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!