Insights Views On Complex Numbers

[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.

 

[Agent Smith]

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).

 

[jbriggs444]

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.

 

[FactChecker]

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.

 

[Agent Smith]

@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.

 

[WWGD]

@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.

 

[Paul Colby]

Is there a way to define complex conjunction without the mapping $\mathbb{C}\rightarrow \mathbb{R}^2$? The involution, $\ast$, always seemed the gateway to $\mathbb{R}^2$.

 

[FactChecker]

If $z=r e^{i\theta}$, then $\bar z = r e^{-i\theta}$. There are other, more general, ways to reflect a vector through a general vector, plane, or other multi-dimensional subspace.

 

[Agent Smith]

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.

Methinks they were aware. Congruence and similarity are based on corresponding angles, sides. Gracias

 

[lekh2003]

I haven't said that this is an incorrect view, except if it is reduced to $\mathbb{R}^2$. I think, it just shouldn't be the first view. But, hey, let's consider it as vector space over the rationals.

I see your argument here, but I feel like its sometimes useful to view the complex numbers instead as endowing a convenient multiplicative structure on #\mathbb{R}^2#. Specifically a multiplicative structure which only really coincidentally happens to be possible because of the complex numbers, quaternions and octonions.

But separately, I find viewing it this way neat because it removes some of the fantastical feeling of viewing complex numbers. It shows that the natural properties of holomorphic functions lead into those for harmonic functions in general for $\mathbb{R}^n$