If an analytic function $f(z)$ has two zeros which are close together, there is a zero of $f^\prime(z)$ which is nearby. This is clear if $f(z)$ is a quadratic polynomial, and intuitive in general from the Hadamard product: $f(z)$ is a quadratic polynomial times a function which is nearly constant in the neighborhood of the zeros.
How rigorous can we make this statement?
Without loss of generality, we can translate so the origin is the midpoint of the two zeros, rotate so they are both real, and dilate so the zeros are $\pm1$, so $f(z)=(1-z^2)g(z)$, with (suppose) $g(z)\ne0$ on $|z|<1$. Thus $\log(g(z))$ is well defined on the disk, and we can write $g(z)=\exp(h(z))$ with analytic $h(z)$.
Thus we seek $z$ (hopefully in $|z|<1$) satisfying $-2z+(1-z^2)h^\prime(z)=0$.
Edit: Based on robjohn's comment, we need to quantify the heuristic above that the 'remainder' of the Hadamard product is roughly constant near the zeros. Thus assume also that $g^\prime(0)$ is 'small', where the proposed solution below makes this precise.