# The zeros of the derivative $f^{\prime}(z)$ are also real

Let $f(z)=e^{\ -\beta \ z^2}g(z)$ where $\beta \geq 0$ and $g(z)$ is a real entire function of genus $p\leq 1$ with real zeros. Prove that the zeros of the derivative $f^{\prime}(z)$ are also real and interlace with the of $f(z)$

suggestion: use hadamard theorem and note that $$\dfrac{f^{\prime}(z)}{f(z)}=-2\beta z+\dfrac{g^{\prime}(z)}{g(z)}$$

But I do not see how to take advantage of this, I hope you can guide me.

• @ hrelmonio: what do you mean by the zeroes interlace? – Robert Lewis Dec 18 '13 at 6:16
• There's a particular hint I'd like to give, but it would be best if you would say what you've tried so far so that I don't say something unhelpful. – Antonio Vargas Dec 18 '13 at 6:17
• How do you define the genius of the function $g$? – Mercy King Dec 18 '13 at 22:42
• – helmonio Dec 19 '13 at 18:04

This is only a partial answer for the case $p=0$
In this case $g(z)$ takes the form $$Cz^m\prod_{n=1}^\infty \left(1-\frac{z}{x_n} \right)$$ where $x_n \in \mathbb R$ are the zeros (multiple zeros being repeated), which satisfy $$\sum_{n=1}^\infty \frac{1}{|x_n|}<\infty.$$ We then have $$f(z)=Ce^{- \beta z^2} z^m \prod_{n=1}^\infty \left(1-\frac{z}{x_n} \right),$$ and as you said $$\frac{f'(z)}{f(z)}=-2 \beta z+ \frac{m}{z}+ \sum_{n=1}^\infty \frac{1}{z-x_n}.$$ Taking the imaginary part of this we find $$\text{Im} \frac{f'(z)}{f(z)}= \left[-2\beta-\frac{m}{x^2+y^2}-\sum_{n=1}^\infty \frac{1}{(x-x_n)^2+y^2} \right] y,$$ where $z=x+iy$. Thus zeros of the derivative $f'(z)$ can only occur when $y=0$, that is when $z$ is real. I will try to develop this solution further, but that's all I've got for now.