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Factorizing a complex number

  1. Dec 28, 2013 #1

    bobie

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    1. The problem statement, all variables and given/known data

    Wolfram gives a step to step solution for
    [tex]\sqrt[3]{-9+46i}[/tex]
    [tex]→ 27+ 54i -36 - 8i →
    27+ 54i -36i^2 + 8i^3 = (3+2i)^3[/tex]
    Can you tell me (where to find) the rules of this derivation?
    Thanks
     
  2. jcsd
  3. Dec 28, 2013 #2
    Uhm, do you actually need to factorize the given expression under the third root, or just find what the value of the root is? If it's the latter, then it's a lot easier to convert the expression under the root into trigonometric form, and use De Moivre's theorem to find the roots (note that any n-th root of a complex number has exactly n different values, so in this case you'll get 3 different roots).
     
  4. Dec 28, 2013 #3

    haruspex

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    Suppose (a+ib)3 = c+id, all integers, and you wish to deduce a and b from c and d.
    Multiply that out. What do you notice about how a and b relate to factors of c and d?
     
  5. Dec 30, 2013 #4

    NascentOxygen

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    It's tantalizingly cryptic, isn't it? But I don't know either. The only heuristic I've come across is the original method of Bombelli himself. It's illustrated in the concluding half of the section subtitled Bombelli's Solution 5 here: http://mathfaculty.fullerton.edu/mathews/c2003/ComplexNumberOrigin.html

    Possibly wolframalpha has a more direct technique for a solution having both integer components? How do you get wolframalpha to show its intermediate steps?
     
  6. Dec 31, 2013 #5

    bobie

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    Either you subscribe at $4 per month, or you can get 2 free-accesses per day opening an account.
    Thanks for the nice link
     
    Last edited: Dec 31, 2013
  7. Dec 31, 2013 #6
    If you think there is an easy solutiion:

    If you compute the norm for (-9+46i)= 9^2+46^2 you get the third power of 13, so the norm of the cube root is 13. It's easy to see that there is only one way to get a sum of two squares of 13, so the cube root must be [itex]\pm[/itex]3 [itex]\pm[/itex]2i or [itex]\pm[/itex]2 [itex]\pm[/itex] 3i.

    If you sketch -9+46i in the complex plane, it's easy to see that the cube root must be in the first quadrant.
     
  8. Jan 1, 2014 #7

    bobie

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    In that very interesting article you can read that Wessel (figure 1.6):
    How can it be? vectors were invented a century later, how did he multiply vectors? do you know the details of that operation?
     
  9. Jan 1, 2014 #8

    BruceW

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    I think Wessel did some of the early work in 'vectors'. also, I think he was mostly talking about a geometrical interpretation of complex numbers, not so much about the Euclidean vectors that we think about in modern physics. I'm not totally sure though, I don't know much about the history of it, although it's pretty interesting.
     
  10. Jan 1, 2014 #9

    bobie

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    Thanks, Bruce could you explain the difference with the 'Euclidean' vectors? why Euclidean if not geometric? How do you multiply 2 Euclidean vectors?
    suppose we have a vector v: x = 3 and y = 1, what is v*v in the two systems?
     
    Last edited: Jan 1, 2014
  11. Jan 1, 2014 #10

    BruceW

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    when I say 'euclidean vector' I mean the modern idea of a vector (but not the abstract mathematical concept of linear vector spaces). so, cross product, dot product, vector differentiation are part of the idea of Euclidean vector. And the 'Euclidean vector' is probably what most physics or engineer undergraduates think of as simply a 'vector'.

    From the very small amount of history I've read, it looks like first was the concept of complex numbers and quaternions (before Euclidean vectors), and then Euclidean vectors developed from these concepts. They all involve some idea of direction, and geometry. But they are slightly different. I think Wessel contributed to the geometric interpretation of the complex numbers, which later others developed further to get to what we know today as the Euclidean vector.
     
