# If $f$ is entire with $f\not\equiv 0$, then $f'/f$ is meromorphic

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 $$Res\left(\frac{f'}{f},a\right)$$?

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

(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 $$f \not\equiv 0$$.)
• 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! Jun 17, 2021 at 10:46
• @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. Jun 17, 2021 at 10:47