# Domain of a complex function

1) Why do the domain of a complex function has to be a disk (circular neighborhood of Zo)? $$|z-z_0|<p$$ 2) Domain is an open connected set. An open set D is said to be connected if every pair of points $z_1$,$z_2$ in S can be joined by a polygonal path that lies entirely in S.

so why are the following sets are not domains?

a) $-1<Im( z )<= 1$

b) $(Re(z))^2 >1$

Thank you

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is (a) open? is (b) connected? – Maesumi Jan 24 '13 at 4:32
It's worth noting that in this post, "domain" is used in two different contexts. In one, it is simply the subset of the complex plane over which a given function is defined. In two, it is (as stated) an open connected set. – Cameron Buie Jan 24 '13 at 4:39

The domain of a complex function does not have to be a disk. It can be any subset of $\mathbb C$.