I'm reading an approximation result from a paper that claims without justifying:
$$\cfrac{d}{d\theta}\text{arg}(B(e^{i\theta}))=\sum_{j=1}^nP(e^{i\theta},a_j),$$
where $z=re^{i\phi}$ and $P(z,a_j)=\cfrac{1-r^2}{1+r^2-2r\cos(\theta-\phi)}$ is the Poisson kernel evaluated at $z$. Also, $B$ here is a Blaschke product of degree $n$ and has zeros $a_j$.
Question: I don't understand what it means to differentiate the argument, much less that of a Blaschke product. How does one see that the equation above is true?
Any help is appreciated. Thanks a lot.
Note that $\log B(e^{i\theta})=\log |B(e^{i\theta})|+i \arg B(e^{i\theta})=i \arg B(e^{i\theta})$ (locally near $e^{i\theta}$ for some branch of the argument as $|B(e^{i\theta})|=1 \ne 0$, while when we take derivatives, the branch becomes irrelevant as two such differ by constants); also since $z=e^{i\theta}$ implies $dz=izd\theta$ or $iz d/dz=d/d \theta$, one has that:
$\cfrac{d}{d\theta}\text{arg}(B(e^{i\theta}))=iz \cfrac{d}{dz}\text{arg}(B(e^{i\theta}))=z\frac{d \log B(z)}{dz}$ evaluated at $z=e^{i\theta}$
But a simple computation shows that for an individual factor $B_j(z)=\frac{z-a_j}{1-z\bar a_j}$ we have $z\frac{d \log B_j(z)}{dz}=z\frac{1-|a_j|^2}{(1-z\bar a_j)(z-a_j)}$ and when we evaluate at $z=e^{i\theta}$ we get
$\cfrac{d}{d\theta}\text{arg}(B_j(e^{i\theta}))=\frac{1-|a_j|^2}{|e^{i\theta}-a_j|^2}=P(e^{i\theta},a_j)$, so by additivity ($\arg, \log$ transform products into sums)
$\cfrac{d}{d\theta}\text{arg}(B(e^{i\theta}))=\sum_{j=1}^n\frac{1-|a_j|^2}{|e^{i\theta}-a_j|^2}=\sum_{j=1}^nP(e^{i\theta},a_j)$