Let $f(z)=\sum\limits_{n=0}^{\infty}a_n z^n$ be an analytic function in $|z|\leq R$ such that $|f(z)|\leq M$. Prove an inequality about $f$'s zero
The Schwarz Lemma is often a good tool to consider when we want to bound (below) the magnitude of a point possessing some property or to bound (above) the derivative at a zero. Our goal here will be to apply the Schwarz lemma to $f$. To do this, we'll apply a couple of transformations to $f$ so that we get a new function $h$ which satisfies the conditions of the lemma, namely
- $h$ is analytic on the open unit disk
- $h(z) < 1$ for all $|z| < 1$
- $h(0) = 0$
By the lemma this will then imply $|h(z)| \leq |z|$ and by our construction of $h$ we will achieve the desired inequality.
First we normalize $f$. Let $$g(z) = \frac{f(Rz)}{M}$$ Since $f$ is analytic on a disk of radius $R$, $g$ is analytic on the unit disk. Similarly the bound of $M$ on $f$ becomes a bound of $1$ on $g$.
Now, we will turn the zero of $f$ at $a_0$ into a zero at the origin. Let $$\phi(z) = \frac{z + z_0/R}{\bar{z_o}z/R + 1}$$
$\phi$ is the Mobius transformation mapping the unit disk into itself, and sending the point $0$ to $z_0/R$. Note that $\phi$ is analytic. It's important to be familiar with Mobius functions. Lars Ahlfors writes in his introductory text that "the rather specialized assumptions of [Schwarz's Lemma] are not essential, but should be looked upon as the result of a normalization". The key to applying Schwarz's Lemma is often normalization by way of Mobius functions.
Finally define $$h(z) = g(\phi(z))$$
Note that $h$ is still analytic on the unit disk and bounded by $1$, since $h$ is the composition of analytic functions on the unit disk, and since $g$ and $h$ have the same image.
We also have now that $h(0) = g(\phi(0)) = g(z_0/R) = f(z_0)/M = 0$
Thus $h$ satisfies the conditions of the Schwarz Lemma, implying $$|h(z)| \leq |z|$$ holds for all $|z| < 1$. Applying this to $z = -z_0/R$, $$|h(-z_0/R)| \leq |z_0|/R$$
On the other hand, by construction $$h(-z_0/R) = g(\phi(-z_0/R)) = g(0) = \frac{f(0)}{M} = \frac{a_0}{M}$$
Putting it all together shows $$\frac{|a_0|}{M} \leq \frac{|z_0|}{R}$$ therefore $$\frac{R|a_0|}{M} \leq |z_0|$$
Noting $M + |a_0| \geq M$ always, we can write $$\frac{R|a_0|}{M + |a_0|} \leq |z_0|$$ which is the inequality slightly weaker inequality you were seeking.
A couple things to note:
- we didn't ever need to use the fact that $z_0$ was a smallest zero of $f$
- in the last equation we see that the inequality you stated is weaker than the inequality we were able to proof (the difference between $M + |a_0|$ and just $M$ in the denominator).
This makes me wonder where the question came from.
I will assume $d<R$, otherwise $$d \geq R > \frac{R|a_0|}{M+|a_0|}$$ is a trivial case. Now, let's look at this function: $$g(z)=\frac{f(z)-a_0}{z}=a_1+\sum_{k=2}^{\infty}a_kz^{k-1} \tag 1$$ which has a removable singularity at $z=0$. Additionally: $$|g(z_0)|\leq |a_1|+\sum_{k=2}^{\infty}|a_k||z_0^{k-1}|=|a_1|+\sum_{k=2}^{\infty}|a_k|d^{k-1}$$
From Cauchy's estimate and Taylor series, for $f(z)$: $$|a_k|=\left|\frac{f^{(k)}(0)}{k!}\right|=\frac{1}{2\pi}\left|\int_{C_{R}} \frac{f(z)}{z^{k+1}} dz\right|\leq \frac{M}{R^k}$$ As a result: $$|g(z_0)|\leq \frac{M}{R}+\frac{Md}{R^2}+\frac{Md^2}{R^3}+...=\frac{M}{R}\left(1+\frac{d}{R}+\frac{d^2}{R^2}+...\right)=\frac{M}{R}\frac{R}{R-d}=\frac{M}{R-d}$$
Then from (1): $$|g(z_0)|=\frac{|a_0|}{d} \Rightarrow d=\frac{|a_0|}{|g(z_0)|}\geq \frac{|a_0|}{\frac{M}{R-d}}=\frac{|a_0|(R-d)}{M}$$ Or $$dM\geq R|a_0|-d|a_0| \Leftrightarrow d(M+|a_0|)\geq R|a_0| \Leftrightarrow d \geq \frac{R|a_0|}{M+|a_0|}$$