Proving well defined (complex variables)

[Milky]

Homework Statement

Let D be a connected domain in R^2 and let u(x,y) be a continuous vector field defined on D. Suppose u has zero circulation and zero flux for any simple closed contour on D.

$$u(x,y) = (u_1(x,y),u_2(x,y))$$

$$\Gamma = \int_{c}(u\circ\gamma)tds = 0$$$$F=\int_{c}(u\circ\gamma)nds$$ = 0[/tex]$$\phi(x,y)=\int_{c}(u\circ\gamma)tds$$$$\psi(x,y)=\int_{c}(u\circ\gamma)nds$$Prove that $$\phi, \psi$$ are well defined.

The Attempt at a Solution

For $$\phi$$:

I think to prove its well defined means to prove that if $$(x,y)=(x_0,y_0), then \phi(x,y)=\phi(x_0,y_0)$$

By Cauchy Formula, the integral of one path is equal to the integral of another:

$$\int_{c}(u\circ\gamma)tds = 0 = \int_{c_2}(u\circ\gamma)tds - \int_{c_1}(u\circ\gamma)tds$$Then, $$\int_{c_2}(u\circ\gamma)tds$$ = $$\int_{c_1}(u\circ\gamma)tds$$So, when $$(x,y)=(x_0,y_0)$$ the integrals are equal as well.
Is this how to prove it is well-defined?

 

[Nino MMP]

For \psi:I think to prove its well defined means to prove that if (x,y)=(x_0,y_0), then \psi(x,y)=\psi(x_0,y_0)By Cauchy Formula, the integral of one path is equal to the integral of another:\int_{c}(u\circ\gamma)nds = 0 = \int_{c_2}(u\circ\gamma)nds - \int_{c_1}(u\circ\gamma)ndsThen, \int_{c_2}(u\circ\gamma)nds = \int_{c_1}(u\circ\gamma)ndsSo, when (x,y)=(x_0,y_0) the integrals are equal as well.Is this how to prove it is well-defined?