Number of zeros using Rouche's Theorem

Question

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?

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 $\mathbb{D}$.