$|f(z) - g(z)| < |f(z)|$ vs $|f(z) - g(z)| < |g(z)|$ in Rouche's theorem?

Question

I was reading Marsden's Basic Complex Analysis and noticed that Rouche's Theorem was formulated as requiring $|f(z) - g(z)| < |f(z)|$ (for all $z$ on closed $\gamma$, etc.), and the statement was that $f(z)$ and $g(z)$ have the same amount of zeros and poles (for simple $\gamma$).

However, I've previously seen the requirement presented as $|f(z) - g(z)| < |g(z)|$ (with the same statement) - and indeed it seems like Marsden uses this version of the theorem in Example 6.2.12 without any justification. So, my question is: how are these two formulations equivalent? Am I missing some simple triangle inequality manipulation?

Thanks in advance!

Answer 1

What's the difference? Assume that the first formulation holds. Now if you have $|f(z)-g(z)|<|g(z)|$, then$$|g(z)-f(z)|=|f(z)-g(z)|<|g(z)|,$$and therefore $f$ and $g$ have the same number of zeros.

Answer 2

Those statements are equivalent, it is just a matter of exchanging the roles of $f$ and $g$.

There is also a “symmetric” version of Rouché's theorem: It suffices to require that $$ |f(z) - g(z)| < |f(z)| + |g(z)| $$ on the boundary of the domain.