(-y,x) is a vector$$f_a(x,y) = \frac {1}{(x^2+y^2)^a}(-y,x)$$
After using the curl or rotation formula i get: for $a = 0$ and $a = 1$ there is constant curl or rotation. For the latter the Curl is 0 but the Function does not have a Potential.
Now my problem is to calculate Potential $u$ and $v$ such that this function is defined in $$\{(x,y)\in R^2| x,y \neq 0\}$$
How do i go about this because as you can see from the values of "$a$" there is no Potential function. Or am i suppose to take a value for $a$ that makes this as a potential function and then work on it? please advice.