0

Question: Let $f\colon D_3(0) \to \mathbb{C}$ be analytic where $D_r(0) = \{z \in \mathbb{C} \colon |z| < r\}$. Let $D = D_2(0)$.

  • (i)Show that $f(D)$ is open and connected in $\mathbb{C}$.
  • (ii) Show that $\partial \overline{f(D)} = f(\partial D)$ if $f$ is injective on $\overline{D}$.
  • (iii) Let $w_0 \in f(D)$ and suppose there exists unique $z_0 \in D$ so that $f(z_0) = w_0$. If $ f'(z_0) =0, f''(z_0) \neq 0$, show that for all $w \in f(D)$, there are exactly two complex numbers $z \in D$ counting multiplicity so that $f(z) = w$.

My attempt:

-(i) Follows directly from open mapping theorem and continuous image of connected set is connected in any metric space.

  • (ii) For the forward inclusion: Let $w \in \partial \overline{f(D)}$ so $w \in \overline{f(D)} \setminus f(D)$ and there exists a sequence of $z_n \in D$ so that $\lim_n f(z_n) = w$. As $D$ is bounded set in $\mathbb{C}$ then there exists subsequence $\{z_{n_k}\}$ so that $z_{n_k} \to z_0 \in \overline{D}$ as $k \to \infty$. By uniqueness of limit, $$w = \lim_n f(z_n) = \lim_k f(z_{n_k}) = f(z_0)$$ so $w \in f(\partial D)$ as $w \notin f(D)$.
  • For the reverse inclusion, suppose $f$ is injective in $\overline{D}$ and let $w = f(z) \in f(\partial D)$. If for a contradiction that $f(z) = w \in f(D)$ then there exists $z_0 \in D$ so that $f(z) = w = f(z_0)$ contradicting the injectivity hypothesis.

As for part (iii), it was hinted to use argument principle. However, I am not sure how to use the argument principle to show that for all $w \in f(D)$ we have $$\frac{1}{2\pi i} \int_{D} \frac{f'(\zeta)}{f(\zeta) - w} \, d \zeta = 2.$$ Is there any way to approach this question without involving the winding numbers?

Any hints/solutions are appreciated.

L-JS
  • 793
  • For (ii) have a look at https://math.stackexchange.com/a/1825237/42969 – Martin R Jul 31 '23 at 16:23
  • for (ii) the fact that $\overline D$ is compact implies that $f_{\overline D}$ is a homeomorphism (assuming its co-domain is restricted to its image) and the result follows-- ref e.g. https://math.stackexchange.com/questions/46353/homeomorphism-of-the-disk . You should be able to visualize this since $f_{\overline D}$ and $f_{\overline D}^{-1}$ are open maps on points in their interior. Of course (ii) follows from the Jordan Curve Theorem as well. – user8675309 Jul 31 '23 at 16:59
  • 1
    I think that (iii) has an error as the Argument Principle works fine to describe local behavior of non-injective $f$ around $z_0$ but not-necessarily for the entire 2-disc. Consider e.g. $f(z)=z^2\cdot (z-2)$ which satisfies the criterion for $z_0=0=w_0$ in but $f(z)+\frac{1}{2}$ has 3 roots in $D$. – user8675309 Jul 31 '23 at 17:46
  • @user8675309 yeah you are right, however according to Robert Ash's notes on complex variables it states the following: Let $\gamma$ be a positively oriented boundary of $\overline{D}(z_), \epsilon)$, let $W_0$ be the component of $\mathbb{C} \setminus (f \circ \gamma)^\ast$ that contains $w_0$ and let $\Omega_1 = D(z_0, \epsilon) \cap f^{-1}(W_0)$, then $f$ is $k$-to-one map of $\Omega_1 \setminus {z_0}$ onto $W_0\setminus{w_0}$ provided $k$ is the order of zeros which $f - w_0$ has at $z_0$.

    Why can't the result follow directly from here?

     – L-JS Aug 01 '23 at 05:41
  • Source: https://faculty.math.illinois.edu/~r-ash/CV/CV4.pdf (Page 15/45) – L-JS Aug 01 '23 at 05:41
  • What you are describing in your comment is the local behavior of a function, i.e. in a sufficiently small neighborhood of $z_0$. There's no reason to think the ball around $z_0$ would have radius anywhere close to $2$ though (iii) is in effect asserting that. – user8675309 Aug 01 '23 at 15:07
  • @user8675309 but the above theorem states that if $w$ lies in the same component as $w_0$ then the pre-image $f^{-1}(w)$ is exactly $k$ counting multiplicity. In that case, from (a), wouldn't all $w \in f(D)$ lie in same component as $w_0$ as $f(D)$ itself is connected. – L-JS Aug 01 '23 at 15:10
  • $f\circ \gamma$ is not a Jordan Curve, so why do you think $\mathbb C-[f\circ \gamma]$ only has two components? If we ignore the unbounded component we can talk instead about $f(D)-[f\circ \gamma]$, which for some reason you seem to think is a single component. Please draw an example curve that look like a figure 8 and convince yourself it has 2 bounded components. – user8675309 Aug 01 '23 at 15:16

0 Answers0