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Better definition for complex number

  1. Aug 22, 2015 #1
    I was me asking why the complex numbers are defined how z = x + i y !? Is this definition the better definition or was chosen by chance?

    In mathematics, some things are defined by chance, for example: 0 is the multiplicative neutral element and your multiplicative inverse (0-) is the ∞. But, 1 is the additive neutral element and it haven't a symbol for its additive inverse (-1), the inverse additive is wrote simply how -1.

    Other example: the conic equation is, actually, the vetorial form of the quadratic equation a x² + b x + c = 0. Therefore, the 'correct' form of write it is: [tex]a_{ij} : \vec{r}^2 + b_{i} \cdot \vec{r} + c = 0[/tex] In matrix form: [tex]
    \begin{bmatrix}
    a_{11} & a_{12}\\
    a_{21} & a_{22}
    \end{bmatrix}:\begin{bmatrix}
    xx & xy \\
    yx & yy
    \end{bmatrix}
    +
    \begin{bmatrix}
    b_1\\
    b_2
    \end{bmatrix}\cdot
    \begin{bmatrix}
    x\\
    y
    \end{bmatrix}
    +c=0[/tex] And not this way:

    01f03b3f0e26d0194a25c59cb3df16f8.png

    https://en.wikipedia.org/wiki/Conic_section#Matrix_notation

    Considering these 'mistakes' and others that I not wrote here, I me asked why the complex numbers are defined how z = x + i y.

    Why i is important? Why? Why emphasize the number i ? Why? i is not important!

    You known that x² - y² = (x + y) (x - y), all right!? And that x² + y² = (x + i y) (x - i y), correct!? But, you know that x³ + y³ = (x + α y) (x + β y) (x + γ y) ? No? You know that α, β and γ are the roots of ³√(-1) ?

    α, β and γ appears in the cubic formula too. Write α, β and γ how the linear combination x + i y always complicates every equation! So, is necessary to define one symbol for α, β and γ too! Or, why not? Why the preconception? Why ²√(-1) has a proper symbol and ³√(-1) haven't? Why ²√(-1) is special?

    I thought wrote the complex number like z = xy, because this notation no emphasize none root to the detriment of other.

    But, linear combination of roots appears be a good way of write complex numbers too. In this case, 1 and i are the best base for the lienar combination?

    And why the complex numbers are bidimensional numbers? If the root of the linear equation is and unidimensional number and if the roots of the quadratic equation are bidimensional numbers, so, the roots of the cubic equation are tridimensional numbers, correct or not!?

    So, what you think about? I'm a little confused and unbeliever...
     
  2. jcsd
  3. Aug 22, 2015 #2

    micromass

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    No it isn't.

    Because the three roots to ##x^3 = -1## are not independent.

    Sure, but I doubt you'll get many people to follow this notation.

    It depends a lot how you define dimension. If you refer to the dimension of ##\mathbb{C}## over ##\mathbb{R}##, then the dimension is ##2##. The cubic roots are not all independent, so the dimension remains ##2##. A more interesting thing is to regard dimension over ##\mathbb{Q}##, in which case you'll get into Galois theory.
     
  4. Aug 22, 2015 #3

    HallsofIvy

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    Frankly, the fact that, in your introductory paragraph, you made one mistake after another about elementary mathematics does not lead to confidence about what the "correct" way to write things is!
     
  5. Aug 22, 2015 #4

    mfb

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    They are not. There are many definitions but this is not one of them.
    That would imply 5*0=5.
    That would imply 0+1=0.
     
  6. Aug 22, 2015 #5

    FactChecker

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    The numbers 1 and i form an orthonormal basis for the complex plane. That is a very good choice.
     
  7. Aug 22, 2015 #6
    Really! :eek:

    This is a good point! 1 and i are orthonormal basis and lidependent linear because 1 were definied like the unit vector of the x-axis and i like the unit vector of the y-axis. If I define ³√(-1) like the unit vector of the y-axis so now any complex number in the form z = x + i y is a linear combination of α + μ β (being α = real number, β = real number and μ = ³√(-1)) and now (1 , μ) is the orthonomal basis for the complex plane and independent linear!

    Or not!?
     
  8. Aug 22, 2015 #7

    micromass

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    Sure, you can invent some system where ##(1,\mu)## is an orthonormal basis. But the system where ##(1,i)## is orthonormal has simpler formulas, so we prefer that.
     
  9. Aug 26, 2015 #8

    martinbn

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    Here is one more way to define the complex numbers: ##\mathbb R[x]/(x^2+1)##
     
  10. Aug 26, 2015 #9

    mathwonk

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    it is ironic that we leave students in the US with the idea that complex numbers are exotic when in actuality it is the real numbers that are much more difficult and subtle to define.
     
