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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
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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

1 Answer 1

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

As for the sets you describe, try identifying these sets in the complex plane and you will see why they do not qualify as domains. The first one is not open and the second one is not connected.

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Thank you, but i believe the first one is not closed as well right? –  mathwannabe Jan 24 '13 at 4:47
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it is certainly not closed. –  Ittay Weiss Jan 24 '13 at 4:48
    
Thank you Ittay –  mathwannabe Jan 24 '13 at 4:58
    
you are welcome @mathwannabe :) –  Ittay Weiss Jan 24 '13 at 5:15

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