta
$$
That the RHS equals $e^{i \theta }$ is Euler's formula.

That's really kind of you to explain what seems to be the relationship between complex numbers $z = x + iy$ and $e^{i \theta}$. What does $e^{i\theta}$ do? Rotate $1$ by angle of $\theta$?

 

[FactChecker]

That's really kind of you to explain what seems to be the relationship between complex numbers $z = x + iy$ and $e^{i \theta}$. What does $e^{i\theta}$ do? Rotate $1$ by angle of $\theta$?

Yes. That would be the position of $e^{i\theta}$ in the complex plane. The rotation is counter-clockwise. In addition, multiplication of any complex number, $z$, by $e^{i\theta}$ will rotate $z$ counter-clockwise by the angle $\theta$.

 

[Agent Smith]

Yes. That would be the position of $e^{i\theta}$ in the complex plane. The rotation is counter-clockwise. In addition, multiplication of any complex number, $z$, by $e^{i\theta}$ will rotate $z$ counter-clockwise by the angle $\theta$.

Is there software online where I can test this?

So $e^{i \theta} \times z$ will rotate $z$ counterclockwise by $\theta$?

 

[FactChecker]

Is there software online where I can test this?

So $e^{i \theta} \times z$ will rotate $z$ counterclockwise by $\theta$?

It looks like Wolfram\Alpha can help, but I am not familiar with it. See this.

 

[fresh_42]

Is there software online where I can test this?

So $e^{i \theta} \times z$ will rotate $z$ counterclockwise by $\theta$?

You can use https://www.wolframalpha.com/ for such calculations but they aren't made for such calculations that are quite basic.

Once you studied/learned/accepted Euler's formula, and the Wikipedia link above has several proofs of the identity
$$
e^{i\varphi }=\cos \varphi + i \sin \varphi
$$
then it is quite easy. Every complex number $z=x+iy$ can be written as $z=|z|\cdot e^{i\varphi }$ where $|z|=\sqrt{x^2+y^2}$ is the absolute value of $z$ that is the distance from the origin $0+i\cdot 0=0$ of the coordinate system, and $\varphi$ the so-called argument of $z,$ the counterclockwise measured angle from the positive real axis, the $x$-axis.

Now we get
$$
e^{i \theta} \cdot z = e^{i \theta} \cdot |z|\cdot e^{i\varphi }=|z|\cdot e^{i\theta + i\varphi }=|z|\cdot e^{i(\theta + \varphi)} .
$$
The result is therefore a complex number of absolute value $|z|$ and an argument $\theta + \varphi$ which is the original $\varphi$ rotated counterclockwise (we always measure from the positive real axis counterclockwise, of course, if the angles are positive) to $\theta + \varphi$ which is just a rotation by $\theta.$

Source: https://de.wikipedia.org/wiki/Komplexe_Zahl#Multiplikation
The pictures in the English version were a bit messy.

 

[Agent Smith]

@fresh_42 I see. I'm amazed that a "simple" expression like $e^{i \theta}$ equals $a + bi$. How??

So if $z = 1$ then $1 = |1| \times e^{i \times 0} = e^0 = 1$?

 

[fresh_42]

@fresh_42 I see. I'm amazed that a "simple" expression like $e^{i \theta}$ equals $a + bi$. How??

By Euler's formula, or if you prefer that, by expressing the same (complex) number $z$ once is Cartesian coordinates $z=(x,y)=x+iy$ and then by polar coordinates $z=r\cdot e^{i \varphi }$ where $r=|z|=\sqrt{x^2+y^2}$ and $\varphi$ determined by $\dfrac{x}{r}=\cos \varphi$ and $\dfrac{y}{r}=\sin \varphi .$ It uses the fact that we can represent complex numbers in a planar coordinate system with a real axis ($x$) and a purely imaginary axis ($i y$).

So if $z = 1$ then $1 = |1| \times e^{i \times 0} = e^0 = 1$?

Yes.

 

[Agent Smith]

@fresh_42

I'm a bit unsure about how $r \cdot e^{i \phi} = z = a + ib, r = \sqrt {a^2 + b^2}$

The "=" sign doesn't seem to have the same meaning as in $2 + 3 = 5$ or $7.5 = 7.5$

 

[Frabjous]

It’s polar coordinates generalized to the complex plane.

