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Sep 7, 2015 at 18:15 history edited Alexandre Eremenko CC BY-SA 3.0
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Sep 7, 2015 at 18:14 comment added Alexandre Eremenko @Simon Henry: you are right. I thought that this was stated in the question but now I see it was not:-)
Sep 7, 2015 at 18:12 comment added Alexandre Eremenko Your sufficient condition is not stated clearly in this remark, but perhaps if you state it clearly it will become correct.
Sep 7, 2015 at 15:00 comment added student I used the expression "number of zeros of the identity in $f(C)$" to denote the winding number of $f(C)$ around zero. Then (it seems to me) the equality I mentioned holds for holomorphic functions, and then one version of the condition I asked for would be (as I said) that $f(C)$ passes through all four quadrants in cyclic order--this is just a condition on the real and imaginary parts of $f(z), z \in C$. Is this indeed a sufficient condition to conclude that the open set bounded by $C$ contains a zero of $f$?
Sep 7, 2015 at 13:14 comment added Simon Henry Of course, this only works if $f$ is defined on all the 'interior' of the curve.
Sep 7, 2015 at 12:56 history answered Alexandre Eremenko CC BY-SA 3.0