Non-existence of a holomorphic map on the disc

Question

Is there a function $f$ that is holomorphic on the punctured unit disc such that $f^{\prime}$ has a pole of order $1$? My answer is: No.

The following post has an answer to my question.

Show that there is no function $f$ that is analytic in punctured unit disc and $f'$ has a simple pole at $0$.

Here's my reasoning:

Let $f$ be such a function. Then $f^{\prime} (z) = \frac{g(z)}{z}$ for some holomorphic function $g$ on the pictured disc and $g(0)$ is NOT equal to $0.$ Now, taking integral both sides along the boundary of a small disc that is contained in the unit disc, we have $0 = g(0),$ which is a contradiction. Is this a valid argument? Any help would be appreciated.

Answer 1

Alternative solution: if the Laurent series of $f$ is $$f(z) = \sum_{n=-\infty}^{\infty}a_n z^n,$$ then the Laurent series of $f'$ will be $$ f'(z) = \sum_{n=1}^{\infty}na_n z^{n-1} + \sum_{n=-\infty}^{-1}na_n z^{n-1} = \sum_{n=0}^{\infty}(n + 1)a_{n+1}z^n + \sum_{n=-\infty}^{-2}(n + 1)a_{n+1}z^n. $$ Also, using integration by parts ("Integration by parts" in complex analysis): $$ 2\pi i\,{\rm res}(f',0) = \int_\gamma f'(z)1\,dz = - \int_\gamma f(z)0\,dz = - \int_\gamma 0\,dz = 0. $$