# Thread: show why domain D is not simply connected.

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

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) = $\displaystyle \frac{1}{z-z_0}$ is analytic on and within C, and by the Cauchy-Goursat theorem $\displaystyle \int_C f(z)dz = 0$
But since here $\displaystyle \int_{|z|=1} \frac{1}{z-z_0}dz \neq 0$ 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!

2. ## Re: show why domain D is not simply connected.

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 $\displaystyle \int_{|z|=1} \frac{1}{z-z_0}dz \neq 0$
Did you mean $\displaystyle \int_{C} \frac{1}{z-z_0}dz \neq 0$ ?

3. ## 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.

4. ## Re: show why domain D is not simply connected.

Originally Posted by punkstart
Yes here C is the curve |z|=1
Well, the result is valid for any $\displaystyle C\subset D$ and your reasoning is essentially right. However, it is sufficient to say: $\displaystyle D$ simply connected implies $\displaystyle \int_{C}\frac{dz}{z-z_0}=0$ by the Cauchy-Goursat Theorem, which contradicts the hypothesis $\displaystyle \int_{C}\frac{dz}{z-z_0}\neq 0$ .