Show $\frac {1}{2\pi i}\int_\gamma\frac {f'(z)}{f(z)}$ is sum of poles and zeroes times their order

Question

Let $f: B(a,r) \to {\Bbb C} \cup \{\infty\}$ be a meromorphic function. For $p$ a zero or pole of $f$, let $\mu(f,p)$ denote its order. Let $\gamma$ be a closed curve in $B(a,r)$. Then

$$\frac {1}{2 \pi i}\int_\gamma \frac {f'(z)}{f(z)} dz = \sum_{p:f(p) = 0} \eta (\gamma , p) \mu (f, p ) - \sum_{p:f(p) = \infty} \eta (\gamma , p) \mu (f, p ) $$

My Attempt: Let $B(a, r')$ be a ball such that $r > r'$. We write

$$f(z) = \Pi_{p:f(p) = 0}(z-p)^{\mu(f,p)}\Pi_{p:f(p) = \infty}(z-p)^{-\mu(f,p)}g(z) $$

such that $g$ analytic everywhere $g$ is nonzero on $B(a,r).$ This implies that $$\frac {f'(z)}{f(z)} =\frac {\frac {d}{dz}(\Pi_{p:f(p) = 0}(z-p)^{\mu(f,p)}\Pi_{p:f(p) = \infty}(z-p)^{-\mu(f,p)}g(z))}{\Pi_{p:f(p) = 0}(z-p)^{\mu(f,p)}\Pi_{p:f(p) = \infty}(z-p)^{-\mu(f,p)}g(z)} $$

Now taking the integral of both sides after some step gives you: *

$$\int_\gamma \frac {f'(z)}{f(z)} dz = 2 \pi i \left ( \sum_{p:f(p) = 0} \eta (\gamma , p) \mu (f, p ) - \sum_{p:f(p) = \infty} \eta (\gamma , p) \mu (f, p ) \right )$$

My question how to do I get to the last line from the second to last. Is it some kind product rule cancellation? Or am I missing it entirely.

Answer 1

Have you met the Argument Principle? This might help! https://en.wikipedia.org/wiki/Argument_principle

Just to write what you've said a little simpler:

"Let $f$ be meromorphic on domain $\Omega$ open.

Let $C \subseteq \Omega $ be a Toy contour (eg triangle circular etc).

Allow $f$ to have a finite number of poles in $C$'s interior but not on $C$.

Let $N$:=number of zeroes of $f$ inside C (including multiplicity)

Let $P$:=number of poles of $f$ inside C (including multiplicity)

then $$\int_C \frac{f'(z)}{f(z)}dz = N-P$$

The Proof is quite staightforward

Proof: We use the Residue theorem!

Since $\frac{f'}{f}$ is holomorphic in $C, \space \forall z \notin \{ N \} \cup \{P\}$

We let $\omega_1, ... ,\omega_s $ be zeroes of $f$

and let $z_1,...,z_k$ be poles of $f$

So

$$\frac{1}{2\pi i}\int_C \frac{f'(z)}{f(z)}dz = \sum \text{Residues} $$

Before we continue we state the following definition :

(Arg Principle) If $f$ holomorphic with a zero of Order N at $z_0$ then we can write

$f(z)=(z-z_0)^N g(z)$ where $\quad g(z)$ holomorphic on $B(z_0,\delta)$ and $g(z)\neq 0 \space \forall z \in B(z_0,\delta)$

We see this because $$\frac{f'(z)}{f(z)} = \frac{N(z-z_0)^{N-1}g(z) +(z-z_0)^N g'(z)}{(z-z_0)^N g(z)} $$ But this reduces to $$\frac{f'(z)}{f(z)} = \frac{N}{z-z_0} + \frac{g'(z)}{g(z)} $$ Since $g \neq 0 \Longrightarrow \frac{g'}{g}$ is holomorphic also

Thus we see $z_0$ is a First order pole of $\frac{f'(z)}{f(z)}$ with residue $N$.

We already know how to deal with poles in the residue theorem so back to our integral

$$\frac{1}{2\pi i}\int_C \frac{f'(z)}{f(z)}dz = \sum \text{Residues} $$

But by the above we can write

$$ \sum \text{Residues} = \sum \text{Multiplicities of the zeroes of $f$} - \sum \text{Multiplicities of the poles of $f$} $$

$$=N-P $$

$$\Longrightarrow \frac{1}{2\pi i}\int_C \frac{f'(z)}{f(z)}dz = N-P$$

Hope this helps!