# show why domain D is not simply connected.

• August 13th 2011, 05:02 AM
punkstart
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 $\int_{|z|=1} \frac{1}{z-z_0}dz \neq 0$

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) = $\frac{1}{z-z_0}$ is analytic on and within C, and by the Cauchy-Goursat theorem $\int_C f(z)dz = 0$
But since here $\int_{|z|=1} \frac{1}{z-z_0}dz \neq 0$ 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!
• August 13th 2011, 07:39 AM
FernandoRevilla
Re: show why domain D is not simply connected.
Quote:

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 $\int_{|z|=1} \frac{1}{z-z_0}dz \neq 0$

Did you mean $\int_{C} \frac{1}{z-z_0}dz \neq 0$ ?
• August 13th 2011, 10:51 PM
punkstart
Re: show why domain D is not simply connected.
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.
• August 14th 2011, 02:41 AM
FernandoRevilla
Re: show why domain D is not simply connected.
Quote:

Originally Posted by punkstart
Yes here C is the curve |z|=1

Well, the result is valid for any $C\subset D$ and your reasoning is essentially right. However, it is sufficient to say: $D$ simply connected implies $\int_{C}\frac{dz}{z-z_0}=0$ by the Cauchy-Goursat Theorem, which contradicts the hypothesis $\int_{C}\frac{dz}{z-z_0}\neq 0$ .