Integral of $\int_0^{2\pi} \frac{e^{-it }dt}{e^{it}-z}$ Sorry if this question seems stupid, but I am confused here:
Does it follow that
$$
I(z)=\int_0^{2\pi} \frac{e^{-it}dt}{e^{it}-z} = 0
$$
For every $z$ with $|z|<1$? I think this is true.
I converted this integral to a line integral around the unit circle $S$:
$$
I(z)=\int_S \frac{d\xi}{i\xi^2(\xi-z)}
$$
I wanted to say the integrand has a primitive, but it seems I need logarithm to express it, which I am not quite sure can be holomorphic on $S$.
 A: Using partial fractions we get: $\dfrac{1}{\xi^2(\xi-z)} = -\dfrac{1/z^2}{\xi}-\dfrac{1/z}{\xi^2}+\dfrac{1/z^2}{\xi-z}$
Therefore, $I(z) = \displaystyle\int_{S}\dfrac{\,d\xi}{i\xi^2(\xi-z)} = -\dfrac{1}{iz^2}\int_S\dfrac{\,d\xi}{\xi} - \dfrac{1}{iz}\int_S\dfrac{\,d\xi}{\xi^2}+\dfrac{1}{iz^2}\int_S\dfrac{\,d\xi}{\xi-z}$. 
The first two of these three integrals are very straight forward to evaluate. 
The third one is a direct application of Cauchy's Integral Formula. (Actually, all three are). 
After evaluating these integrals, you do indeed get $0$ as the final answer. 
A: Another take on it:
We have:
$I(z) = \int_0^{2\pi} \dfrac{e^{-it}dt}{e^{it} - z} = \int_0^{2\pi} \dfrac{e^{-2it}dt}{1 - e^{-it}z}, \tag{1}$
and since
$\vert e^{-it}z \vert = \vert e^{-it} \vert \vert z \vert = \vert z \vert < 1, \tag{2}$
we also have
$\dfrac{1}{1 - e^{-it}z} = \sum_0^\infty (e^{-it}z)^n = \sum_0^\infty e^{-int}z^n; \tag{3}$
the series on the right-hand side converges absolutely and uniformly by virtue of (2); thus we may exploit term-by-term integration in the evaluation of 
$I(z) = \int_0^{2\pi} \dfrac{e^{-it}dt}{e^{it} - z} = \int_0^{2\pi} \dfrac{e^{-2it}dt}{1 - e^{-it}z} = \int_0^{2\pi} e^{-2it}(\sum_0^\infty e^{-int}z^n)dt$
$= \int_0^{2\pi} (\sum_0^\infty e^{-i(n + 2)t}z^n)dt = \sum_0^\infty \int_0^{2\pi}(e^{-i(n + 2)t}z^n)dt , \tag{4}$
and since
$\int_0^{2\pi}(e^{-i(n + 2)t}z^n)dt = z^n \int_0^{2\pi} e^{-i(n + 2)t}dt =  \dfrac{z^n}{-i(n + 2)}(e^{-i(n + 2)t} \mid_0^{2\pi}$
$ = \dfrac{z^n}{-i(n + 2)}(e^{-2(n + 2)\pi i} - e^0) = \dfrac{z^n}{-i(n + 2)}(1 - 1) = 0; \tag{5}$
it follows that $I(z) = 0$.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
