Views On Complex Numbers

[fresh_42]

It's in the Bible and everything - be fruitful and multiply!

The concept being discussed is called closure. If I have c = a ⊗ b where ⊗ is addition, subtraction, multiplication or division, and a and b are both real, so is c. If a and b are both complex, so is c. If a and b are purely imaginary, c might or might not be - it is not closed.

This makes purely imaginary numbers less useful.

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
$$
\mathbb{C} \neq \mathbb{R}^2.
$$

 

[FactChecker]

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
$$
\mathbb{C} \neq \mathbb{R}^2.
$$

The geometry of the complex plane is very important in many applications of complex analysis. IMO, a lot is lost when it is only considered algebraically.

 

[fresh_42]

The geometry of the complex plane is very important in many applications of complex analysis. IMO, a lot is lost when it is only considered algebraically.

It is a tool, and shouldn't be the central view that it often unfortunately is. The complex plane supports the perspective of a two-dimensional real vector space. You can throw complex analysis in the trash with this limited view (sounded better in German).
$$
\begin{pmatrix}0\\1\end{pmatrix}\cdot \begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}-1\\0\end{pmatrix}
$$
is crucial!

 

[martinbn]

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.

It is not reducing, it is using it.

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.

Why is it not helpful? It definitely is for some things.

This article was all about
$$
\mathbb{C} \neq \mathbb{R}^2.
$$

This is not true without any context. The opposite is true.

Ps. Do you have a similar objection for the Gaussian integers?

 

[fresh_42]

I am simply of the opinion that the complex numbers are primarily a field and not a real vector space. Putting the perspective as a vector field in front is in my opinion stupid. You can be of different opinion. I already know that old habits die hard, regardless of how questionable they are.

The Gaussian integers are primarily a ring and not a $\mathbb{Z}$-module. I guess, I have the same objections.

 

[fresh_42]

This [$\mathbb{C}\neq \mathbb{R}^2$, ed.] is not true without any context. The opposite is true.

What??? This is a clear misinformation!
$$
\underbrace{\mathbb{C}}_{\nearrow \text{ field }} \quad \neq \quad \underbrace{\mathbb{R}^2}_{\text{ no field }\nwarrow}
$$

 

[dextercioby]

Yes, but he probably meant vector spaces over $\mathbb R$.

 

[fresh_42]

Yes, but he probably meant vector spaces over $\mathbb R$.

This is at the same level as saying they are both additive groups. $\mathbb{C}$ and $\mathbb{R}^2$ have a common understanding as, resp. field, two-dimensional real vector space. Of course, one can always right them as $\mathbb{C}=\left(\mathbb{C},s_1,\ldots,s_n\right)$ or $\mathbb{R}^2=\left( \mathbb{R}^2, t_1, \ldots , t_m \right)$ where $s_i , t_j$ represent the countless structures they can carry but nobody does this.

 

[FactChecker]

Complex analysis is usually a required course for Engineers. The geometry of analytic functions is critical to their use. IMO, approaching complex analysis from an abstract algebra perspective would be a mistake. I would follow Ahlfors.

 

[Agent Smith]

Then what is the difference from the real numbers?

Square roots of negative numbers are imaginary numbers e.g. $\sqrt {-3}$

 

[Agent Smith]

You can if you wish. The first thing you should find out is that these are simply called imaginary numbers.


I am not sure what you mean by an 'imaginary π' but the number $i \pi$ has a very interesting property.


You can see $\LaTeX$ Math expressions in some of the posts in this thread. There is a tutorial at https://www.physicsforums.com/help/latexhelp.


When we want to think about something in Mathematics the first thing we need to do is define exactly what it is we want to think about. I am not sure what you think you mean by an 'imaginary circle' but I do think that you will find it difficult to find a definition that works.

That interesting property being $e^{i \pi} + 1 = 0$?

We've already defined imaginary numbers as ... square roots of negative numbers, right?

As far as I know, an imaginary circle should have imaginary dimensions.

P.S. Danke for the advice.

 

[FactChecker]

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.

 

[martinbn]

Square roots of negative numbers are imaginary numbers e.g. $\sqrt {-3}$

If you ignore multiplication and consider only addition, the imaginari numbers are indistiguishable from the reals.

 

[Hans de Vries]

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
$$
\mathbb{C} \neq \mathbb{R}^2.
$$

The point is that one should see $\mathbb{C}$ rather as a set of operators: ( 1, $i$, $\ast$ ) on $\mathbb{R}^2$
The total number of independent operators is 2x2=4 with $i\ast$ as the fourth one. (see post #26)

In case of $\mathbb{R}^4$ the set of operators contains two subgroups of quaternians: ( 1, $j_x, j_y, j_z$ and $i_x, i_y, i_z$ )
The total number of independent (orthormal) operators is 4x4=16, the 7 given above plus the nine quaternian products $j_x~i_x$ through $j_z~i_z$.

Now $\mathbb{R}^4$ determines the physics of the world we live in and you will always see mathematicians trying to explain physical laws by using $\mathbb{H}$, but one will never get the complete picture as long as the complete set of 4x4=16 opererators as given above is not used.

So in this case $\mathbb{H} = \mathbb{R}^4$ throws away 12 of the 16 orthonormal operators on $\mathbb{R}^4$

 

[FactChecker]

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.

 

[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 “=“.