  11. Aug 26, 2015 #10

    HallsofIvy

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    I've mentioned this so often that I am a little reluctant. We can (and some textbooks do) define the complex numbers as pairs of real numbers, (a, b) with addition defined by (a, b)+ (c, d)= (a+ c, b+ d) and multiplication by (a, b)*(c, d)= (ac- bd, ad+ bc). The real numbers themselves can be identified with the pairs (a, 0) and "i" with (0, 1) so that a+ bi is (a, b).
     
  12. Aug 31, 2015 #11

    lavinia

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    Given two non-colinear vectors in the plane, any other vector is some linear combination of them. Thus one can write any complex number as a linear combination of two fixed vectors that do not lie on the same line. From this point of view there is nothing special about choosing 1 on the positive x-axis and 1 on positive y-axis. Any two vectors will do.

    For each choice one would get a different law of multiplication and perhaps a good reason to use 1 and i is that the law of multiplication is particularly simple. and also that the coefficients of a linear combination are just perpendicular projections onto the x and y axes.

    Complex multiplication can be described as adding the angles to the x-axis and multiplying the radii. If one measures angles from the positive x-axis then angle addition is easily computed from the laws for the sine and cosine of the sum of two angles.

    One can certainly use linear combinations of two roots of a polynomial. Try to work out the law of multiplication using two cubic roots of unity.
     
    Last edited: Aug 31, 2015
  13. Sep 1, 2015 #12
    This is a good argument!
     
  14. Sep 2, 2015 #13

    gill1109

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    You can *define* the complex numbers as pairs of real numbers with a certain rule for addition, and a certain rule for multiplication. Now you can *define* i = (0, 1) and now you can *prove* i^2 = -1.

    But there are plenty of other possible definitions, because there are plenty of theorems which *characterise* the complex numbers. Of the kind: suppose you have a set and you have an addition and a multiplication operation satisfying ..., then what you have got is: the complex numbers. AFAIK all such theorems and all such definitions take it for granted that we already know what the real numbers are.
     
  15. Dec 23, 2015 #14

    Demystifier

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    Even if real numbers are harder to define formally and rigorously, they are quite intuitive to most humans (including US students). Continuum is quite intuitive. Complex numbers are not that intuitive.
     
  16. Dec 23, 2015 #15

    WWGD

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    Ultimately, like many other objects in Mathematics (and likely in other areas) Complex Numbers have many different types of structures:

    as fields (And a field extension of the Reals), as metric spaces, etc. So depending on the type of structure, once you fix a choice, you will have a suitable definition.
     
  17. Dec 23, 2015 #16

    micromass

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    I think the majority of people living before 1900 would want to disagree with you on that. What constitutes a number has been debated for millenia. They're only intuitive to us now because we're taught real numbers at an early age. In reality, a real number is an amazingly difficult object.
     
  18. Dec 23, 2015 #17

    mathwonk

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    if the argument is that the real numbers are more intuitive because they are represented geometrically as points on a line, aren't the complex numbers just as intuitive when represented as the points of a plane? if the algebraic part is what seems mystifying, try defining rigorously even the product of two arbitrary reals. by contrast, multiplication by i is just a rotation of the plane. How e.g would you explain multiplication of real numbers to a class, like one I had, that firmly believed multiplication to be repeated addition?

    And if you have not had my experience teaching calculus to US students, you may be surprised to find that many of them actually assume all real numbers are integers, when interpreting a general statement about them. I.e. after reading that the derivative is linear, e.g that (af+bg)' = af'+bg' for all real numbers a and b, my students only thought of a and b as being whole numbers, certainly not reals like e or π. Lack of intuition about reals may stem from thinking all numbers appear in the display of a calculator. I.e. other students think all reals are finite decimals., or that π = 22/7, or 3.1416. Even among professional mathematicians there is no clear concept of the real number continuum until maybe Dedekind, as micromass suggests.
     
    Last edited: Dec 23, 2015
  19. Dec 23, 2015 #18

    WWGD

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    Notice the discussion on 0.9999...=1 and the inability to understand the definition of a number as a rep. of a class, and not as an object to start with. Then other things add up to the difficulty. The dense ordering implying the lack of a successor does not help either, as I can't think of a case outside of the Reals where this happens.
     
  20. Dec 23, 2015 #19

    jbriggs444

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    The standard ordering of the rationals is a dense ordering as well.
     
  21. Dec 23, 2015 #20

    WWGD

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    Your right, and this follows from Rationals being Reals, what I meant to say (but did not) is that this is not seen outside of Mathematics, AFAIK.
     
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