$2\pi i\beta = \int\limits_{|z|=r} \frac{dz}{f(z)-z}$ 
Let $\beta \in \mathbb{C^{*}}$ and $f(z) = z+ z^{k+1} - \beta z^{2k+1}$. Show if if r is small enough then
  $$2\pi i\beta = \int\limits_{|z|=r} \frac{dz}{f(z)-z}$$

This is my input: 
I have to resolve it using residues aplications, so 
$$ \int\limits_{|z|=r} \frac{dz}{f(z)-z} =  \int\limits_{|z|=r} \frac{dz}{z+ z^{k+1} - \beta z^{2k+1}-z} = \int\limits_{|z|=r} \frac{dz}{ z^{k+1} - \beta z^{2k+1}} = \int\limits_{|z|=r} \frac{dz}{z^{k+1}(1-\beta z^{k})} $$ 
My question is, I have a lot of methods to resolve integrals by residues, in this case, I think I have to resolve it on unit disk for $z = e^{i \theta}$ 
I have to calculate his residue, but what are the singularities of $f(z)$ and what is the best method to resolve the exercise? 
Can you help me, please?
 A: If $\beta |z|^k < 1$  then
$$
 \frac{1}{f(z)-z} = \frac{1}{z^{k+1}(1-\beta z^k)}
$$
can be developed into a Laurent series:
$$
 \frac{1}{z^{k+1}}(1 + \beta z^k + \beta^2 z^{2k} + \ldots)
= \frac{1}{z^{k+1}} + \frac{\beta}{z} + \beta^2 z^{k-1} + \ldots
$$
and the residuum at $z=0$ is the coefficient of $z^{-1}$:
$$
\frac{1}{2 \pi i}\int\limits_{|z|=r} \frac{dz}{f(z)-z} =  \operatorname{Res}(\frac{1}{f(z)-z}, 0) = \beta \, .
$$
A: You could try Cauchy's integral formula and the hint that $r$ is small enough, thus, the only singularity is $z=0$.
$$g^{(k)}(0)=\frac{k!}{2\pi i} \int\limits_{|z|=r}\frac{g(z)}{z^{k+1}}dz$$
where $g(z)=\frac{1}{1-\beta z^k}$
$$g'(z)= \frac{\color{red}{\beta\cdot k} \cdot z^{k-1}}{(1-\beta z^k)^2}$$
$$g''(z)=\frac{\color{red}{\beta\cdot k(k-1)} \cdot z^{k-2} (1-\beta z^k)^2-\beta kz^{k-1}\left((1-\beta z^k)^2\right)'}{(1-\beta z^k)^3}$$
by the time you reach $k$-th derivative, $g^{(k)}(0)=k!\cdot\beta$ (left as an exercise) and you are done.
A: We have
$$ 
\begin{split}
\int\limits_{|z|=r} \frac{dz}{f(z)-z} & = \int\limits_{|z|=r} \frac{dz}{z^{k+1}(1-\beta z^{k})}\\
&=\int\limits_{|z|=r} \frac{1}{z^k}\frac{dz}{z^{k+1}\left( \dfrac{1}{z^k}-\beta\right)}\\
(\text{by using the}& \text{ change of variables }\zeta= z^{-k})\\
&=\int\limits_{|\zeta|=r^{-k}} \frac{\zeta d\zeta}{\zeta-\beta}= 2\pi i\beta.
\end{split}
$$ 
The last equality can be obtained by using the Cauchy integral formula, or the residue theorem, whichever you like more, since if $r$ is small enough such that the disc $B(r^{-k})=\big\{z\in\Bbb C\:|\:|z|<r^{-k}\big\}$ includes the point $\beta$ in its interior you can apply both procedures.
A: Hint:  Use the limit formula for higher order poles to compute the residue at $0$.
Better hint:  Use the Laurent series. 
