Finding The Zeros Of Sum And Product Of Functions

Question

Let $z_0$ be zero of order $n$ of $f(z)$ and zero of order $m$ of $g(z)$ what can we say about the order of $z_0$ of $f(z)+g(z)$ and $f(z)\cdot g(z)$

Let assume WLOG that $n<m$ so for $[f(z_0)+g(z_0)]^{(m)}=f(z_0)^{(m)}+g(z_0)^{(m)}$ so $z_0$ is of order $m$

for $[f(z_0)\cdot g(z_0)]^{(m+n)}=f(z_0)^{(m+n)}g(z_0)+g(z_0)^{(m+n)}f(z_0)$ so the order is $m+n$

Answer 1

If $z_0$ is a zero of order $n$ of $f(z)$ then $z_0$ is repeated $n$ times as a root of $f(z)$ i.e., $$f(z)=(z-z_0)^nf_1(z)$$ where $f_1(z_0)\neq 0$. Similarly if $z_0$ is a zero of order $m$ of $g(z)$ then $$g(z)=(z-z_0)^mg_1(z)$$ where $g_1(z_0)\neq 0$. So in $$f(z)+g(z)=(z-z_0)^nf_1(z)+(z-z_0)^mg_1(z),$$ $z_0$ has order $\min(n, m)$. Similar solution for $f(z)g(z)$ shows that order of $z_0$ is $m+n$.

Answer 2

The first should be $n$ not $m$, e.g. $f(x):= x$, $g(x):=x^2$.

The second is correct.

An easy way to see this is to consider the local Taylor series of $f$ and $g$ about $z_0$ (assuming they're holomorphic in an open region containing $z_0$, which from your question it appears is the case).