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Complex numbers linear algebra

  1. Feb 8, 2008 #1
    [SOLVED] Complex numbers linear algebra

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

    Let w, z be complex numbers. Solve the linear equation wx=z; in other words find all x (of the set of complex numbers) such that wx=z. (hint: You need to distinguish 3 cases)

    2. Relevant equations



    3. The attempt at a solution

    If w and z are complex numbers equal to i then if x was i the the power 0 it would equal one and still satisfy the equation. A second case is if x = -i^2 as this also equals one. I can't think of a third case, but to be honest i really don't think i have understood this question.

    Can anyone clarify what i have to do here please??/
     
  2. jcsd
  3. Feb 8, 2008 #2

    NateTG

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    I suspect there are really 4 cases to consider:
    [tex]z \neq 0, w \neq 0[/tex]
    [tex]z = w = 0[/tex]
    [tex]z =0, w \neq 0[/tex]
    [tex]z \neq 0, w=0[/tex]
     
  4. Feb 8, 2008 #3

    HallsofIvy

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    For this particular problem, there is really nothing special about complex numbers. The answer is essentially the same as if they were asking about real numbers.
     
  5. Feb 10, 2008 #4

    So i have to identify 3 values of x that make wx=z. I don't see how i do this though because all i know about w and z is that they are complex. Any hints please?
     
  6. Feb 10, 2008 #5

    Dick

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    You must be tempted to write that the solution set to wx=z is just {z/w}, right? That's great if w and z are non-zero. Now just start thinking about how you have to qualify that statement if for example w is zero and z is nonzero. Or if both w and z are zero.
     
    Last edited: Feb 10, 2008
  7. Feb 10, 2008 #6
    if w and z are zero then x must be zero, the question states that x is in the set of complex numbers though so does x =0 count as a solution? If either w or z are zero, then that would also make x zero in that case too? I am confused, possibly because i am new to linear algebra and havent seen a question like this before. Am i missing something i need to know to answer the question? thanks
     
  8. Feb 10, 2008 #7

    Dick

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    If w AND z are zero, then 0*x=0 is true for ANY x, right? This isn't really a 'linear algebra' problem. As Halls pointed out the complex numbers in a problem like this are just like the ordinary real numbers. What about trying to solve 0*x=z for z not equal to zero?
     
  9. Feb 10, 2008 #8
    Ok so a solution would be x is a set of the complex numbers (indicating it has any value) when w and z are zero. If z is [tex]\neq[/tex] 0, then i would think 0*x is inconsistent?
     
  10. Feb 10, 2008 #9

    What am i missing here anyone, please. Thanks
     
  11. Feb 10, 2008 #10

    Dick

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    Yes, it's inconsistent. So there are no solutions. So w=z=0 has all complex numbers as a solution, w and z not equal to zero has a single solution and w=0, z not equal to zero has no solutions. Any other cases you need to worry about?
     
  12. Feb 10, 2008 #11
    No, that's a great help Dick for clarifying that. The next part of the question is to solve the following system of linear equations:

    (5i)x1 + 3x2 = 12+i (1)

    x1 + (2i)x2 = 3-i (2)

    I treated it just like simultaneous equations, by multiplying (2) by 5i and cancelling i^2 terms to make (3). Then Doing (1)-(3) to get

    -7x2+14i-7=0 to find x2=2i-1. Then substituting to get x1=(3/i) -1
     
  13. Feb 10, 2008 #12

    Dick

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    That's the right idea. But what did you get for (3)? I get 13x2+14i-7=0. Check the work for simple mistakes.
     
  14. Feb 10, 2008 #13
    Also for w and z not zero i put x is the set of complex numbers excluding zero as a solution. When writing either w or z = 0, but not both, can i write this:

    when exclusively w[tex]\cup[/tex]z=0 then there are no solutions.
     
  15. Feb 10, 2008 #14

    Yes, you're right, thanks again. I then get x2= 7/13 - 14i/13 and x1 = 11/13 - 27i/13
     
  16. Feb 10, 2008 #15

    Dick

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    Why would you exclude x=0 from the solutions in the case w,z not zero? And the cases w=0 or z=0 with the other being nonzero are quite different. I think you should think this over with a clear head. Because I'm not sure what your "when exclusively" phrase even means.
     
    Last edited: Feb 10, 2008
  17. Feb 10, 2008 #16

    Dick

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    The only possibilities are everything is a solution, nothing is a solution and there is one unique solution. Figure how to split these possibilities up depending on whether w and z are or are not zero. If you are solving linear equations in two variables you can probably handle that.
     
  18. Feb 10, 2008 #17
    I thought that if w and z were not zero then x couldn't be either?

    For the case when w or z are zero but not both, (that is what i mean the exclusive union of the set of w and z) then there are no solutions. I think iwill just write it more simlply as my answer. Thanks again
     
  19. Feb 10, 2008 #18

    Yes i think ive got it now. Thanks
     
  20. Feb 6, 2011 #19
    Re: [SOLVED] Complex numbers linear algebra

    Hey, i got the same questions for my exercise, but when solving part b i got x1= -1-3i and x2=-1+2i. Not sure if that's right but i substituted it into each of the equations. But thanks for the help on part a! i'm new to linear algebra too
     
  21. Feb 6, 2011 #20

    Dick

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    Re: [SOLVED] Complex numbers linear algebra

    If you are talking about the problem in post 11, that's not the right solution.
     
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