Residue theorem with nth power of singular points I am solving some exercises from the Complex Analysis book of Busam, Freitag. There is a question related to the Residue theorem however I could not come with a solution. The curve is centered at $1$ and has a radius of $1$.
$$\int_{\alpha _{1;1}}\left(\frac{z}{z-1}\right)^n\,\mathrm{d}z,\quad n\in\mathbb{N}$$
The $n$-th power is what makes me confused about the solution. Provided answer by the book is $2\pi in$
The second question is:
$$\int_{\alpha _{0;r}}\frac{1}{(z-a)^n(z-b)^m}\,\mathrm{d}z,\quad\lvert a\rvert <r < \lvert b\rvert,\quad n,m\in\mathbb{N}$$
And it has a complex answer related to binomials as:
$$2\pi i(-1)^m {n+m-2\choose n-1}\frac{1}{(b-a)^{n+m-1}}$$
I do not have any idea about how to solve them, How are these questions solved?
 A: In the first one you have a pole of order $n$ at $z=1$ so the relevant residue is
$$\frac{1}{(n-1)!} \left. \frac{d^{n-1}}{dz^{n-1}} \left ( (z-1)^{n-1} f(z) \right ) \right |_{z=1}=\frac{1}{(n-1)!} \left. \frac{d^{n-1}}{dz^{n-1}} \left ( z^n \right ) \right |_{z=1}.$$
Here the derivatives are easily computed.
In the second one you have a pole of order $n$ at $z=a$ so the relevant residue is
$$\frac{1}{(n-1)!} \left. \frac{d^{n-1}}{dz^{n-1}} \left ( (z-a)^{n-1} f(z) \right ) \right |_{z=a}=\frac{1}{(n-1)!} \left. \frac{d^{n-1}}{dz^{n-1}} \left ( (z-b)^{-m} \right ) \right |_{z=a}.$$
Here the derivatives are a little bit more annoying to compute for general $n,m$ due to the need to do some fiddling with factorials.
A: For the first part, note that
$$\left( \frac{z}{z-1} \right) ^n = \left( \frac{1}{z-1} + 1 \right)^n = \sum_{i=0}^n (z-1)^{-i} \binom{n}{i}$$
From this, it is clear that the residue at $1$ is $\binom{n}{1} = n$.
For the second part, we only need to check the residue at $a$ since it is the only pole inside the contour.
You can do a similar argument but it is a bit more painful, so you are probably better off using the usual formulas for residues in terms of derivatives.
