# Thinking more intuitively about complex analysis, in particular regarding Schwarz's Lemma

In a proof of Schwarz's Lemma (Sarason, "Complex Function Theory", pp. 91-92) the function $g$ is defined in the disk by

$$g(z) = \begin{cases}\frac {f(z)}z&\mbox{ for } 0 < |z| < 1\\ f'(0)&\mbox{ for } z = 0.\end{cases}$$

The proof goes on to say that for $0 < r <1$, $g$ is bounded by $\frac {1}{r}$ on the circle $|z| = r$ and thus has the same bound in the disk $|z| \le r$ by the maximum modulus principle.

This being true for all $r$ in $(0,1)$, $g$ is bounded in absolute value by $1$.

This makes perfect sense formally.

Here are my questions:

1) Intuitively when I see something is bounded by $\frac {1}{r}$ and $r$ has values in $(0,1)$, I would naturally think of $r$ being close to $0$, and $\frac {1}{r}$ getting quite large. Then along comes the max. modulus principle and shuts it all down to a bound of $1$.

How can I think intuitively about complex analysis to see how my naive instinct does not apply here?

2) In the statement of the Lemma, we are given that $f(0) = 0$. It seems to me that this comes into play in defining $g$.

Again, how can I get an intuitive insight as to the power this stipulation has in providing the criterium for the conclusion of the Lemma.

And on a larger scale, how do I train myself to "think more complex"?

Thanks

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You'd like to say $|g(z)| \le 1$ on $|z|=1$, except that $g$ is only defined for $|z| < 1$, so you have to use $r$ slightly less than 1, then take the limit as $r \to 1-$. Bigger $r$'s produce better bounds.

Of course, $f(0)=0$ is needed for the conclusion of the Lemma to be true. There is a slightly more general version of the Lemma where you assume $|f(z)| \le 1$ for $|z| < 1$ and $f(c) = 0$ for some $c$ with $|c| < 1$, and conclude $|f(z)| \le \left| \frac{z-c}{1-\overline{c}z}\right|$ (this follows from the ordinary Lemma after composing $f$ with a fractional linear transformation). On the other hand, if $f(z)$ is never 0 for $|z| < 0$, I don't know what kind of a bound you could hope for: e.g. $f$ could be constant.

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Thanks for your help. – TheBirdistheWord Sep 15 '11 at 23:12