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Domain of z = ln (x^2 + y^2 )

  1. Oct 12, 2016 #1
    1. The problem statement, all variables and given/known data

    for the domain of ln (x^2 + y^2 ) , it it given in my notes that the ans is x ≠ 0 and y ≠ 0

    IMO , it's wrong to give x ≠ 0 and y ≠ 0 , because the meaning of x ≠ 0 and y ≠ 0 is that x and y cant be 0 all the times so just leave the ans (x^2 + y^2 ) > 0 , will do ?
    2. Relevant equations


    3. The attempt at a solution
    when x = 0 , y not = 0 , z is defined , when y = 0 , z not = 0 , z is also undefined , So , x and y can be 0
     
  2. jcsd
  3. Oct 12, 2016 #2

    andrewkirk

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    You are correct. The domain is the number plane excluding only the point (0,0) (the 'origin').
     
  4. Oct 13, 2016 #3

    Mark44

    Staff: Mentor

    The equation ##x^2 + y^2 = 0## can be thought of as a degenerate circle whose center is at (0, 0) and whose radius is 0. In other words, the point (0, 0). The graph of the solution set of the inequality ##x^2 + y^2 > 0## is all of the points in the plane outside that degenerate circle.
     
  5. Oct 13, 2016 #4
    so the ans cant be x ≠0 , y≠0 ?
    the correct ans is either (x^2) + (y^2) >0 or (x,y) ≠(0,0) which means x any y cant be 0 at the same time ?
     
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