# An Alternate Formulation of Rouche's Theorem?

The general form of Rouche's Theorem states:

For any two complex-valued functions $$f$$ and $$g$$ holomorphic inside some region $$K$$ with closed contour $$\partial K$$ if $$|g(z)| < |f(z)|$$ on $$\partial K$$, then $$f$$ and $$f + g$$ have the same number of zeros inside $$K$$, where each zero is counted as many times as its multiplicity. This theorem assumes that the contour $$\partial K$$ is simple, that is, without self-intersections.

Is the following inequality a correct formulation of Rouche's theorem as well?

If $$|f + g| < |f| + |g|$$ for all $$z \in \partial K$$, where $$f$$ and $$g$$ are analytic functions on $$K$$, then $$f$$ and $$g$$ have the same number of zeros inside $$K$$.

I am trying to prove this statement: $$|f + g| < |f| + |g| \Rightarrow$$ same number of zeros for $$f$$ and $$g$$. But my gut feeling is telling me that it is different from the statement in Rouche's theorem, and it may not be correct to begin with. I have tried to come up with a counter-example, but could not make one that always satisfices the strict inequality.

If it is not correct, please give a small explicit counter-example, and if it is correct please point out why it's true and how can it be shown that it is equivalent to its general form (i.e. the statement from Wikipedia)

Update

As discussed in the comment section, this is a valid formulation of Rouche's theorem too. It is called its symmetric version. Can someone please provide a simple, self-contained argument as to why the symmetric version implies the general version? (by general, I mean the quoted format of the theorem from Wikipedia. I am being informal by calling it the general form, but I hope you understand what I mean).

• just look up the symmetric version of Rouche en.wikipedia.org/wiki/Rouch%C3%A9%27s_theorem#Symmetric_version Commented Mar 5, 2023 at 23:04
• I see! Can you please provide a simple argument as to why the symmetric version implies the general form? The article in Wikipedia is skipping over details in my opinion.
– Josh
Commented Mar 6, 2023 at 1:20
• Because you have a homotopy $h_t=tf-(1-t)g, 0 \le t \le 1$ from $f$ to $-g$ which doesn't vanish on the unit circle (by the hypothesis since if $tf(w)=(1-t)g(w)$ then $|(f+g)(w)|=|f(w)|+g(w)|)$, so by the homotopy invariance of the integral $\int h_t'/h_t$ which counts the zeroes of $h_t$ in the unit disc we get that $h_t$ have same number of zeroes in the unit disc Commented Mar 6, 2023 at 1:38

Proof: If $$|f|<|g|$$ in $$\partial K$$, then: $$|(f+g)-g|<|g|\leq|g|+|f+g|$$ Hence, by the symetric Rouche, $$f+g$$ and $$g$$ have the same number of zeros. $$\square$$
If $$|f|<|g|$$ on the border $$\partial K$$, then $$|f/g|<1$$ and $$|(f+g)/g-1|<1$$. Because we are in a circle centered at $$1$$ failing to meet the origin, the logarithm is well defined and hence:
$$0=\int_\gamma \frac{d}{dz}\log\left(\frac{f+g}{g}\right) dz=\int_\gamma \frac{g}{f+g}\frac{(f+g)'g+g'(f+g)}{g^2}dz=\int_\gamma \frac{(f+g)'}{f+g} +\frac{g'}{g}dz$$
Hence by the argument principle, if we have meromorphic functions, $$Z_{f+g}-P_{f+g}=Z_g-P_{p}$$. In the particular case $$f$$ and $$g$$ are holomorphic it follows that the number of zeros coincide.