What is meromorphic differentials like on Riemann Sphere? [closed]

Question

There is a proposition that every meromorphic differential on Riemann Sphere (or $\mathbb{P}^1 = \mathbb{C} \cup \{ \infty \}$) can be written as $f dz$ where $f$ is a meromorphic function on $\mathbb{P}^1$, and that $f dz$ is meromorphic if and only if $f$ is holomorphic on $\mathbb{C}$ and $z^2 f(z)$ tends to a finite limit as $z \rightarrow \infty$.

This is Exercise 6.3. on Page 178 of Frances Kirwan's Complex algebraic curves. I know how to prove the former half of the proposition (every meromorphic differential can be written as $f dz$), but I cannot find a proof of the second half. In fact, I'm not sure whether the propostion is true, because I seem to find a meromorphic differential which is not holomorphic on $\mathbb{C}$ (e.g. $1/z$ $dz$).

Does anyone can answer my question?

Answer 1

The second half of the statment is not true, but if we replace "meromorphic" by holomorphic, it becomes a true proposition:

$f(z)dz$ is holomorphic if and only if $f(z)$ is holomorphic on $\mathbb{C}$ and $z^2f(z)$ tends to a finite limit as $z \rightarrow \infty$.

This is easy to prove because $$f(z_1)dz_1=-f(1/z_2)/z_2^2 dz_2$$ when changing local coordinate, and $$\lim\limits_{z_1\to\infty}f(z_1)z_1^2 = \lim\limits_{z_2\to 0}f(1/z_2)/z_2^2$$

In fact, there is no non-zero holomorphic function $f(z)$ that satisfies the condition above, so there is no non-zero holomorphic differential on $\mathbb{P}^1$.