Question about holomorphic function

Question

Let $f$ and $g$ be two holomorphic functions defined from :

$D=\{ z \in \mathbb{C} \mid |z| < 1 \}$ to $U \subset \mathbb{C}$, where $U$ is a simply connected set. Suppose that $g:D\to U$ is bijective and for some $z_0 \in D$, we have $f(z_0) = g(z_0)$ and $f'(z_0) = g'(z_0)$. We want to show that $f = g$ on $D$

I don't know how to approach this question, we have $g'(z_0)\neq 0$, then $f'(z_0)\neq 0$, which means that $f$ is locally invertible.

Answer 1

As followed from the hint in comment, consider $h=g^{-1}\circ f.$

Consider, if needed, the composition with the following Möbius transformation: $$\phi_{z_0}(z)=\frac{z_0-z}{1-\overline{z_0}z}$$ by $h=\phi_{z_0}(z)\circ g^{-1}\circ f\circ \phi_{z_0}(z).$ This will make $h(0)=0$. So $h$ is now a holomorphic function from $D$ to $D$.

And, check that $h'(0)=1$ by using the chain rule. (The commputation here may be a little bit complicated, and you may need some facts about Möbius transformations)

Thus, by Schwarz lemma, $h(z)=e^{i\theta}z$ for some $\theta\in\mathbb R.$

Finally, since $h'(z)=e^{i\theta}$ and $h'(0)=1$, it is not hard to conclude $h(z)=z$ for all $z\in D.$ Hence we obtain $g=f$ on $D$.

Answer 2

Consider $h := g^{-1} \circ f : D \to D$ as per Greg Martin's hint in the comments.

Then, we have: $$h(z_0) = g^{-1}(f(z_0)) = g^{-1}(g(z_0)) = z_0$$ and: $$h'(z_0) = (g^{-1} \circ f)'(z_0) = f'(z_0) \cdot \frac{1}{g'(g^{-1}(f(z_0)))} = g'(z_0) \cdot \frac{1}{g'(z_0)} = 1$$

However, it just so happens that a holomorphic function $\varphi : \Omega \to \Omega$ with $\Omega$ open, connected and bounded satisfying $\varphi(w_0) = w_0$ and $\varphi'(w_0) = 1$ for some $w_0 \in \Omega$ is necessarily the identity (see Holomorphic function $\varphi$ with fixed point $z_0$ such that $\varphi'(z_o)=1$ is linear? or A condition implying that a holomorphic function is the identity map for example), and since $\varphi = h$ and $\Omega = D$ respect those conditions, $h$ is the identity, therefore: $\forall z \in D,\,\,h(z) = z$.

But since $h = g^{-1} \circ f$, this becomes $\forall z \in D,\,\, f(z) = g(z)$, and we have proven the claim.

Answer 3

Here is an answer via the Argument Principle. Consider analytic
$h:D\longrightarrow \mathbb C$ given by $h(z)=g^{-1}\big(f(z)\big) - z$

If an analytic map from $D\to D$, like $g^{-1}\circ f$, has (at least) two distinct fixed points then it is the identity map, ref e.g. Schwarz's Lemma, fixed points question. This is equivalent to saying for distinct $w,w'\in D$ that $h(w)=0=h(w')\implies h=0$. And in this case we can almost 'eyeball' this to be true since $h(z_0)=0=h'(z_0)$ would seem to imply a winding number of at least $2$ around $z_0$.

To formalize this, suppose for contradiction that $h\neq 0$.
Then $h(z_0)=0$ and $h'(z_0)=0$; by principle of isolated zeros for small enough $r\gt0$ we know $h(z)\neq 0$ for $z \in\overline B(z_0,r) -\big\{z_0\big\} \subseteq D$. With $\gamma(t) := z_0 +r\cdot\exp\big(2\pi i\cdot t\big)$ for $t\in[0,1]$, we have winding number $n\big(h\circ \gamma,0\big)=m\geq 2$ by the Argument Principle.

Define $h_k(z) := \frac{1}{k}z_0 + \frac{k-1}{k}g^{-1}\big(f(z)\big) – z $, checking
(a) $h_k\big(z_0\big)= \frac{1}{k}z_0 + \frac{k-1}{k}\cdot z_0 – z_0 = 0$ and
(b) $h_k'\big(z_0\big)= 0+\frac{k-1}{k}\cdot 1 – 1 = -\frac{1}{k}\neq 0$
So $z_0$ is a simple zero of $h_k$ for all $k\in \mathbb N$. However $h_k$ converges uniformly to $h$ in $\overline B(z_0,r)$ hence it has $m$ zeros in $B(z_0,r)$ for all $k$ large enough [Hurwitz]-- i.e. $h_k(z)$ has at least 2 distinct zeros in $B(z_0,r)\subseteq D$ for $k\geq K$. This implies that the map $z\mapsto \frac{1}{k}z_0 + \frac{k-1}{k}g^{-1}\big(f(z)\big) = z$ for all $k\geq K$, which is an obvious contradiction.
(e.g. observe this implies $h_k=0$ for $k\geq K$ which converges uniformly to $h\neq 0$).

Conclude $h=0\implies g^{-1}\big(f(z)\big) = z\implies f(z) = g(z)$ for all $z\in D$ as desired.