Preliminaries:
Proposition: Let $\gamma\colon [\alpha,\beta]\to\mathbb{C}$ be a smooth path to pieces and let $\phi\colon\gamma([0,2\pi])\to\mathbb{C}$ be continuos.Let $U=\mathbb{C}\setminus{\gamma([0,2\pi])}$ be and $f\colon U\to\mathbb{C}$ given by $$f(z)=\int_{\gamma}{\dfrac{\phi(\xi)}{(\xi-z)}d\xi}.$$ Then f supports serial development of powers in U and therefore $ f \in \mathcal {H} (U). $
Corollary: Let $U\subseteq{\mathbb{C}},$ $f\colon U\to\mathbb{C}$ be.If $f\in\mathcal{H}(U),$ then $f^{(n)}\in\mathcal{H}(U),\phantom{a}\forall n\in\mathbb{N}.$ Even more, if $B[a,r]\subseteq{U}$ and $\gamma\colon [0,2]\pi\to\mathbb{U}$ given by $\gamma(t)=a+re^{it},\phantom{a}t\in[0,2\pi],$ then $$\dfrac{f^{(n)}(a)}{n!}=\dfrac{1}{2\pi i}\int_{\gamma}{\dfrac{f(\xi)}{(\xi-a)^{n+1}}d\xi},\phantom{a}\forall n\in\mathbb{N}.$$
Let $m,n\in\mathbb{N}$ be and let $a,b\in\mathbb{C}$ be, with $$|a|<1<|b|.$$ Determine the value of the integral $$\int_{\gamma}{\dfrac{d\xi}{(\xi-a)^{m}(\xi-b)^{n}}},$$ with $$\gamma(t)=e^{it},\phantom{a}\forall t\in [0,2\pi].$$
Attempt: Let $f\colon \gamma([0,2\pi])\to \mathbb{C}$ be given by $$f(\xi)=\dfrac{1}{(\xi-a)^{m}}.$$ Notice that f is continuos. Let $U=\mathbb{C}\setminus{\gamma([0,2\pi])}.$ Let $F\colon U\to\mathbb{C}$ be given by $$F(z)=\int_{\gamma}{\dfrac{f(\xi)}{(\xi-z)}d\xi}.$$ You can prove that the F function supports the development of power series and, therefore, $ F \in \mathcal {H} (U)$ (by the proposition and corollary above.) Then $F^{(n)}\in\mathcal{H}(U),\phantom{a}\forall n\in\mathbb{N},$ and \begin{align*} \dfrac{F^{(n-1)}(b)}{(n-1)!}&=\dfrac{1}{2\pi i}\int_{\gamma}{\dfrac{f(\xi)}{(\xi-b)^{n}}d\xi}\\ &=\dfrac{1}{2\pi i}\int_{\gamma}{\dfrac{d\xi}{(\xi-a)^{m}(\xi-b)^{n}}d\xi}. \end{align*} Then $$\int_{\gamma}{\dfrac{d\xi}{(\xi-a)^{m}(\xi-b)^{n}}d\xi}=2\pi i\dfrac{ F^{(n-1)}(b)}{(n-1)!}.$$
Is the argument correct?
Any hints would be appreciated.
The Residue theorem would be a great way to do this simply, but since you are not that far along yet, you'll have to get deeper in the details.
First, use the method of partial fractions to separate the roots:
$\dfrac 1{(z-a)^m(z-b)^n} =\dfrac{P(z)}{(z-a)^m} + \dfrac{Q(z)}{(z-b)^n}$ for polynomials $P,Q$ of degree less than $m,n$. (I'll leave the task of determining $P, Q$ to you.)
Now the thing to note is that $b$ is outside of the unit circle $\gamma$, so inside the circle and on the circle itself, $\dfrac{Q(z)}{(z-b)^n}$ is analytic. Therefore by the Cauchy integral theorem, $$\oint_\gamma \dfrac{Q(\xi)}{(\xi-b)^n}\,d\xi = 0$$
Since $a$ is inside the circle, the same trick does not work for it. You don't indicate whether you seen the Cauchy integral formula (that is "formula", not "theorem") yet. If so, it will easily handle the integral about $a$.
If not, then draw a second circle $\beta$ around $a$, small enough to be entirely inside $\gamma$. And let $\alpha$ be a curve which follows $\gamma$ fully around ccw, then follows the line segment from $1$ to a point on $\beta$, then follows $\beta$ fully around cw, and finally follows the line segment again backe to $1$. $\alpha$ is a closed curve, and the itegrand is analytic in its interior, so by Cauchy's theorem, its integral is $0$. Since the two intgrations along the line segment are the same, but in opposite directions, they cancel out. leaving $$\oint_\alpha = \oint_\gamma - \oint_\beta$$ Therefore $$\oint_\gamma \dfrac{P(\xi)}{(\xi-a)^m}\,d\xi - \oint_\beta \dfrac{P(\xi)}{(\xi-a)^m}\,d\xi= 0\\\oint_\gamma \dfrac{P(\xi)}{(\xi-a)^m}\,d\xi = \oint_\beta \dfrac{P(\xi)}{(\xi-a)^m}\,d\xi$$
The reason for this argument is that since $\beta$ is a circle about $a$, the integral along $\beta$ will be easier to calculate. If $\beta(t) = a + \epsilon e^{it}$, the integral becomes
$$\oint_\beta \dfrac{P(\xi)}{(\xi-a)^m}\,d\xi = \int_0^{2\pi} \dfrac{P(a + \epsilon e^{it})}{e^{imt}}\epsilon i e^{it}\,dt$$