Zeros of the derivative of a function are real

Question

I'm reviewing Complex Analysis from Ahlfors' book and stuck at this question.

"Show that if $f(z)$ is of genus $0$ or $1$ with real zeros, and if $f(z)$ is real for real $z$, then all zeros of $f'(z)$ are real. Hint: Consider $Im (f'(z)/f(z))$."

I feel like this is hard even with the simple case. Say $f$ has genus 0. THen the book proved that it must be of the form $$Cz^m\prod_1^\infty (1-\frac{z}{a_n})$$ with $\sum 1/|a_n|<\infty$. I have the form already, but I don't see how I can use the other hypothesis and see anything about the zeros of $f'$. So I'm stuck even in this case.

Answer 1

An entire function of genus $0,1$ is of the form $f(z)=cz^ke^{bz}\Pi_1^{\infty}{(1+t_kz)e^{-pt_kz}}$, where $p \in$ {$0,1$} is such that $\sum {|t_k|^{p+1}} < \infty$ (we rename the non zero roots as $z_k=-\frac{1}{t_k}$ as it will be useful for notation simplicity and if we have only finitely many roots, we take $t_k=0$ after that so those products are $1$ and do not change anything) and the genus is $1$ precisely when $b \ne 0$ or $p=1$.

Let also $Q_m(z)=cz^ke^{bz}\Pi_1^{m}{(1+t_kz)e^{-pt_kz}}$, so $Q_m \to f$ uniformly on compact sets; note that $Q_m$ is a polynomial precisely in genus $0$

Now using the hypothesis that $f(z)$ real when $z$ real and that $t_k$ real, we get $b,c$ real (as $ce^{bz}$ is then real for all $z$ real non-root etc)

Let $s_m=b-p(t_1+t_2+..+t_m)$ and $|n_m| > m|s_m|+2m, n_m \to \infty $

The crucial result we use is that if $P_m(z)=cz^k(1+\frac{s_mz}{n_m})^{n_m}(1+t_1z)...(1+t_mz)$ then $P_m(z)-Q_m(z) \to 0$ uniformly on compacts sets, so $P_m \to f$ uniformly on compact sets.

It follows that $P_m' \to f'$ locally uniformly. However as $P_m$ has real coefficients and has only real roots, $P_m'$ has only real roots by Rolle theorem, so by Hurwitz theorem, $f'$ has only real roots and we are done!

Note that for any $a \ge 0, (1-\frac{az^2}{m})^{m}$ has only real roots too and converges locally uniformly to $e^{-az^2}$ so we can extend the result to functions of the type $f(z)=cz^ke^{-az^2+bz}\Pi_1^{\infty}{(1+t_kz)e^{-pt_kz}}$ with the the other parameters as above.

Then the polynomial approximation theorem holds - $f$ entire (not identical zero to avoid trivialities) can be locally approxmiated by polynomials with real coefficients and only real roots if and only if it is as above (with $c \ne 0$)

Answer 2

We have

$$ f(z) = Cz^m e^{\alpha z} \prod_{n=1}^\infty \left(1 - \frac{z}{a_n} \right) e^{z/a_n}. $$

The condition $f(z)$ is real for real $z$ requires that $\alpha$ be real.

Taking the logarithmic derivative, we have

$$\frac{f'(z)}{f(z)}= \frac{m}{z}+ \alpha + \sum_{n=1}^\infty \frac{1}{z-a_n} + \frac{1}{a_n}. $$

Taking the imaginary part, we have

$$\text{Im} \frac{f'(z)}{f(z)}= -\left(\frac{m}{|z|^2}+\sum_{n=1}^\infty \frac{1}{|z-a_n|^2} \right) \text{Im} z.$$

The term inside the parentheses is strictly positive. Thus zeros of the derivative $f'(z)$ can only occur when $z$ is real.