Definitions of a pole :
- The isolated singularity $z_0$ is a pole, of order $k$ $\iff$ $a_{−k}≠0$ and $a_n=0$ for all $n<−k$.
- $f(z)$ has a pole at $z_0$ if $z_0$ is an isolated singularity and $\lim_{z\to z_0}|f(z)|=\infty$
Is anyone could explain to me in details why both definition are equivalent?
It simplifies the notation to translation the entire plane so that $z_0$ is at $0$. You have forgotten to mention a very important property of $f$ that I am not going to provide for you.
Your description in 1 is too abbreviated. The singularity of the [something goes here] function, $f$, at $0$ is a pole of order $k$ if and only if the Laurent expansion of $f$ centered at $0$ has nonzero $-k^\text{th}$ coefficient and zero $n\text{th}$ coefficients for $n<-k$. This means the Laurent expansion of $z^k f(k)$ is a power series with nonzero constant term. This is enough to show that $0$ is isolated from the other poles of $f$ and that the limit of interest is of $a_{-k} \, z^{-k} \rightarrow \infty$ (but you need to say a bit more than I have).
To go the other way, realize that the given limit implies $1/|f| = \left| \frac{1}{f} \right| \rightarrow 0$. Because $f$ is [something] and we have our other hypothesis, we can write $1/f$ as a power series with leading coefficent zero. Equivalently, $1/f(z) = z^k \hat{f}(z)$ for some positive integer $k$ and some $\hat{f}$ having nonzero constant coefficient. Then by division, you get the statement you wrote as 1.