$ 5^{* *}$. Given these two statements:
i) $\forall \gamma \in \mathbb{C}\backslash \{0,1\} $ where $\gamma$ is a closed path, it holds that: $\displaystyle{\int_{\gamma} \frac{d\eta}{\eta(\eta-1)}=0}$
ii) $\forall \gamma \in \mathbb{C}\backslash[0,1]$, where $\gamma$ is a closed path, it holds that: $\displaystyle{\int_{\gamma} \frac{d\eta}{\eta(\eta-1)}=0}$
which one is true ?
My thoughts:
In i) also points between 0 and 1 are allowed, only 0 and 1 aren't. In ii) the piece with between 0 and 1 is not allowed. I guess that the first statement is true, but the second isn't, because there are more points not allowed in the second one.
How does one prove i) and contradict ii)? Does anybody see a way? Please do tell!
Proof attempt:
i)
Let $\gamma = 0.5e^{it}+1 , t\in [0,2\pi]$, this is a closed path with the point 1 in it but not 0. From the Cauchy Goursat Theorem and Cauchy Integral formula it follows that with:
$\eta_{0}=0 , \eta_{1}=1$ only $\eta_{1}$ lies within this disc, so one has:
$\displaystyle{\int_{\gamma} \frac{\frac{1}{(\eta)}}{\eta-1}}= 2\pi i (\frac{1}{\eta|_{\eta=1}})= 2\pi i $
So i) can not be true.
ii)
Only both singularities can be within the circle or outside. If they are outside, then the Integral is 0. If they are both inside, then with the Cauchy Goursat theorem and the Cauchy integral formula: we can write them as the sums in the form:
$\displaystyle{\int_{\gamma}\frac{1}{\eta(\eta-1)}d\eta = \int_{\gamma} \frac{\frac{1}{\eta-1}}{\eta}}d\eta + \int_{\gamma}\frac{\frac{1}{\eta}}{\eta-1}d\eta = -2\pi i + 2\pi i = 0 $
Is this proof correct so far, I believe one also needs to show that only these two cases are possible and not any other. How to do that?