If I have $f(z) = 1$ and $g(z) = \frac{1}{z}$ and I am looking for a minimal surface on $\mathbb{C} \backslash \{0\}$ using the Weierstraß-Enneper representation of minimal surfaces. Now I was wondering if there is any restriction on how I have to choose the path in the line integral due to the fact that my domain is now simply connected ? I mean, depending on how I choose my path, the result may be different.
If anything is unclear, please let me know.
This can be either the catenoid or the helicoid, depending on the chosen convention for the Weierstrass-Enneper parametrization. With the convention chosen by Wikipedia $$X=\frac12\operatorname{Re}\int f(1-g)^2\,dz, \quad Y=\frac12\operatorname{Re}\int if(1+g^2)\,dz, \quad Z=\operatorname{Re}\int fg\,dz$$ it's the catenoid. You would get a helicoid with the above formulas, if $g(z)=i/z$.
The first two integrals are single valued in your example: you can (and should) write down an explicit antiderivative, thus showing that the result of integration does not depend on the path. The choice of the initial point of integration is not important: it contributes just an additive constant, which means translating the surface without changing its shape.
The third one involves $\int\frac1z \,dz$, which is multivalued (complex logarithm). But the real part is single-valued here: $Z=\log|z|$. So you end up with single-valued parametrization anyway.
It's more interesting to consider $g(z)=i/z$ in this context. Then $Z= \arg z$, where the argument can take infinitely many values separated by integer multiples of $2\pi$. Now the parametrization really depends on the path; specifically on how the path winds around the origin. There are two approaches you can take here. One is to allow any paths (no passing through $0$), accept the multivaluedness of parametrization, and get the entire helicoid. The other is to make a branch cut (say, along the negative semi-axis), prohibiting the paths to go across the cut. This makes the parametrization single-valued, but you get only one turn of helicoid.