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An incomplete (maybe wrong) proof about R^2

  1. Aug 10, 2012 #1
    1. The problem statement, all variables and given/known data[/b]
    If z is a complex number, prove that there exists an r > 0 and a complex number w with |w|=1 such that z=rw. Are w and r always uniquely determined by z?


    2. Relevant equations
    n/a


    3. The attempt at a solution
    As C is a set closed under multiplication, we can define z=rw where r=(a,0) and w=(x,y). Hence z=(a,0)(x,y)=(ax,ay).
    We can choose a w such that |w|=1; sqrt{(x+yi)(x-yi)}=1 = x^2+y^2=1 and clearly, this is the equation of the unit circle. Therefore, w must be determined uniquely by z so that they could intersect.
    I don't know if my proof is valid and I'm aware of that I didn't say anything about r but I think it is related to z=(a,0)(x,y)=(ax,ay) somehow. Help please...
     
  2. jcsd
  3. Aug 10, 2012 #2

    HallsofIvy

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    Perhaps the difficulty is that you think this has something to do with R^2. It doesn't. What do you get if you write z in polar form?
     
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