Holomorphically simply connected implies simply connected

In my book on complex analysis a "Holomorphically simply connected" set is defined as a set where for any holomorphic function $f$ and any closed path $\gamma _1$ we have that $\int_{\gamma _1 } f(z) dz =0$

My question is:

Given two closed paths $\gamma _1$ and $\gamma _2$ in a holomorphically simply connected set, how can i construct a path homotopy between them, and also show that they are path homotopic to the constant path, so that the set is simply connected?

• Isn't the value of the integral over a curve constant in the homotopy class of the curve, and conversely, if the integral over two curves has the same value, then $0=\int_{\gamma_1} f=\int_{\gamma_2}f \rightarrow \int_{\gamma_1-\gamma_2}f=\int_0 f$? – user99680 Jul 1 '14 at 6:37
• The idea would be that if the set is not simply connected, then there is a loop that cannot be contracted. Not being contractable means there is a hole in the inside of the loop, so that we can use the standard $1/z$ trick to get a holomorphic (on our set) function that has nonzero integral, and thus isn't simply connected. Notably, this is a proof by contradiction. I don't know of a constructive proof or method, and I would be surprised if one existed. – Pax Kivimae Jul 1 '14 at 7:12