# Using Rouche's Theorem to find the number of solutions of $f(z)=z$ in the open unit disc [duplicate]

How many roots does the equation $$f(z)=z$$ have in the circle $$|z|<1$$ if for $$|z|\leq 1$$, $$f(z)$$ is analytic and satisfies $$|f(z)|<1$$?

My idea: I figured I could do this pretty easily using Rouche: Consider $$|z|=1$$, and let $$g(z)=z$$, then $$|f(z)-g(z)|=z-z=0<1=|g(z)|$$. So, since $$g$$ has only $$1$$ root in $$|z|<1$$, then so does $$f$$.

I just feel like I am missing something. In particular, can I define $$g$$, and use it, in the way I did? Any thoughts are greatly appreciated! Thank you.

• How do you get $f(z)-g(z) = 0$???? The equality $f(z)=z$ only holds at the solution of the equation $f(z)=z$. Commented Jul 1, 2020 at 21:56
• ahhhh, okay, I see. So, maybe Rouche isn't the way to go about this? Commented Jul 1, 2020 at 21:58
• I'd still think Rouche's theorem is the only thing available, but you may need to be more careful/precise. Commented Jul 1, 2020 at 22:04
• No, absolutely not. When I read it, I thought about Rouche right away, but, any method is acceptable. Commented Jul 1, 2020 at 22:14
• But then what is the point of the assumption $|f(z)|<1$ when $|z|\leq 1$? Commented Jul 1, 2020 at 22:21

Let $$h(z) = f(z)-z$$, $$g(z) = z$$, then for $$|z|=1$$ we have $$|h(z)+g(z)| = |f(z)| < 1 = |g(z)|$$, hence $$h,g$$ have the same number of zeroes inside the circle. Since $$g$$ has exactly one zero, so does $$h$$.

• Always nice when I am this stuck on a 2 line proof :) Thank you, @copper.hat! Much appreciated, as always! Commented Jul 1, 2020 at 22:52
• @User7238 Thank you. Your appreciation is the reward :-). Commented Jul 8, 2020 at 7:12

Here is an alternative solution without using Rouché's Theorem. While it is lengthier than cooper.hat's great answer, it determines exactly what the fixed points of $$f$$ are.

We can consider $$f$$ as a holomorphic function on $$\overline{\mathbb{D}}:=\big\{z\in\mathbb{C}\,\big|\,|z|\leq 1\big\}$$. Note that the image of $$f$$ lies in $$\mathbb{D}:=\big\{z\in\mathbb{C}\,\big|\,|z|< 1\big\}$$. We first claim that the equation $$f(z)=z$$ has at most one solution $$z\in\mathbb{D}$$.

Suppose on the contrary that $$f(z)=z$$ has two solutions $$z=z_i$$ for $$i\in\{1,2\}$$ in $$\mathbb{D}$$. Then, we can find a Möbius transformation $$\mu:\overline{\mathbb{D}}\to\overline{\mathbb{D}}$$ that maps $$z_1\mapsto 0$$. Write $$w:=\mu(z_2)$$. Let $$\phi:=\mu\circ f\circ \mu^{-1}$$. Then, $$\phi:\overline{\mathbb{D}}\to\overline{\mathbb{D}}$$ is such that $$\phi(0)=0$$ and $$\phi(w)=w$$. By Schwarz's Lemma, $$\phi(z)=z$$ for all $$z\in\overline{\mathbb{D}}$$. Thus, $$f(z)=z$$ for all $$z\in\overline{\mathbb{D}}$$. This contradicts the assumption that the image of $$f$$ lies in $$\mathbb{D}\subsetneq \overline{\mathbb{D}}$$. Therefore, $$f$$ has at most one fixed point.

Now, define $$f^{\circ 1}:=f$$ and $$f^{\circ k}:=f\circ f^{\circ (k-1)}$$ for $$k=2,3,4,\ldots$$. Take $$I_k$$ to be the image of $$f^{\circ k}$$ for each positive integer $$k$$. Note that $$I_1\supseteq I_2\supseteq I_3\supseteq \ldots$$. Because $$f$$ is a continuous function and $$\overline{\mathbb{D}}$$ is a compact set, we can easily see that each $$I_k$$ is a compact set. Due to Cantor's Intersection Theorem, the set $$I:=\bigcap_{k=1}^\infty\,I_k$$ is nonempty. By our previous paragraph, $$I=\{\zeta\}$$ for some $$\zeta\in\mathbb{D}$$. Clearly, $$f(\zeta)=\zeta$$. From this result, it follows that $$\zeta=\lim_{k\to\infty}\,f^{\circ k}(z)$$ for any $$z\in\overline{\mathbb{D}}$$.

• This is cool - I like it. To @Batominovski and copper.hat, I can't thank you two enough! Commented Jul 2, 2020 at 1:24