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- #1

Let $f$ be analytic on the punctured disc $\mathbb{D} - \{0\}$.

$$

f(z) = \frac{g(z)}{h(z)}

$$

such that $h(z)\neq 0$.

Then

$$

f'(z) = \frac{g'(z)h(z) - g(z)h'(z)}{(h(z))^2}

$$

So $f'(z)$ has only poles of order 2. Therefore, $f'$ can't have a simple pole at 0.

How is this?