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Concept of complex plane

  1. Jul 11, 2012 #1
    I'm having difficulty understanding the nature of a plane in complex space.

    Specifically, I have two complex [itex] (N \times 1) [/itex] vectors, [itex]\underline{u}_1 [/itex] and [itex]\underline{u}_2 [/itex], which are orthonormal:
    [tex]\underline{u}_1^H\underline{u}_2 = 0 \\
    \Vert \underline{u}_k \Vert = 1 \ \ \ (k=1,2)[/tex]
    So, [itex]\underline{u}_1 [/itex] and [itex]\underline{u}_2 [/itex] span a 2D subspace (plane) and any vector in that plane can be written in the form:

    [tex]\underline{v} = a_1\underline{u}_1 + a_2\underline{u}_2[/tex]
    (for some complex scalars [itex]a_1[/itex] and [itex]a_2[/itex]). What I want to know is whether certain properties of [itex]a_1[/itex] and [itex]a_2[/itex] can tell us something about the nature of [itex]\underline{v}[/itex] (in the context of [itex]\underline{u}_1 [/itex] and [itex]\underline{u}_2 [/itex]). Specifically, what if [itex]a_1, a_2[/itex] are purely imaginary (i.e. [itex]\Re e\{a_1\}=\Re e\{a_2\}=0[/itex])? I feel that such a construction must have specific (visualisable/intuitive) properties, but I can't see what they might be. Can anyone shed any light on this?
     
    Last edited: Jul 11, 2012
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  3. Jul 11, 2012 #2

    chiro

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    Hey weetabixharry.

    The only thing I would have to offer merely as a suggestion is to interpret the multiplications themselves of complex numbers visually, which can be interpreted as rotations and scalings corresponding to the polar magnitudes and the angle descriptions.

    In terms of linear combinations though, they are just sums of complex numbers which have the standard vector space, arrow-like behaviour but in terms of the multiplication parts, the geometric interpretation in the polar form is probably a good way to answer your question.
     
  4. Jul 14, 2012 #3
    Thanks for the response. I've done some more thinking (and reading) but am not convinced that I've really got to the bottom of the issue.

    From what I read, one interesting perspective is to interpret the magnitude of the inner product of two (unit-length) complex vectors as being the cosine of the "angle", [itex]\theta[/itex], between them: [tex]cos(\theta) \triangleq \frac{\vert \underline{w}_1^H\underline{w}_2 \vert}{\Vert \underline{w}_1 \Vert \Vert\underline{w}_2 \Vert}[/tex]
    Then, the argument/phase of the inner product represents the "phase", [itex]\varphi[/itex], between the two vectors:[tex]\varphi \triangleq arg \{\underline{w}_1^H\underline{w}_2\}[/tex]
    It seems that this interpretation has the useful property that scaling either/both vector by a complex number (scalar) does not change the "angle" between them.

    Returning to my original post, I suppose my problem is that I can't see the role of this "phase" between the two vectors. Specifically, if the phase is zero, then what does that actually tell us? Does this allow us to, in some sense, study the geometry as if it were in real space (since the differences in the two vectors' directions are fully characterised by the real angle [itex]\theta[/itex])?
     
  5. Jul 14, 2012 #4

    chiro

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    If the phase is 0 or pi, then the vectors are linearly dependent (or non-zero. scalar multiples of one another).

    Phase is the first way to establish orientation in geometry. With complex numbers, the phase gives the rotation and establishes orientation since the multiplication of vectors in the complex space adds the phases and multiplies the lengths.

    The orientation part is critical and has a few interpretations. One interpretation is the lie group interpretation where you can form a group with a unit length set of complex vectors.

    You can't do this normally with real numbers because of division by zero and also because of the situations involved with dealing with large numbers, so if we need to form a group, we need to remove zero. The unitary complex vectors with the group operation as multiplication provides a way of relating complex numbers, multiplication and geometry.

    The other thing is that it fits nicely with existing geometric theory that is based around two important things: distance and angle. The angle corresponds to an inner product and the distance corresponds to either a norm or a metric. This is why complex numbers are kind of a basic relation to geometry in the form of having both distance and angle (real numbers have no orientation and orientation is through angle).

    So in a sense multiplying two unit complex numbers tells you the geometric relative orientation between those vectors. Think of it like for example having a way of saying how 'similar' two vectors are. The angle does this and you will always get cos(angle) between -1 and 1 inclusive where 1 means 'directly similar' and -1 means 'completely opposite'.

    This one attribute obtained from the inner product gives a relation not only with respect to the relative orientation, but for the relative length as well. So in this sense we get a kind of 'combined' indicator for relativity that is a 'length-phase' measure in same way that we use a formulation of 'space-time' to describe a distance in a 4-dimensional entangled geometry that entangles space with time.

    Think about for example two vectors that are dis-similar. When the inner-product is less than 0 we know that they are pointing in somewhat different directions. We could for example keep the value the same by changing the orientation, but then changing the length. Similar case for the positive inner product value. When the inner-product is 0 however, we know that either a vector is zero or the phase is pi/2 and this means no matter how you scale the length of a vector, it doesn't change it's inner product which is useful as well.
     
