Show that $\frac{f'(z)}{f(z)}$ has a pole if and only if $f(z)$ has a zero or a pole. In our analytic number theory course, our instructor told us that $\zeta(s)$ has a zero or a pole at $z_0$ if and only if $\frac{\zeta'(s)}{\zeta(s)}$ has a pole at $z_0$. In fact, this is true for any meromorphic functions. One direction is easy to see. If $f(z)$ has a pole or zero at $z_0$ then $\frac{f'(z)}{f(z)}=\pm \frac{n}{z-z_0}+G(z)$ for some non vanishing holomorphic function $G$ at $z_0$ (where $n$ is the order of the zero or the pole). From here we can see that $\frac{f'(z)}{f(z)}$ has a simple pole at $z_0$. But the other direction does not look so obvious. Because if $\frac{f'(z)}{f(z)}$ has a pole at $z_0$ of order $n$ then we have $$\frac{f'(z)}{f(z)}=\frac{c_{-n}}{(z-z_0)^n}+\frac{c_{-(n-1)}}{(z-z_0)^{n-1}}+\ldots+\frac{c_{-1}}{z-z_0}+\sum\limits_{k=0}^\infty c_k(z-z_0)^k,$$ from here we cannot say immediately anything. I want a little help to proceed from here onwards. I think I am making some mistake as I did like this, $\frac{f'(z)}{f(z)}$ has a pole at $z_0$ if and only if $\frac{f(z)}{f'(z)}$ has a zero at $z_0$ if and only if $f(z)$ has a zero or $|f'(z)|\to \infty$ as $z\to z_0$. But from the second case I don't see any natural way.
 A: By definition, a function is meromorphic in a domain $D$ if

*

*There is a (possibly empty) set $P \subset D$ which is isolated in $D$, i.e. $P$ has no limit points in $D$,

*$f$ is holomorphic in $D \setminus P$, and

*$f$ has a pole at each point in $P$.

Now set $P' = P \cup Z$ where $Z = \{ z \in D \mid f(z) = 0 \}$ is the set of zeros of $f$. Then

*

*$P'$ is isolated in $D$,

*$f'/f$ is holomorphic in $D \setminus P'$ as the quotient of two holomorphic functions with a non-zero denominator.

*$f'/f$ has – as you calculated – a simple pole at each point $z_0 \in P'$.

This shows that $f'/f$ is meromorphic in $D$ with (simple) poles exactly at the points in $P'$, that is where $f$ has a pole or a zero.
In short: If $f(z_0) \ne 0, \infty$ then $f'/f$ is holomorphic in a neighbourhood of $z_0$, and therefore does not have a pole at $z_0$.
A: If $f$ has a zero or a pole at $z=z_0$, then $f$ is of the form $f(z)=(z-z_0)^kg(z)$, where $k\in\mathbb Z\setminus\{0\}$ and $g$ analytic in a neighborhood of $z_0$. In such case
$$
\frac{f'(z)}{f(z)}=\frac{k}{z-z_0}+\frac{g'(z)}{g(z)},
$$
and hence $f'/f$ has a simple pole at $z=z_0$.
Nevertheless, if $f(z)=\exp(z^{-n})$, then $f$ has an essential singularity at $z=0$ and
$$
\frac{f'(z)}{f(z)}=-\frac{n}{z^{n+1}},
$$
which is a pole of order $n+1$ at $z=0$.
In general, if $f'/f$ is expressed in a neighborhood of $z=z_0$ as
$$
\frac{f'(z)}{f(z)}=\sum_{k=-\infty}^{-2}c_k(z-z_0)^k
+\frac{\ell}{z-z_0}+g(z)
$$
where $\ell$ is an integer and $g$ is analytic in a neighborhood of $z=z_0$, then $f$ is of the form
$$
f(z)=h(z)(z-z_0)^\ell \exp\left(\sum_{k=-\infty}^{-2}\frac{c_k}{k+1}(z-z_0)^{k+1}\right),
$$
where $h$ is analytic and $h(z_0)\ne 0$. If $\ell$ is not an integer, then there is no such $f$ - Only a Riemann surface! For example, if $f'/f=1/(2z)$, then $f$ is the Riemann surface corresponding to the $\sqrt{z}$.
A: Let be any holomorphic function on $\Omega \setminus \{z_0\}$. Then $f'/f$ has a pole with order $1$ at $z_0$ with integer residue if and only if $f$ has a zero or a pole at $z_0$.
You proved the « if » part. Here is the « only if » part. I take $z_0=
0$ for simplicity.
If $f'/f$ has a simple pole at $0$ with residue $r \in \mathbb{Z} \setminus \{0\}$, then the function $g$ defined by
$$g(z) = \frac{f'(z)}{f(z)}-\frac{r}{z}$$
has a false singularity at $0$, so it has an holomorphic extension on some ball $B(0,\rho)$, and has some primitive $G$ on the same ball.
\begin{eqnarray*}
\frac{d}{dz}\Big(\frac{f(z)e^{-G(z)}}{z^r}\Big) 
&=& \frac{z^r(f'(z)-f(z)g(z))-rz^{r-1}f(z)}{z^{2r}}e^{-G(z)} \\
&=& \Big(f'(z)-f(z)g(z)-(r/z)f(z)\Big)\frac{e^{-G(z)}}{z^r} \\
&=& 0
\end{eqnarray*}
As a result $f(z)e^{-G(z)}/z^r$ is a non-null constant on $B(0,\rho) \setminus \{0\}$, so $f$ has a zero if $r>0$, a pole if $r<0$, with order $|r|$.
