1. Complex Analysis, Schwarz lemma

$D=\left \{ z:\left | z \right |<1 \right \}$

Let $f$ and $g$ be an analytical functions $f, g \rightarrow D " alt="f, g \rightarrow D " />

and also $f(0)=g(0), f'(0)=g'(0)$

Prove that $f(z)=g(z), \forall z\in D$.

Thanks!

2. Originally Posted by sinichko

$D=\left \{ z:\left | z \right |<1 \right \}$

Let $f$ and $g$ be an analytical functions $f, g \rightarrow D " alt="f, g \rightarrow D " />

and also $f(0)=g(0), f'(0)=g'(0)$

Prove that $f(z)=g(z), \forall z\in D$.

Thanks!
Not true, for example $f(z) = z^2,\ g(z) = z^3.$

3. Originally Posted by sinichko

$D=\left \{ z:\left | z \right |<1 \right \}$

Let $f$ and $g$ be an analytical functions $f, g \rightarrow D " alt="f, g \rightarrow D " />

and also $f(0)=g(0), f'(0)=g'(0)$

Prove that $f(z)=g(z), \forall z\in D$.

Thanks!
That statement is false, choose:

$f(z)=z^2,\;g(z)=z^3$

Fernando Revilla

Edited: Sorry, I didn't see Opalg's post.

4. See if you have the statement right.

Meanwhile, a good strategy for any Schwarz problem is to construct a function satisfying the hypotheses of the lemma. Then everything frequently falls nicely onto your lap.

You need an analytic function $h \to D" alt="h \to D" /> sending 0 to 0.

Try $h(z) = \frac{1}{2}\left[f(z) - g(z)\right].$.
(Question: why did I scale by one half?)

This may not lead where you want (especially since at the moment the conclusion you seek is unclear), but constructing something like this is usually a good first step.

Verify that $h$ satisfies the hypotheses of the lemma and apply it. See what happens.