  6. Jul 14, 2012 #5
    If I understand correctly, this is the "angle" which I defined as [itex]\theta[/itex], above. Therefore, where you talk about "orientation", I assume this refers to the thing I'm having difficulty understanding (I defined it as [itex]\varphi[/itex]).

    I agree with your discussion of inner products (similar things can be said about real-valued vectors). However, I suppose my question is about orientation. Specifically, when the orientations of two complex vectors are the same

    [tex]arg \{\underline{w}_1^H\underline{w}_2\} = 0[/tex]

    and whether this simplifies analysis in some way.
     
  7. Jul 14, 2012 #6

    chiro

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    When you use w^h is this vector a conjugate?
     
  8. Jul 14, 2012 #7
    Yes, sorry, a superscript [itex](.)^H[/itex] denotes the complex conjugate transpose (and all my vectors are column vectors).
     
  9. Jul 14, 2012 #8

    chiro

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    Well if this is the case then it means both vectors have the same phase since a conjugate inverts the phase with respect to 2pi (i.e. conjugate basically calculates a vector with phase 2pi - normal-phase) but it keeps the length the same.

    Does this help?
     
  10. Jul 14, 2012 #9
    Yes - I think that is what I was trying to say with this equation:
    [tex]arg \{\underline{w}_1^H\underline{w}_2\} = 0[/tex]When this is the case, does it help/simplify the analysis in any way? I feel as though it means that the two complex vectors somehow lie in the same real 2D subspace (?).
     
  11. Jul 14, 2012 #10
    Pardon me if I'm recounting the obvious, but we're not working in the complex plane here--there are no two orthonormal complex numbers in the plane, just as there are no two nonzero real numbers that aren't scalar multiples of one another. The span of two linearly independent vectors in a complex vector space, sometimes called "a complex plane", has four real dimensions. "The complex plane" is used to refer to the set of complex numbers, which (like every field) is a 1-dimensional vector space over itself. But one complex dimension is two real dimensions, and since we think in real dimensions, we traditionally call that a plane, even though "complex line" is more appropriate, and is often used when talking about a one (complex) dimensional subspace of a complex vector space (which has two real dimensions and so "looks" like a plane).

    So, to get back to your specific question: the complex span of [itex]u_1,u_2[/itex] is a four-dimensional real vector space. By complex span I mean [itex]\{a_1u_1+a_2u_2\mid a_1,a_2\in\mathbb{C}\}[/itex]. Now if you constrain [itex]a_1,a_2[/itex] to be pure imaginary, you have that [itex]ia_1,ia_2[/itex] are real, and so the span can be written as [itex]\{x_1(iu_1)+x_2(iu_2)\mid x_1,x_2\in\mathbb{R}\}[/itex], and so the geometric interpretation is that your subspace has two real dimensions. Or, you could write the span as [itex]\{x(iu_1)+(yi)u_2\mid x+yi\in\mathbb{C}\}[/itex] to make it clear that it's also a one-complex-dimensional subspace.
     
  12. Jul 14, 2012 #11
    Ah yes, sorry - it seems that I got the terminology wrong. You are correct that I'm not talking about the complex plane (as in a real axis and an imaginary one). I'm talking about a two-dimensional complex vector space.

    The rest of what you have written seems to have arrived at what I guessed at in my last post. I'll have to take some time to think about the details, but I think I'm getting there!

    Thanks very much for taking the time to respond!
     
  13. Jul 14, 2012 #12
    You didn't get it wrong--"complex plane" is certainly correct terminology for what you're working with! It's just that it's also used in the other sense, which can be confusing. I thought that probably you had it right, but I wasn't 100% sure, and anyway it's good to have it clarified for others who will come across this thread later. Good luck with your investigation.
     
  14. Aug 24, 2012 #13
    if the function f(z*) is analytic if the function f(z) is analytic

    Hi, there
    I ask if the function f(z*) is analytic if the function f(z) is analytic. Here z* is the complex conjugate of z and f: A---C the complex plane.
     
  15. Aug 24, 2012 #14

    chiro

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    Re: if the function f(z*) is analytic if the function f(z) is analytic

    Being analytic means that you have to consider some neighbourhood and only in trivial cases can this be generalized for both cases without any other specific information.

    Do you know where f(z) is analytic?
     
  16. Aug 24, 2012 #15
    Re: if the function f(z*) is analytic if the function f(z) is analytic

    Hint: what if f is the identity map?
     
  17. Aug 24, 2012 #16
    Do you know where f(z) is analytic?

    Yes, f(z) is analytic in the domain: 0<Re(z)<1. Re(z) is the real part of z.
     
  18. Aug 24, 2012 #17

    chiro

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    It should also be analytic there for the conjugate as well since the conjugate doesn't change the real part.
     
  19. Aug 24, 2012 #18
    Thank you very much Chiro
     
  20. Aug 24, 2012 #19
    Can one tel me about this problem: Let us consider the series (sum an/n*s) if the coffecient an is depend on s, then it is true to consider this serie as a Dirichlet serie or not.
     
  21. Aug 24, 2012 #20

    Bacle2

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    It may help you to remember that a plane is one-dimensional as a complex subspace,

    but two-dimensional over the reals. Not a brilliant insight, but it helped me unconfuse

    things a bit.
     
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