# bounded domain, meromorphic functions

• Jan 17th 2010, 12:31 PM
hilbertcube182
bounded domain, meromorphic functions
Let $D$ be a bounded domain, and let $f(z)$ and $h(z)$ be meromorphic functions on $D$ that extend to be analytic on $\partial D$. Suppose that $|h(z)|< |f(z)|$ on $\partial D$. Show by example that $f(z)$ and $f(z)+h(z)$ can have different numbers of zeros on $D$. What can be said about $f(z)$ and $f(z)+h(z)$? Prove your assertion.

I do not see what example I can use here. I think what we can say is that they have the same number of zeros minus poles. I just don't see an example here. Thanks.
• Jan 19th 2010, 05:40 PM
Jose27
Quote:

Originally Posted by hilbertcube182
Let $D$ be a bounded domain, and let $f(z)$ and $h(z)$ be meromorphic functions on $D$ that extend to be analytic on $\partial D$. Suppose that $|h(z)|< |f(z)|$ on $\partial D$. Show by example that $f(z)$ and $f(z)+h(z)$ can have different numbers of zeros on $D$. What can be said about $f(z)$ and $f(z)+h(z)$? Prove your assertion.

I do not see what example I can use here. I think what we can say is that they have the same number of zeros minus poles. I just don't see an example here. Thanks.

How about $f(z)=\frac{2}{z^2}$ and $g(z)=\frac{1/2-z}{z^3}$ with $D=\{z\in \mathbb{C} : |z|< 1 \}$