# Properties of analytic functions with no real roots?

Suppose one has an entire complex analytic function $$f(z)$$ having no zeros on the real axis. Is it possible to find an analytic function $$g(z)$$ such that the coefficients of the power series expansion of $$f(z)g(z)$$ in $$z$$ are always real and positive (so that, e.g. "Descartes' rule of signs" can be used to "explain" the absence of zeros)? How can one find $$g(z)$$?

This is attempting be a kind-of generalization of the question Property of a polynomial with no positive real roots to complex analytic functions. The goal is to "explain" why $$f(z)$$ has no real roots, by studying the properties of $$f(z)g(z)$$ (or obtaining insights into $$f(z)$$ by studying the properties of $$g(z)$$). Are there (non-trivial) theorems that apply to complex analytic functions with no real zeros? Some Fredholm-alternative-like thingy?

• What are "the coefficients" of an analytic function? Are you talking about the power series at some specific point? Also, on what domains are these analytic functions defined? If $f$ is analytic at $x_0$ and $f(z_0)\ne0$, then on a small enough disc around $z_0$, one can choose $g$ such that $f(z)g(z)$ is any analytic function you want. Dec 26, 2018 at 7:27
• It's entire; the power series at $z=0$, as "usual". I'm not trying to be tricky, here. I don't want "any" analytic function; I want to understand why there are no zeros on the real axis. Dec 26, 2018 at 7:58

Let $$g(z)=\overline{f(\overline z)}$$. Then, regardless of whether $$f$$ has any zeroes on the real axis, the function $$fg$$ is real on the real axis, so all its derivatives at the origin are real.
I'm mystified what it might mean to explain why a function has no zeroes on the real axis, or what sort of "explanation" you're hoping for. But the existence of $$g$$ so that all the coefficients of $$fg$$ are real has absolutely nothing to do with whether $$f$$ has any zeroes on the real axis.
• Yes, you are right. My question arises from some late-night insanity on my part; I guess the correct protocol is to accept your answer. What I actually have is a function that has no zeros on any ray that is a rational multiple of $2\pi$. I was imaging that I could rotate each ray so that it lies along the real axis ... and deduce something clever. Dec 26, 2018 at 17:19