# Analytic Functions with Isolated Zeros of Order k

• POTW
• Euge
In summary, an analytic function with isolated zeros of order k is a complex-valued function that is differentiable at every point in its domain, except at a finite number of isolated points where it has a zero of order k. This is different from regular zeros, which occur continuously throughout the domain. The order of a zero directly affects the behavior of an analytic function, with higher order zeros resulting in a more "rounded" function near that point. An analytic function can only have a finite number of isolated zeros, and they are used in various areas of mathematics and science for understanding functions and solving equations and modeling physical systems.
Euge
Gold Member
MHB
POTW Director
Suppose ##f## is analytic in an open set ##\Omega \subset \mathbb{C}##. Let ##z_0\in \mathbb{C}## and ##r > 0## such that the closed disk ##\mathbb{D}_r(z_0) \subset \Omega##. If ##f## has a zero of order ##k## at ##z = z_0## and no other zeros inside ##\mathbb{D}_r(z_0)##, show that there an open disk ##D## centered at the origin such that for all ##\alpha\in D##, ##f## takes on the value ##\alpha## exactly ##k## times, counting multiplicity.

Greg Bernhardt and topsquark
Is the final disk really centered at 0 and not ##z_0##?

We wish to show that

\begin{align*}
f (z) - \alpha = 0
\end{align*}

happens exactly ##k## times, counting multiplicity inside ##\mathbb{D}_r (z_0)## for all ##\alpha## such that ##|\alpha| < R## for some ##R >0##.

Rouche's theorem:

"Let ##f## and ##g## be analytic in a simply connected domain ##U \in \mathbb{C}##. Let ##C## be a simple closed contour in ##U##. If ##|f(z)| > |g(z)|## for every ##z## on ##C##, then the functions ##f(z)## and ##f(z) + g(z)## have the same number of zeros, counting multiplicities, inside ##C##."

Take ##C## to be the circle centred at ##z_0## with radius ##r##. Note ##|f(z)| \not= 0## on ##C##. Let ##R = \min_C |f(z)|## and define an open disk ##D## about the origin of radius ##R##. For ##\alpha \in D##, write ##g(z) = - \alpha##. Then ##|f(z)| > |g(z)|## for every ##z## on ##C##. By Rouche's theorem ##f(z)## and ##f(z) + g(z)## have the same number of zeros, counting multiplicities, inside ##C##. Therefore, ##f## takes on the value ##\alpha## exactly ##k## times, counting multiplicity, inside ##\mathbb{D}_r (z_0)##.

Last edited:

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