# why don't imaginary numbers make so much sense?

Discussion in 'Mathematics and Physics' started by PG1995, Aug 29, 2011.

1. ### misterTWell-Known MemberMost Helpful Member

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I'm talking about a field in the context of number theory. A field that satisfies field axioms. Stop being stupid on purpose. Show me some references to back up your "rotational operator" definition of i. Otherwise you have lost this game.

Last edited: Sep 14, 2011
2. ### RatchitWell-Known Member

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misterT,

All right, you referred to number theory. How does it explain jω or ωj algebraically?

Both Eric Gibbs and I have shown you references which you ignore. I have also explained that you need rotation to get orthogonality. So far you have not explained why that is not so. You instead refer to a vague field concept in "number theory", as if that answers something.

Ratch

3. ### misterTWell-Known MemberMost Helpful Member

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What is there to explain? jω is a point on complex plane (0,ω). It has a real part of 0 and imaginary part ω. You can think ω was rotated 90 degrees if you want and get to the same conclusion.

None of the references prove that i is an operator. Treating it as an operator works, but none of the references show why i is an operator.

Yes you have, but you are wrong. You don't need rotation to get orthogonality. You can just define orthogonality.

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5. ### RatchitWell-Known Member

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misterT,

Being able to map something on a plane does not prove anything. I can map a complex number on a curved surface and show the the lines defining the points are not orthogonal. What is the significance of j number of items each ω in length, and how does that make orthogonality? You only get orthogonality only if you consider j as a rotational operator.

The references explain the orthogonality of number operated on by j. Considering j to be a constant does not, and only works because of conformal similarity. That is why you cannot explain the algebraic significance of "j number of items each ω in length". You can only point to the result and say "it works".

You have to have a basis for any definition. Things cannot just be defined because you want them to be in existence. They also have to agree with other relationships to which they are coupled. In the case under discussion, the results are correct when defining j as a constant, but the algebraic term does not make sense. When defining j as a operator, the results are still correct, but j is not an algebraic constant, so there is no conflict. Orthogonality is obtained by rotation, not by definition.

Ratch

6. ### misterTWell-Known MemberMost Helpful Member

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That question is just some ******** you keep ranting about.

".. algebraic significance of 'j number of items each ω in length'."

what a ******** question. stop embarrassing yourself.

No they don't. The references do not even include the word "orthogonality".

That is not true.. you need to show some proof on that one!

I agree, i (imaginary unit) cannot be defined as an operator just because you like it that way.

What is the conflict you refer to? And do not talk some nonsense about orthogonality and rotations. You need to show some proof to keep that argument alive.

As I said: I let you call j as an operator (in the context of electrical engineering). But don't tell me that: "The symbol 'i' does not mean √-1. 'i' is a mathematical operator, not a finite value." You know it is ********.

Last edited: Sep 14, 2011
7. ### misterTWell-Known MemberMost Helpful Member

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Try googling imaginary unit. It will bring up more convincing results.

First three results when googling "j operator" are:
- some forum discussion
- wikipedia (Peter Landin's J operator, not related to complex numbers)
- Electrical engineering handout.

First three results when googling "imaginary unit" are:
- wikipedia (about imaginary unit, saying i^2 = -1)
- wikipedia (about imaginary numbers, saying i^2 = -1)
- wolfram mathworld (saying i^2 = -1)

Last edited: Sep 14, 2011
8. ### RatchitWell-Known Member

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misterT,

You are the one using j as an algebraic constant, not me. Therefore it should be you that explains it.

They don't have to do so. Obviously they are referring to "imaginary" quantities as being angled 90° with respect to each other. That is how orthogonality is defined.

It is too simple to prove. Obviously you get a orthogonal direction if you rotate something 90°. That is the very definition of orthogonality.

Right, it is defined as a rotational operator in order to give the imaginary quantity orthogonality.

The algebraic meaning of jω.

As I said before, that is too simple to prove. Rotate some quantity by 90° to get a orthogonal direction. Can't get more basic than that.

Not so, I can work problems (as I already proved in post #19) by considering j as a rotational operator. That would not be possible if the concept was incorrect. Your method of considering j a complex constant gets correct answers by its fortunate equality relationship with rotation. But your concept of considering j a constant is wrong.

Ratch

9. ### misterTWell-Known MemberMost Helpful Member

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Show me the proof then. Saying so does not make it so. As you said: "Things cannot just be defined because you want them to be in existence."

Let me get this right. What you are saying is:
If I have objects p and q so that p||q, then p⊥jq always holds (for all objects {p,q | p||q})? Here j is an operator defined as you say it is (rotate by 90°).

Last edited: Sep 14, 2011
10. ### RatchitWell-Known Member

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misterT,

I told you, I cannot prove it because it is too basic. Every mathematician knows that the simplest things are the hardest to prove. However, what I said is consistent with its mathematical concepts.

What does {p,q | p||q} mean? Perhaps it would be clearer if you gave a verbal description of what you mean.

