Find all functions holomorphic in a punctured unit disk such that $zf' + 2f = 0$.

Question

I think all such functions are of the form $c/z^2$ where $c$ is a constant. I'd like to expand it as a taylor series about $0$ and then compare coefficients of $f$ and $f'$ but I don't know whether it is holomorphic at $0$.

Answer 1

Note first that $f(z) \ne 0$, if $z \ne 0$ since otherwise if $m \ge 1$ is the order of $f$ at such a $z_o$, $f'$ has order $m-1$ and we get a contradiction after simplifying $(z-z_0)^{m-1}$.

Hence we get $\frac{f'}{f}=-\frac{2}{z}$ on $D^{*}$, so on any domain that consists of $D^{*}$ from which we cut a segment going from the origin to the boundary (which makes it simply connected), $f(z)=\frac{c}{z^2}$ with $c$ apriori depending on the segment removed; however since any two such domains have non-empty open intersection, the constant must be the same by the identity principle, so we are done and indeed $f(z)= \frac{c}{z^2}$ on the whole $D^{*}$

Answer 2

Consider the function $g(z) = z^2 f(z)$. $g$ is holomorphic in the punctured disk, with $$ \color{red}{g'(z)} = 2zf(z) + z^2f'(z) = z(2f(z) + zf'(z)) \color{red}{= 0 \, .} $$ It follows that $g$ is constant: $$ g(z) = c \implies f(z) = \frac{c}{z^2} \, . $$