Let $A(z) = \sum_{n=0}^{+\infty}a_n z^n$ and $B(z) = \sum_{n=0}^{+\infty}b_n z^n$ be two formal power series with complex coefficients. The Hadamard product of $A$ and $B$ is the formal power series $A\star B$ defined by $$ (A\star B)(z) = \sum_{n=0}^{+\infty}a_n b_n z^n. $$ If $r_A$ (resp. $r_B$) is the radius of convergence of $A$ (resp. $B$), it is clear that the radius of convergence $r_{A\star B}$ of $A\star B$ is greater than or equal to $r_A \cdot r_B.$

Inside the disk $D(0,r_A \cdot r_B)$, what do we know concerning the zeros of $A \star B$ and their relations with the zeros of $A$ and $B$ ?

Note that if $f$ (resp. $g$) is a holomorphic function defined by $A$ (resp. $B$) on the disk $D(0,r_A)$ (resp. $D(0,r_B)$) and if $r \in (0,r_A)$, then \begin{align*} \sum_{n=0}^{+\infty}a_n b_n z^n & = \sum_{n=0}^{+\infty}\left(\frac{1}{2i\pi}\int_{C(0,r)^+}\frac{f(\zeta)}{\zeta^{n+1}}\, d\zeta \right) b_n z^n\\ & = \frac{1}{2i\pi}\int_{C(0,r)^+} f(\zeta) \left(\sum_{n=0}^{+\infty} b_n \left(\frac{z}{\zeta} \right)^n \right) \frac{d\zeta}{\zeta}\\ & = \frac{1}{2i\pi}\int_{C(0,r)^+} f(\zeta)g\!\left(\frac{z}{\zeta}\right) \frac{d\zeta}{\zeta}, \end{align*} for any $z \in D(0,r \cdot r_B)$.