2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Just apply what you know to each of the points $a_i$ to see that $g$ is differentiable at each point. $\endgroup$ – Kabo Murphy Feb 9 at 23:59
1
$\begingroup$

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.