Thread: show why domain D is not simply connected.

Suppose we are given a domain D containing a simple closed contour C. Suppose that in addition we are given a point Zo with Zo NOT in D and

Explain why this ensures that D is not simply connected.

My attempt:
If C is contained in D, but point Zo is not in D then there are two options:

a) Zo is outside C and outside D, or
b) Zo is inside C, but in a region inside C that is not in D ( here i picture D as an annulus of two circles, C within the annulus)

For a) since Zo is outside C, f(z) = is analytic on and within C, and by the Cauchy-Goursat theorem
But since here it cannot be that Zo is outside C, hence option b) must b true.

Therefore it is the case that Zo is inside C, but in a region inside C that is not in D, which guarentees that D is not a simply connected domain.

I'm not convinced this is correct. I especially wonder if my assumption of only two cases a) and b) are correct.It makes sense to me but i feel i might be overlooking something. Would appreciate any comments!

Originally Posted by punkstart

Suppose we are given a domain D containing a simple closed contour C. Suppose that in addition we are given a point Zo with Zo NOT in D and

Did you mean ?

Yes here C is the curve |z|=1, not sure why C was given, surely it is not necessary to know that C is |z|=1 to solve this problem ? But for the sake of including all information i retype exactly my problems as they are in the text.

Originally Posted by punkstart

Yes here C is the curve |z|=1

Well, the result is valid for any and your reasoning is essentially right. However, it is sufficient to say: simply connected implies by the Cauchy-Goursat Theorem, which contradicts the hypothesis .