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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?

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$N(r, 0, F) - Ñ(r, 0, F)$ 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.

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