# Thread: Complex Analysis, Schwarz lemma

1. ## Complex Analysis, Schwarz lemma

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

Let $\displaystyle f$ and $\displaystyle g$ be an analytical functions$\displaystyle f, g \rightarrow D$

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

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

Thanks!

2. Originally Posted by sinichko

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

Let $\displaystyle f$ and $\displaystyle g$ be an analytical functions$\displaystyle f, g \rightarrow D$

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

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

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

3. Originally Posted by sinichko

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

Let $\displaystyle f$ and $\displaystyle g$ be an analytical functions$\displaystyle f, g \rightarrow D$

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

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

Thanks!
That statement is false, choose:

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

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 $\displaystyle h \to D$ sending 0 to 0.

Try $\displaystyle h(z) = \frac{1}{2}\left[f(z) - g(z)\right].$.
(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 $\displaystyle h$ satisfies the hypotheses of the lemma and apply it. See what happens.