I am trying to find how many zeros of the given function f are located in the unit disk $|z|<1$:

$f(z) = z - \phi(z)$, where $\phi(z)$ is analytic and $|\phi(z)|<1$ whenever $|z|<1$.

Using Rouche's Theorem, which states that if both $f$ and $g$ are holomorphic inside and on some closed contour $C$, and $|f(z)| > |g(z)|$ for all $z \in C$, then $f$ and $f+g$ have the same number of zeros inside of $C$.

I started by letting $f(z) = z$ and $g(z) = \phi(z)$. My problem is how to figure the answer out using Rouche's Theorem. Can someone assist?


1 Answer 1


Let $f_1:=\textrm{id}_{\mathbb{C}}$ and $f_2:=f$. One has that $f_1,f_2$ are holomorphic functions in a open neighborhood of $\mathbb{D}:=\{z\in\mathbb{C}\textrm{ s.t. }|z|<1\}$ the unit disk of $\mathbb{C}$. Moreover, one has: $$\forall z\in\partial\mathbb{D},|f_1(z)-f_2(z)|=|\Phi(z)|<1=|f_1(z)|.$$ Using Rouché's theorem, one has that $f_1$ and $f_2$ have the same number of $0$ in $\mathbb{D}$ counting with multiplicity. Since, $f_1$ has only a simple zero in $\mathbb{D}$, $f_2$ has only a simple zero in $\mathbb{D}$. Finally, $f=\textrm{id}-\Phi$ has one zero in $\mathbb{D}$.

  • $\begingroup$ could id mean an identity function here? $\endgroup$
    – J.R.
    Dec 1, 2015 at 23:44
  • $\begingroup$ Yes, it is. One has: $\textrm{id}_{\mathbb{C}}:z\in\mathbb{C}\mapsto z$. $\endgroup$
    – C. Falcon
    Dec 1, 2015 at 23:45
  • $\begingroup$ can we also use the theorem I stated in the problem above where $|f(z)|>|g(z)|$ and define $f(z)=z$ and $g(z)=\phi(z)$? Would that also help? $\endgroup$
    – J.R.
    Dec 2, 2015 at 1:22
  • $\begingroup$ I used the Rouché's theorem I know, but if you let $f_1(z)=z$ and $f_2(z)=-\Phi(z)$, since for all $z\in\partial\mathbb{D}$ $|f_1(z)|=1>|f_2(z)|$, your Rouché's theorem would do the job. $\endgroup$
    – C. Falcon
    Dec 2, 2015 at 1:25
  • 1
    $\begingroup$ One has $\partial\mathbb{D}=\{z\in\mathbb{C}\textrm{ s.t. }|z|=1\}$, $|f_1(z)|=|z|$ and your hypothesis is $|f_2(z)|<1$. $\endgroup$
    – C. Falcon
    Dec 2, 2015 at 1:30

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