 

[Agent Smith]

It’s polar coordinates generalized to the complex plane.

Sorry, couldn't parse that. I know that graphically, $r \cdot e^{i \phi} \overset{?} = z = a + ib$. I've seen the vector/complex number plot in the complex plane, they're identical. So am I supposed to interpret the equality graphically, as representing the same vector (the same complex number)?

 

[Agent Smith]

$e^{i \varphi}$ is the line segment $(0, 0) \text{ to } (1, 0)$ rotated counterclockwise by an angle $\varphi$?

 

[Frabjous]

Sorry, couldn't parse that. I know that graphically, $r \cdot e^{i \phi} \overset{?} = z = a + ib$. I've seen the vector/complex number plot in the complex plane, they're identical. So am I supposed to interpret the equality graphically, as representing the same vector (the same complex number)?

re^iφ is a complex number.
We know that any complex number can be written as a+ib.
So “normal” equality holds.

 

[Frabjous]

$e^{i \varphi}$ is the line segment $(0, 0) \text{ to } (1, 0)$ rotated counterclockwise by an angle $\varphi$?

While one can apply “vector like operations” to complex numbers, they are more than that. That is the major point of the Insight.

 

[Agent Smith]

re^iφ is a complex number.
We know that any complex number can be written as a+ib.
So “normal” equality holds.

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 $5 = 5$? These are representations of the same mathematical object (complex numbers) in 2 different systems (polar coordinate system and the usual Cartesian coordinate system). It's more like saying that love (English) = Ishq (Urdu) than saying love = love, right?

 

[Agent Smith]

@jbriggs444 , , beginner here.

 

[Frabjous]

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 $5 = 5$? These are representations of the same mathematical object (complex numbers) in 2 different systems (polar coordinate system and the usual Cartesian coordinate system). It's more like saying that love (English) = Ishq (Urdu) than saying love = love, right?

No. You want to view a+ib as a vector and re^iφ as a scalar which makes “=“ ambiguous. They are both complex numbers with a clean definition of “=“.

 

[Agent Smith]

No. You want to view a+ib as a vector and re^iφ 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?

 

[Frabjous]

Wikipedia describes a couple of different proofs.
https://en.wikipedia.org/wiki/Euler's_formula#Proofs

 

[FactChecker]

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.

 

[fresh_42]

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 $z=(a,b)= r\cdot e^{i \varphi }.$

$r=|z|=\sqrt{a^2+b^2}$ (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: $(a,b)=a+ i b.$

 

[WWGD]

Some people call certain objects "comex curves", other people call them "Riemann surfaces".

Ah, the commercial exchange. They got great curves ;).

 

[WWGD]

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.

 

[FactChecker]

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.

 

[fresh_42]

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
$$
\mathbb{R}[T]/ \bigl\langle T^2+1 \bigr\rangle \neq \mathbb{R}[T]/ \bigl\langle T^2 \bigr\rangle
$$

 

[WWGD]

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 $\mathbb R^{2n}$.

 

[fresh_42]

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 $(i\mathbb{R})\cdot (i\mathbb{R}) \subseteq \mathbb{R}.$

 

[MidgetDwarf]

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.

 

[FactChecker]

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.

 

[Agent Smith]

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")

 

[Agent Smith]

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 $z=(a,b)= r\cdot e^{i \varphi }.$

View attachment 348698​

$r=|z|=\sqrt{a^2+b^2}$ (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.

 

[martinbn]

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.

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.

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.

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.

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.

 

[FactChecker]

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.

 

[WWGD]

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 $\mathbb Q \subset \mathbb R \subset \mathbb C$
?

 

[WWGD]

Never mind. Tower theorem makes the intermediate extensions impossible.

 

[fresh_42]

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.

 

[martinbn]

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?

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!

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.

 

[fresh_42]

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.

 

[martinbn]

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?

 

[FactChecker]

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.

 

[Agent Smith]

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 $\varphi$.

 

[WWGD]

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.

 

[FactChecker]

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.

 

[FactChecker]

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, $r$.

 

[Agent Smith]

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?

 

[WWGD]

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.

 

[FactChecker]

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.

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".

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.