# Calculating the integrated counting function for a certain transcendental entire function

For a transcendental entire function $$f$$, set $$h(z) = z + \frac{f(z)}{f'(z)} \quad \text{and} \quad F(z) = {(z-a)}{f(z)}.$$ Let $$E = \{p: f(p)=0\} \cup\{h(p): h'(p)=0\}$$ and suppose that $$a\notin E$$. Then how can I prove that $$N(r, a, h) ≤ N(r, 0, F') - [N(r, 0, F) - Ñ(r, 0, F)] \;?$$ Where $$N(r, a, h)$$ is the number of $$a$$-points of $$h$$ counting multiplicities, $$Ñ(r, 0, F)$$ denotes the number of zeros of $$F$$ without counting multiplicities while $$N(r, 0, F)$$ is the number of zeros of $$F$$ counting multiplicities; on $$|z| ≤ r$$.

As per my thinking, since $$F'(z) = 0$$ whenever $$h(z) = a$$ , so $$N(r, a, h) = N(r, 0, F')$$ But how can I prove the inequality from here?

$$N(r, 0, F) - Ñ(r, 0, F)$$ is the counting function of the multiple zeros of $$F$$, where a $$k$$-fold zero ($$k \ge 2$$) is counted with multiplicity $$k-1$$. It follows that $$N(r, 0, F') - [N(r, 0, F) - Ñ(r, 0, F)]$$ is the counting function of the zeros of $$F'$$ which are not zeros of $$F$$.
Therefore one has to show that every zero of $$h-a$$ is a zero of $$F'$$ (with at least the same multiplicity) but not a zero of $$F$$.
$$F'(z) = (z-a)f'(z) + f(z) = f'(z) (h(z) - a)$$ shows that a zero of $$h-a$$ is a zero of $$F'$$ with at least the same multiplicity. So it remains to show that $$h(z) = a$$ implies $$F(z) \ne 0$$.
If $$F(z) = 0$$ then $$z=a$$ or $$f(z) = 0$$, but not both (since $$f(a) \ne 0$$ is assumed). In both cases $$h(z) \ne a$$, and that finishes the proof.