Check my answer - complex analysis, using residue and rouche's theorem I was asked the following questions and I am unsure of my solutions, any advice would be appreciated, maybe there is a better way of doing this.
Question:
We are given $f(z)=2z-\sinh (z)$ defined on the circle $|z|=1$. Our goal is to calculate 
$\int_{|z|=1}\frac{1}{f(z)}dz$
What I did:
The first thing we need to take care of is finding how many roots does $f(z)$ have in the given domain. If we can prove that $|2z-f(z)| \leq |2z|$ in the circle then from Rouche's theorem they have the same number of roots, and we know that $2z$ has a root only at $z=0$.
Here I also used a known inequality: $|\sinh(z)| \leq |z| \cosh (|z|)$:
$|2z-f(z)|=|\sinh(z)|\leq |z|\cosh(|z|=\cosh(1)=\frac{e+e^{-1}}{2}<2=|2z|$ in the circle.
We proved the inequality, so from rouche's theorem, $f(z)$ has only $1$ root in the circle, and we can see that $f(0)=0$ is the only root, and $f'(0) \neq 0$ so the "rank" of the root is $1$.
We can use taylor series to separate $f$ to a function with roots multiplied by a function without roots (in the circle ofcourse):
$$f(z)=2z-\sinh(z)=2z-\sum_{n=0}^{\infty}\frac{z^{2n+1}}{(2n+1)!}=z(2-\sum_{n=0}^{\infty}\frac{z^{2n}}{(2n+1)!})=z(1-\sum_{n=1}^{\infty}\frac{z^{2n}}{(2n+1)!})$$
Let $\psi(z)=1-\sum_{n=1}^{\infty}\frac{z^{2n}}{(2n+1)!}$, so overall we have $f(z)=z\psi(z)$ Notice that $\psi(0)=1$ (that's the residue) and that $\psi$ has no roots in $|z|=1$ (otherwise, $f(z)$ would have had more roots, and we already showed that $0$ is its only root).
So to calculate the integral:
$$\int_{|z|=1} \frac{1}{f(z)}dz=\int_{|z|=1} \frac{1}{z}\frac{1}{\psi(z)}dz=\frac{2i\pi}{\psi(0)}=2i\pi$$ from residue theorem.
Is this result correct?

As @Mhenni Benghorbal said, you can do this a simpler way; that is, you don't need to use Rouche's Theorem. To find the poles, we only need to determine when
$$
2z - \sinh(z) = 0.
$$
Using the series representation of $\sinh(z) = \sum_{n=0}^{\infty}\frac{z^{2n+1}}{(2n+1)!}$, we have
$$
2z - \sum_{n=0}^{\infty}\frac{z^{2n+1}}{(2n+1)!} = z\biggl[2 - \sum_{n=0}^{\infty}\frac{z^{2n}}{(2n+1)!}\biggr] = z\biggl[2 - 1 - \sum_{n=1}^{\infty}\frac{z^{2n}}{(2n+1)!}\biggr]
$$
As you have found, the only simple pole occurs when $z=0$ which is inside $\lvert z\rvert = 1$. By the residue theorem, we have
\begin{align}
\oint_{\gamma}\frac{dz}{2z-\sinh(z)}&=2\pi i\sum\text{Res}\\ 
&= 2\pi i\lim_{z\to 0}(z-0)\frac{1}{z\Bigl[1 - \sum_{n=1}^{\infty}\frac{z^{2n}}{(2n+1)!}\Bigr]}\\
&=\lim_{z\to 0}\frac{2\pi i}{1 - \sum_{n=1}^{\infty}\frac{z^{2n}}{(2n+1)!}}\\
&= 2\pi i
\end{align}
