This is where the problem arises. The function

is not single valued, since is not a one-one-function. In fact, if we consider

then it is one-one, onto and the corresponding inverse function is called the principle branch of . Thus, in order to make the above calculations possible, we can decide to select the principle branch of . But then, this function is not everywhere continuous. In order to set a valid solution, we can modify our domain, as shown in the next example.

Consider

i.e., consists of the space minus the plane consisting of negative part of the -axis (including 0) and the -axis. Define

Then is the required potential for