# Elimination of zeros of an entire function

Let be $$f$$ an entire function and $$a_1,a_2,...,a_N$$ the zeros of $$f$$ (i.e. $$f(a_k)=0$$). A $$g$$ function as: $$g(z)=\begin{cases} \frac{f(z)}{(z-a_1)(z-a_2)...(z-a_N)}, \qquad \text{for }z\neq a_k, \\ \lim_{z\rightarrow a_k}g(z), \qquad \text{for }z=a_k \end{cases}$$ where those limits exist. Show that $$g$$ is an entire function. I know that $$\frac{f(z)-f(a)}{z-a}$$ is an entire function.

• Just apply what you know to each of the points $a_i$ to see that $g$ is differentiable at each point. – Kabo Murphy Feb 9 at 23:59

Obviously the only points in which we may have problems are $$a_1,...,a_N$$. But it is easy to see that all these points are just removable singularities. For example, $$a_1$$ is zero of $$f$$ and by a known theorem in complex analysis we can write $$f(z)=(z-a_1)h(z)$$ when $$h$$ is holomorphic in the neighborhood of the point $$a_1$$. Can you finish from here?