f(x) = g(x) + i h(x) if g(x) is real even function, h(x) is real odd function how about the root of f(x)=0

Question

I want to know whether f(x)=0 always has complex root if real part of f(x) is a real even function and imaginary part of f(x) is a odd function.

Thank you very much!

Answer 1

Correct me if the following assumptions are wrong:

• $g,h : \mathbb{R} \rightarrow \mathbb{R}$
• $f: \mathbb{R} \rightarrow \mathbb{C}$

Otherwise, I don't quite know what definition of even / odd you are using since parity is a rather strange concept in its extension to higher dimensions.

You're question is asking if you are guaranteed a $c \in \mathbb{R}$ such that $$f(c) = g(c) + i h(c) = 0$$

This clearly boils down to a system: $$\cases{g(c) = 0 \\ h(c) = 0}$$

The problem with this is that it's now relatively easy to generate a counter-example where $g$ is even, $h$ is odd, yet they share no $c \in \mathbb{R}$ where they are both simultaneously 0. Specifically, consider $g(x) = x^2 + 1$ and $h(x) = x^3$.

So if you want some guarantees on an existence of a zero, you will need some more conditions. Specifically, if you required it to be holomorphic, that might actually give you what you want.