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

1. Oct 12, 2016

### chetzread

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. Oct 12, 2016

### andrewkirk

You are correct. The domain is the number plane excluding only the point (0,0) (the 'origin').

3. Oct 13, 2016

### 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.

4. Oct 13, 2016

### chetzread

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 ?

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted