2

let $ f:\mathbf{D}\rightarrow \mathbf{D}$ be a holomorphic function. And suppose that it is Bijective on ${D}\setminus \left \{ 0 \right \}\rightarrow \mathbf{D}\setminus \left \{ 0 \right \} $. Can we conclude that
f(0) = 0?

I can conclude that $\begin{vmatrix}f(0)\end{vmatrix} < 1$ because other wise if $\begin{vmatrix}f(0)\end{vmatrix} = 1$ then by maximum modulus principle f(z) is constant but that cannot happen because the function is a bijection. but from this is it possible to deduce that f(0)=0 ?

I'm referring the correct answer of conformal self maps on punctured disk

Community
  • 1
Charith
  • 1,616
  • If $f(0)\neq 0$, then $f(z)=a$ would have precisely two solutions for precisely one value of $a$. I feel like Rouche's theorem should show this is a contradiction, but it might be a bit of an overkill. – Wojowu Nov 04 '16 at 09:23
  • Intuitively, I'm pretty sure continuity should be enough to prove this. – Mees de Vries Nov 04 '16 at 09:23
  • Hi. Check the link I posted on the content. It said once by images of points z near 0 and once by images of points... can you please elaborate it – Charith Nov 04 '16 at 09:33
  • It's bijective on $D\setminus{0}$, and $f(0)\not 0$ implies $f(0)\in D\setminus{0}$. So $f(0)$ has a preimage in $D\setminus{0}$. – Cave Johnson Nov 04 '16 at 09:52

2 Answers2

1

Let $f(0) = a$ with $0<|a|<1.$ As $f:\mathbb D /\{0\} \mapsto \mathbb D /\{0\} ,$ So there exist a point $b\in \mathbb D /\{0\}$ so that $f(b)= a.$ Consider the function $g: \mathbb D \mapsto \mathbb D$ defined by $g(z)= f(z)- a.$ The function $g$ has a zero at $b$ and $0.$ Let order of the zero of $g$ at $b$ and $0$ be $m_1, m_2$ respectively. Note that $m_1,m_2 \geq 1.$

So there exist $\epsilon_1,\epsilon_2 >0$ and a $\delta>0$ so that each point $w$ in $B(0,\delta),$ the equation $g(z)=w$ has exactly $m_1$ simple root in $B(b,\epsilon_1)$ and $m_2$ simple root in $B(o,\epsilon_2).$ Note that $\delta$ can be taken so small so that corresponding $\epsilon_1,\epsilon_2$ will have the property that $B(o,\epsilon_2) \cap B(b,\epsilon_1) = \emptyset. $ In that case bijectivity of $f$ in $\mathbb D/\{0\}$ will be contradicted.

Hence $f(0)$ must be $0.$

Timon
  • 2,401
0

I'm Referring the book Function Theory of One Complex Variable By Robert Everist Greene, Steven George Krantz..

Supppose that

$f(0) \neq 0 $ then $ \exists z` \in \mathbb{D} \setminus \begin{Bmatrix}0\end{Bmatrix} $ such that $f(0) = z`$. Also note that $ \exists z_{1} \in \mathbb{D} \setminus \begin{Bmatrix}0\end{Bmatrix} $ such that $f(z_{1}) = z`$.

Now by using Theorem 5.2.2 we have $ \delta_{0}, \varepsilon _{0}, \delta_{1}, \varepsilon _{1} $ such that $ \forall z \in D(z`,\varepsilon _{0}) \setminus \{z` \} $ has exactly one pre image on $ D(0,\delta_{0}) $.

and similarly $ \forall z \in D(z`,\varepsilon _{1}) \setminus \{z` \} $ has exactly one pre image on $ D(0,\delta_{1}) $.

now by considering the intersection of two open balls $D(z`,\varepsilon _{0}) \setminus \{z` \}$ and $D(z`,\varepsilon _{1}) \setminus \{z` \}$ which is nonempty we can contradict with the fact that the function is one to one on the punctured disc.

Charith
  • 1,616