  12. Jan 1, 2014 #11

    bobie

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    Where can I find how to multiply two such vectors? I tried "vector space" at wiki but did not find that
     
  13. Jan 1, 2014 #12

    BruceW

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    http://en.wikipedia.org/wiki/Euclidean_vector gives the 'usual' concept of a vector which is most commonly known.

    http://en.wikipedia.org/wiki/Vector_space gives the abstract (and more general) concept of a linear vector space, which is maybe less widely known.

    edit: and complex numbers you know, the magnitudes multiply and the arguments add up.

    edit: I should probably mention, the general linear vector space does not necessarily have 'multiplication' (i.e. inner product), this is an additional structure. http://en.wikipedia.org/wiki/Inner_product_space
     
    Last edited: Jan 1, 2014
  14. Jan 1, 2014 #13

    bobie

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    Thanks, Bruce, but in both links there is only scalar ,dot and cross product, no trace of product of vectors the way Wessel intended for complex numbers where the length of the vector is squared and angle is doubled.
     
  15. Jan 1, 2014 #14

    BruceW

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    yeah, none of these links are related to Wessel's work. Sorry, I thought you were looking for links for definitions of Euclidean vectors and linear vector spaces. Uh, for Wessel's work, I'm not sure, I can't find much easily. It looks like his ideas of adding complex numbers was similar to our modern way. I don't know about multiplication.
     
  16. Jan 2, 2014 #15

    bobie

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    I'd like to know if you can multiply 2 Euclidean vectors (or any vectors), say va (3,1) * vb (8,6) if you can, what is the result?
     
  17. Jan 2, 2014 #16

    NascentOxygen

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    It may be possible to deduce the steps. Could you post WA's intermediate steps for a handful of other examples, e.g., cube root of (65+142i), (236+115i), (286-259i)
     
  18. Jan 2, 2014 #17

    bobie

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    (65+142i)= 125+150i - 60 - 8i = 125 + 150i + 60i^2 + 8i^3

    (236+115i) =241/2 + 615√3/8 + 75i/8 + 123i√3/2 + 231/2 -615√3/8+ 845i/8 -123i√3/2
    = ( 964 + 615√3 + 75i/8 + 492i√3 + 924 - 615√3+ 845i - 492i√3 )/8
    ( 964 + 615√3 + 75i/8 + 492i√3 - 924i^2 + 615√3i^2 - 845i^3 + 492i^3√3 )/8
    (4+5√3)^3 + 3(4+5√3)^2 i (4√3-5) +3(4+5√3) ( i (4√3-5))^2) + (i (4√3-5))^3) /8
    and that is the cube of (4+5√3 + i (4√3-5))/2
    Good luck!
     
    Last edited: Jan 2, 2014
  19. Jan 2, 2014 #18

    NascentOxygen

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    I liked your first comment better, "Not easy".

    EDIT: For (236+115i) I thought Wolframalpha would find this cube root (-4+5i), but instead it came up with this one (4+5√3 + i (4√3-5))/2 !
     
    Last edited: Jan 2, 2014
  20. Jan 2, 2014 #19

    Mentallic

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    The second solution has a smaller argument in the complex plane and WA seems to give a higher priority to those.

    Similarly, you'll find [itex]\sqrt[3]{-1}[/itex] turns out to be [itex]e^{i\pi/3}[/itex] as opposed to [itex]-1=e^{i\pi}[/itex]
     
  21. Jan 3, 2014 #20

    ehild

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    A complex number has three cubic roots. Now you want that cubic root of Z=65+i142 which has integer components. Z=(a+bi)3.

    Writhe the equation for the square of the magnitudes: |Z|2=20164=(a2+b2)3.

    Taking the cubic root of both sides : 29 = a2+b2.

    Find a pair of numbers so the sum of squares add up to 29.

    These give the possible magnitudes of a,b. Check which choice with what signs is appropriate. :biggrin:

    ehild
     
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