# Entire function as product of meromorphic and entire functions

Let $\{a_n\},\{b_n\}$ be complex numbers such that $|a_n|\rightarrow\infty$ as $n\rightarrow\infty$. Let $g$ be an entire function with simple zeros at the $a_n$'s, and let $h$ be a meromorphic function with poles at the $a_n$'s and the corresponding singular part $\dfrac{b_n}{z-a_n}$.

Why is $f=gh$ an entire function with $f(a_n)=b_n$?

I know we can write $h(z)=\sum_n\left[\dfrac{b_n}{z-a_n}-p_n(z)\right]+q(z)$, where $p_n(z)$ are polynomials and $q(z)$ is entire. How can I know $f$ is entire and $f(a_n)=b_n$?

## 1 Answer

Around $a_n$ we can write the Laurent series of $h$, it is $$h(z)=\frac{b_n}{z-a_n}+c_0+c_1(z-a_n)+c_2(z-a_n)^2+\dots$$ At the same time, the Taylor series of $g$ around $a_n$ goes like $$g(z)=0+u_1(z-a_n)+u_2(z-a_n)^2+\dots$$ with $u_1= g'(a_n)$.

So, the Laurent series of $f=gh$ around $a_n$ is $$f(z)=b_n\, u_1+\ (c_0u_1+b_nu_2)\,(z-a_n)+(c_1u_1+c_0u_2+b_nu_3)\,(z-a_n)^2+\dots$$ As with other points there can be no problem, this shows that $f$ is indeed entire function, but we arrived at $$f(a_n)= b_n\cdot g'(a_n)\,.$$