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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?

Thanks in advance!

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  • $\begingroup$ 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 $0=\int_{\gamma_1} f=\int_{\gamma_2}f \rightarrow \int_{\gamma_1-\gamma_2}f=\int_0 f$? $\endgroup$ – user99680 Jul 1 '14 at 6:37
  • $\begingroup$ Is there any other theorem available - it is hard to "construct" such a homotopy. Does your book mention other characteristics of simply connected domains (existence of primitives etc.)? $\endgroup$ – Blah Jul 1 '14 at 6:58
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    $\begingroup$ 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. $\endgroup$ – Pax Kivimae Jul 1 '14 at 7:12
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    $\begingroup$ OK I understood the proof in my book now. The whole complex plane is simply connected. If the set is not the whole of the complex plane, then we apply the Riemann mapping theorem. Then the set is conformally equivalent to the unit disc, and it is easy to see that the unit disc is simply connected. So my approach where were not the right one! $\endgroup$ – Alexander Jul 1 '14 at 7:28

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