# Homework Help: Find harmonic conj of u(x,y)=ln(x^2+y^2)

1. Jun 7, 2014

### DotKite

1. The problem statement, all variables and given/known data
Find harmonic conjugate of u(x,y)=ln(x^2+y^2) and specify the region it is defined
then show u has no harm conj on C\{0}

2. Relevant equations

3. The attempt at a solution
Ok so i found the harmonic conj by converting to polar and found it to be v(r,Θ) = Θ.

I am having trouble finding out where v is defined. Also it seems our analytic function f = u + iv
is log(z) since u(r,Θ) = ln|r|. Isn't that defined everywhere but zero? Why would u not have a harmonic conj at C\{0}?

2. Jun 7, 2014

### Staff: Mentor

The complex log can be defined everywhere but zero, but then you cannot make it continuous.
The same problem comes up with v=Θ.

3. Jun 7, 2014

### DotKite

If you dont mind, could you explain where the problem with v = Θ comes in?

Also, since v = Θ is not continuously defined on C\{0} would that mean that it doesn't have a partial, and therefore cannot satisfy cauchy riemann, which in turn is the reason why u cannot have a harm conj on C\{0}?

4. Jun 7, 2014

### Staff: Mentor

What happens if go around the unit circle, for example?

I don't know what you mean here.

5. Jun 7, 2014

### DotKite

Θ will go from 0 to 2pi

I am trying to find the reason why v(r,Θ) not continuous on C\{0} implies u does not have a harmonic conjugate on C\{0}. Is it because if it is not continuous then it's partial derivative will not exist on C\{0}, thus implying that that Cauchy Riemann equations will not be satisfied on C\{0}?

6. Jun 8, 2014

### Staff: Mentor

And what happens at the transition "close to 2 pi -> 0"?

At least not everywhere. Right.

7. Jun 8, 2014

### DotKite

When theta gets to 2pi it repeats. So at 0 and 2pi the function is multivalued thus not continuous?

8. Jun 8, 2014

Right.

9. Jun 8, 2014

### DotKite

Thanks for all the help these last couple of days mfb