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Suppose $f$ an entire function with $f'(z)\ne0$ for all $z\in\Bbb C$. How can I see that $f'/f$ is meromorphic?

I know that if $f$ and $g$ are holomorphic with $g\not\equiv0$, then $f/g$ is meromorphic, so is it enough to prove that $f'$ is holomorphic? Or is there a different way to approach this problem?

Also, which would be the poles of $f'/f$? And if we consider $a$ a pole of $f'/f$, what would be the value of $Res\left(\frac{f'}{f},a\right)$?

Thanks in advance!

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    $\begingroup$ Hint: Zeros of $f$, if any, are all of order $1$. $\endgroup$ Jun 17, 2021 at 7:56
  • $\begingroup$ Oh, so the last part would be like this? $Res\left(\frac{f'}{f},a\right) = \lim_{z\to a} (z-a)\frac{f'(a)}{f(a)}$. $\endgroup$ Jun 17, 2021 at 8:23

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(I don't use the fact that $f'(z) \neq 0$ for all $z \in \Bbb C$ until the last Edit.)

It is a standard theorem that if $f$ is holomorphic, then $f$ is analytic. In turn, $f'$ is holomorphic. Thus, if $f \not\equiv 0$, then $f'/f$ is meromorphic. This answers your first part.

Now, since $f'$ is entire, the zeroes of $f$ are the only possible poles of $f'/f$. We now show that these are actually poles. In fact, we show that all of these are simple poles.

Let $a$ be a zero of $f$. Let us analyse $f'/f$ in a neighbourhood of $a$. Suppose that $k \ge 1$ is the order of the zero. Then, we have $$f(z) = (z - a)^kg(z)$$ for some holomorphic $g$ such that $g(a) \neq 0$.

Then, we have $$f'(z) = k(z - a)^{k - 1}g(z) + (z - a)^k g'(z).$$ Thus, we get $$\frac{f'(z)}{f(z)} = \frac{k}{z - a} + \frac{g'(z)}{g(z)}.$$ Since $g(a) \neq 0$, we see that $g'/g$ is holomorphic in a neighbourhood of $a$ and thus, $a$ is a simple pole of $f'/f$ with residue $k$.


Edit. Now, let us assume that $f'(z) \neq 0$ for all $z \in \Bbb C$. From this, it follows that $k = 1$. Indeed, recall that the order of the zero is the smallest positive integer such that $f^{(k)}(z) \neq 0$.

To conclude:

  • $f'/f$ is a meromorphic function.
  • The poles of $f'/f$ are precisely the zeroes of $f$.
  • All the poles of $f'/f$ are simple.
  • The residue at each pole is $1$. (Only this assumes that $f' \neq 0$. The earlier ones just assume that $f \not\equiv 0$.)
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  • $\begingroup$ From the comment of Kavi we know that the order of the zeros of $f$ is 1, so $k=1$, isn't it? For the rest, I don't see anything wrong. Thanks! $\endgroup$ Jun 17, 2021 at 10:46
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    $\begingroup$ @GreekCorpse: Yes, and that is the only place where we use that $f'(z) \neq 0$ for all $z \in \Bbb C$. In hindsight, I should add that. $\endgroup$ Jun 17, 2021 at 10:47

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