Ratch

11. ### misterTWell-Known MemberMost Helpful Member

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The set of p and q such that p and q are parallel.
It tells that we are only concerned about objects p and q that are parallel (with each other).

So, If objects p and q are parallel, then p⊥jq always holds, if j is an operator defined as you say it is? (p⊥jq means that objects p and jq are orthogonal)

Last edited: Sep 14, 2011
12. ### misterTWell-Known MemberMost Helpful Member

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I did not have a signature before.. but that is an instant classic!

Last edited: Sep 14, 2011

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misterT,

Huh?

Ratch

14. ### misterTWell-Known MemberMost Helpful Member

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Sorry for posting twice. Page break hides the previous one.. see the post #50 for reply to your question.. actually here it is repeated:

The set of p and q such that p and q are parallel.
It tells that we are only concerned about objects p and q that are parallel (with each other).

So, If objects p and q are parallel, then p⊥jq always holds, if j is an operator defined as you say it is? (p⊥jq means that objects p and jq are orthogonal)

15. ### RatchitWell-Known Member

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misterT,

I would agree with the above. Continue.

Ratch

16. ### misterTWell-Known MemberMost Helpful Member

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Ok, so lets assume that p and q are 2-dimensional real vectors ( p,q ∈ ℝ² ) and parallel (p||q), say:
p = [1 2]
q = [2 4]

What, in your opinion, is jq?

1) If j is an operator that rotates q by 90 degrees, then jq = [-4 2] (and it holds that p is orthogonal to jq).

2) If jq = [2j 4j], then it follows that p is not orthogonal to qj.
in this case the dot product of p and jq is: p·jq = 2j+8j = 10j ≠ 0

Last edited: Sep 14, 2011
17. ### RatchitWell-Known Member

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misterT,

It is orthogonal to both p and q.

Agreed.

jq ≠ [2j 4j]
jq = j([2,4] =j (2,j(4)) = j(2) + j(j(4)) = j(2) - 4 = [-4,2]

dot(p,j(q)) = 1*-4+2*2 = 0

dot(q,j(q)) = 2*-4 +4*2 = 0

Ratch

18. ### misterTWell-Known MemberMost Helpful Member

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Show me a book, or a reference, or anything, where it is said that a real value vector [2 4] multiplied by j equals [-4 2], and does not equal [2j 4j].
That example proves the "rotation operator" concept wrong.

The first option is incorrect because jq ≠ [-4 2]
The second option is correct: jq = [2j 4j], but it is not orthogonal with p = [1 2]

The rotation operator concept just does not work.

Last edited: Sep 14, 2011
19. ### RatchitWell-Known Member

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misterT,

First of all, to be pedantic, complex numbers are not truly vectors, but the quantities they represent can be described by orthogonal vectors. Now, the number in your notation was [2 4]. It must be assumed that it means a real component of 2 and a orthogonal component of 4. In my rotational notation, that would be [2 j(4)]. All I did was rotate both components of the number, which is a legitimate operation. So the real part 2 ===> j(2) and the orthogonal part j(4) ===> -4 . The final rotational sum is [-4 j(2)] or [-4 2] by your notation.

I beg to differ. You made a mistake in your calculations. The second number 4 in your notation [2 4] has an implied rotation which you did not consider.

Ratch

Last edited: Sep 14, 2011
20. ### misterTWell-Known MemberMost Helpful Member

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I agree. The vectors in my numerical example vere not representations of complex numbers. They were just 2-dimensional vectors. And because complex numbers are not vectors, the multiplication is done componentwise: [2 4]j = [2j 4j]

Now you are just making stuff up. I defined the vectors to be purely real ( p,q ∈ ℝ² ). Show me any example, a web page or a book, that shows the kind of calculations that you just did.. just plain wrong. A vector multiplied by imaginary unit i, or j operator, or a scalar, is done componentwise.

I would also like to know how you would calculate a three dimensional vector multiplied by j: [1 2 3] j = ?

Last edited: Sep 15, 2011
21. ### RatchitWell-Known Member

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misterT,

Yes, component wise AND direction wise. You are representing a vector by a complex number. Therefore [2 4] implies [2 j(4)], which is why I get a correct answer.

You defined the components of each direction to be real. But, don't forget, vectors have a direction, too.

Sure, it's easy. Setting some coordinate orthogonal axes we get [x y z], OK? Then we can assign x to be the real axis, y to be orlthogonal to x, and z to be orthogonal to both x and y. So, a j rotation means rotate everything in the x-y plane about the z-axis, a k-rotation would mean rotate everything is the x-z plane about the y-axis. Applying this to your line coordinate, we rotate the x-coordinate to the y-axis, and the y coordinate to the x axis. The z-coordinate is unaffected because it is orthogonal to the x-y plane of rotation. Therefore j:[1 2 3] becomes [-2 1 3].

Oh, don't expect the dot product to be zero for that rotation. That particular line has infinitely many solutions which are represented by a plane.

Now, show me you understand the concept by doing k:[1 2 3]

Ratch

Last edited: Sep 16, 2011