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 $n<−k$.
- $f$ has a pole of order $m$ at $z_0$ if there exists a function $g$ that is holomorphic in a neighbourhood of $z_0$ such that $ (z-z_0)^m f(z) = g(z) $ with $g(z_0) \not= 0$
Is anyone could explain to me in details why both definition are equivalent?
Your definition (2) is incorrect. It must also include $g(z_0) \neq 0$. Let $f$ be a complex function with pole of order $m$ at $0$. Let $\sum_{j=-\infty}^\infty a_j z^j$ be its Laurent series expansion at $0$ with inner radius $0$. I made the centre of the power series to be $0$ for convenience.
Definition (1) $\implies$ Definition (2): If Definition (1) holds with order $m$, we have $$ f(z) = \sum_{j=-\infty}^\infty a_j z^j = \sum_{j=-m}^\infty a_j z^j $$ We can define $g(z) = \sum_{j=0}^\infty a_{j-m} z^j$. Evidently $z^m f(z) = g(z)$, and since $g(z)$ is a power series, it is holomorphic within its radius of convergence, which is not zero since the outer radius of $f$ is not zero.
Definition (2) $\implies$ Definition (1): Let $g(z) = \sum_{j=0}^\infty a_j z^j$. Then $$ f(z) = \frac{1}{z^m} g(z) = \sum_{j=-m}^\infty a_{j+m} z^j $$ Since $g(0) \neq 0$, we have $a_0 \neq 0$, and all terms below $j = -m$ are zero.