# Thread: analytic functions, bounded domain

1. ## analytic functions, bounded domain

Let $f(z)$ and $g(z)$ be analytic functions on the bounded domain $D$ that extend continuously to $\partial D$ and satisfy $|f(z)+g(z)|<|f(z)|+|g(z)|$ on $\partial D$. Show that $f(z)$ and $g(z)$ have the same number of zeros in D, counting multiplicity.

The hint here says that neither $f(z)$ nor $g(z)$ has zeros on $\partial D$, so each has at most finitely many zeros in $D$. Estimate shows $\frac{f(z)}{g(z)} \not \in (-\infty, 0]$ for z near $\partial D$, so $\text{Log}(f(z)/g(z))$ is continuously defined near $\partial D$. Increase in $\text{arg} f(z)/g(z)$ around any closed path near $\partial D$ is zero.

We have covered Rouche's Theorem too. I do not see how to prove this right now. I need some pointers on where to go. Thanks.

2. Originally Posted by poincare4223
Let $f(z)$ and $g(z)$ be analytic functions on the bounded domain $D$ that extend continuously to $\partial D$ and satisfy $|f(z)+g(z)|<|f(z)|+|g(z)|$ on $\partial D$. Show that $f(z)$ and $g(z)$ have the same number of zeros in D, counting multiplicity.

The hint here says that neither $f(z)$ nor $g(z)$ has zeros on $\partial D$, so each has at most finitely many zeros in $D$. Estimate shows $\frac{f(z)}{g(z)} \not \in (-\infty, 0]$ for z near $\partial D$, so $\text{Log}(f(z)/g(z))$ is continuously defined near $\partial D$. Increase in $\text{arg} f(z)/g(z)$ around any closed path near $\partial D$ is zero.

We have covered Rouche's Theorem too. I do not see how to prove this right now. I need some pointers on where to go. Thanks.
Your hint gives that f and g have a finite number of $0$'s inside $D$, so pick a curve $C$ that encloses them all (simple and closed) such that $d(c,d)$ (where $c\in C$ and $d\in \partial D$ ) ensures that $h(z)=\frac{f(z)}{g(z)} \notin [0,\infty )$ (not the negative interval since what we have is $|h(z)+1|<|h(z)|+1$ ) and we define $w(z)=\ln (h(z))$ in this $C$ (maybe what could cause problems in a rigorous proof is to build such a $C$ because $\partial D$ can be really complicated, but lets assume you can, or consider $\partial D$ to be a simple closed curve) then $w'(z)= \frac{f'(z)}{f(z)} -\frac{g'(z)}{g(z)}$ and then $\int_{C} w'=0$ but by the argument principle this integral is equal to $Z_f-Z_g$