Derivative of Blaschke product

Question

Let $z_n$ be a Blaschke sequence in $\mathbb{D}$ and let $B$ be the Blaschke product defined by $$B(z)=z^m\prod_{n=1}^{\infty}\frac{|z_n|}{z_n}\frac{z_n-z}{1-\bar{z}_nz}$$ I'm trying to show the following relationship is true for any $n$. $$(1-|z_n|^2)|B'(z_n)|=\prod_{m=1,m\neq n}^{\infty}\left|\frac{z_n-z_m}{1-\bar{z}_n z_m}\right|$$

By simply taking the derivative of the product, we have $$B'(z) =mz^{m-1}\prod_{n=1}^{\infty}\frac{|z_n|}{z_n}\frac{z_n-z}{1-\bar{z}_nz}+z^m\prod_{n=1}^{\infty}\frac{|z_{n}|}{z_{n}}\frac{(\bar{z}_{n}z-1)+(z_{n}-z)\bar{z}_{n}}{(1-\bar{z}_{n}z)^{2}}$$

But I'm having trouble seeing how the right hand side of the equation can be derived from here.

Answer 1

Your computation of the derivative seems to be wrong because apparently you differentiated the product termwise. Also you are using the variable $m$ for two different purposes.

One can use the definition of the derivative directly instead. $B(z_n) = 0$, therefore $$ \begin{align} B'(z_n) &= \lim_{z \to z_n} \frac{B(z)}{z-z_n}\\ &= \lim_{z \to z_n} z^m \left(\frac{|z_n|}{z_n}\frac{-1}{1-\bar z_n z}\right) \cdot \prod_{k \ne n}\frac{|z_k|}{z_k} \frac{z_k-z}{1-\bar z_k z} \\ &= \frac{z_n^m}{|z_n|^2-1 }\frac{|z_n|}{z_n}\prod_{k \ne n} \frac{|z_k|}{z_k}\frac{z_k-z_n}{1-\bar z_k z_n} \end{align} $$ and then the desired conclusion follows by taking the absolute value.

Answer 2

From mu comment:

By brute force $$ B'(z)= mz^{m-1}\prod_{n\in\mathbb{N}}\frac{|z_n|}{z_n}\frac{z_n-z}{1-\overline{z_n}z_n} + z^m\sum_{n\in\mathbb{N}}\frac{|z_{n}|}{z_{n}}\frac{(\bar{z}_{n}z-1)+(z_{n}-z)\bar{z}_{n}}{(1-\bar{z}_{n}z)^{2}}\prod_{k\neq n}\frac{|z_k|}{z_k}\frac{z_k-z}{1-\overline{z_k}z} $$ So $$ B'(z_{n_0})=z^m_{n_0}\frac{|z_{n_0}|}{z_{n_0}}\frac{|z_{n_0}|^2 -1}{(1-|z_{n_0}|^2)^2}\prod_{k\neq n_0}\frac{|z_k|}{z_k}\frac{z_k-z_{n_0}}{1-\overline{z_k}z_{n_0}} $$

(To obtain $\Big(\prod_nf_n(z)\Big)'$ one can use $\log$ (formally first, and then rigorously) to get $$\Big(\prod_nf_n(z)\Big)' = \Big(\prod_n f_n(z)\Big)\Big(\sum_n\frac{f'_n(z)}{f_n(z)}\Big)$$

A smarter solution in this particular problem is to notice that $B(z_n)=0$ for each $n$ (as in Martin R's answer) and then use definition of derivative at $z_n$ directly.