# Upper bound for zeros of holomorphic function

I'd appreciate some help with the following problem form Conway's book on functions of one complex variable:

Let $f$ be analytic in $\overline B (0;R)$ with $|f(z)|\le M$ for $|z|\le R$ and $|f(0)|=a>0$. Show that the number of zeros of f in $B(0;R/3)$ is less than or equal to $$\frac{1}{\log(2)}\log\left(\frac M a\right)$$

I know that the number of zeros is given by $$n = \frac 1 {2\pi i}\int_{|z|=R/3} \frac{f'}{f} \, dz$$

And there is a hint to look at $g(z) = f(z) \prod_{k=1}^n (1-z/z_k)^{-1}$, where the $z_k$ are the zeros of $f$. I have given it some time now, but don't seem to get anywhere. In particular I don't see how the logarithm, $M, a$ come into play. The problem is in the chapter on the maximum modulus theorem, if that's of any help.

Might someone maybe give me a hint?

Cheers, S.L.

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The function $g$ is holomorphic in $B(0,R)$ and continuous on $\bar{B}(0,R)$. For $|z|=R$ we have $\left|\frac{z}{z_k}\right|\geq 3$ and one obtains $|g(z)|\leq 2^{-n}M$. From the maximum principle one can infer that this also holds for $|z|<R$, in particular for $z=0$. Now solve for $n$.