Four Ways of Rouché's Theorem, how are they equivalent?

Question

In Conway, they expose like that the theorem

$f$ and $g$ are meromorphic in a neighborhood of $B[a,r]$ with no zeros or poles on the circle $\gamma =z: |z-a|=r$. If $Z_{f},Z_{g}$ $(P_{f},P_{g})$ are the zeros (poles), of $f$ and $g$ inside $\gamma$ counted according their multiplicies and if

$|f(z)+g(z)|<|f(z)|+|g(z)|$ on $\gamma$ then $Z_{f}-P_{f}=Z_{g}-P_{g}$.

On Spigel we have the following

If $f$ and $g$ are analytic on a set bounded by a simple closed curve $C$, if $|f(z)|<|g(z)|$ over $C$, then $f+g$ and $f$ has the same number of zeros.

how can these two be equivalent? looking for the answer i found another versions of the theorem, some use the Spigel version and say that we can conclude too that $|f+g|<|f|$. I found too a version of the Conway but with

$|f(z)-g(z)|<|f(z)|+|g(z)|$ intead of $|f(z)+g(z)|<|f(z)|+|g(z)|$ , how can these all be equivalent? i can't see that

Answer 1

As Berci said in a commentary, the equivalence between the $|f + g| < |f| + |g|$ and $|f - g| < |f| + |g|$ versions is obvious, but...

According to Wikipedia and The symmetric versions of Rouché's theorem via $\partial$-calculus, the Conway (symmetric) version is stronger than the Spigel (asymmetric) version. Quotes from the links:

A stronger version of Rouché's theorem was already known to Theodor Estermann by 1962. It states: let $K\subset G$ be a bounded region with continuous boundary $\partial K$. Two holomorphic functions $f,g\in {\mathcal{H}}(G)$ have the same number of roots (counting multiplicity) in $K$, if the strict inequality $$|f(z)-g(z)|<|f(z)|+|g(z)|\qquad\left(z\in \partial K\right)$$ holds on the boundary $\partial K$. The original version of Rouché's theorem then follows from this symmetric version applied to the functions $f + g,f$ together with the observation that $f(z)+g(z)\neq 0$ when $z$ is on $\partial K$.

Section 1 we unveil a class of examples that show that the symmetric version of Rouchés theorem actually is a stronger result than the non-symmetric one.