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Sufficiency of n eqs. for n unkwns.

  1. May 7, 2004 #1

    turin

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    I remember from way back in jr high learning that a set of n independent equations is sufficient to determine n unkowns. Lately, I've been a little confused when it comes to complex variables.

    My confusion is particularly manifest when dealing with magnitudes and complex conjugation.

    I just wanted to get a second oppinion. For example, given two complex valued unkowns w and z, it seems like one should actually consider these as four unkowns, two for the real parts and two more for the imaginary parts. But then again, they can be determined by only two equations. But then again, each equation seems to have two compoents. I don't know how I should think about this.
     
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  3. May 7, 2004 #2

    NateTG

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    Consider that each complex equation is really two equations, one for the imagiary part, and one for the real part, so if you seperate into components, you still get a matching number of equations and unknowns.
     
  4. May 7, 2004 #3

    turin

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    Score 1 for 2 eqs. required per complex variable.
     
  5. May 8, 2004 #4

    HallsofIvy

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    Two real equations (i.e. with real coefficients) per complex variable.

    As NateTG said, that is still one complex equation per variable.

    The fact that n "independent" equations in n unknowns has a unique solution is pretty much the definition of "independent".
     
  6. May 9, 2004 #5

    turin

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    Here's an example.

    |z| + |w| = 1
    z + w* = 0

    This system of equations pins down the magnitudes but leaves the phases arbitrary. Does this show that these equations are not independent?
     
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