Talk:Open mapping theorem (complex analysis)

This write-up is 10000000 times more clear than what's in my book! Thank you!

This proof needs modification to deal with the case in which $$z_0$$ is not a simple zero.

uhm, i don't think so. However it seems that the connection hypothesis is unnecessary and, indeed, unused. 151.45.69.228 (talk) 15:43, 21 August 2012 (UTC)

This is, indeed, total bogus. For non connected U, even you can easily construct a counterexample by choosing f to be the zero-function on a connection component of U, while f(z) = 1 for all other z in U. This is, indeed, holomorphic and nonconstant, but the image of f, {0,1}, is clearly not open. ~ batman