If replacing i by -i doesn't change a complex valued function, then the function output is purely real - What's proof/intuition behind this?

Question

Consider below complex function $$H(\omega) = \dfrac{1}{i\omega} \left(e^{i\omega} - e^{-i\omega}\right)$$

If I replace $i$ by $-i$ in $H(\omega)$, I get back the same $H(\omega)$.
Easy to see that $H(\omega) = \dfrac{2}{\omega}\sin(\omega t)$.

Long back I heard somebody claim this, but I couldn't pursue further... Now in another topic(signals and systems), this exact property is being used again.

I feel this has something to do with flip transform. Like, f(-x) flips the graph of f(x) around y axis. Since the functions see the opposite x values, so the graphs of them flip around y axis. If f(-x) = f(x), then the function is symmetrical around y axis and we call it an even function. Hmm... I couldn't connect this to the complex domain. Help appreciated..

EDIT : Here $\omega$ is a real number (angular frequency)

Answer 1

Simply observe in general that

$$z=x+iy$$

$$\bar z=x-iy$$

therefore

$$z=\bar z \implies y=0$$

In your example, assuming $\omega \neq 0$ to be real and defining

$$H(\omega)=F(i\omega)$$

we have

$$F(i\omega) = \dfrac{1}{i\omega} \left(e^{i\omega} - e^{-i\omega}\right)=\overline{F(i\omega)}=F(\overline{i\omega})=F(-i\omega)$$

therefore $H(\omega)=F(i\omega)$ is real.

Answer 2

Suppose you have an arbitrary complex valued function $F : A \to \mathbb{C}$ where $A$ is the domain of $F$, such that replacing $i$ by $-i$ doesn't change the value of $F$. Since every complex number can be written in the form $a+ib$ for some real numbers $a$ and $b$, for each $x \in A$ there are two real numbers $y_1, y_2$ such that $$F(x) = y_1 + iy_2$$ but replacing $i$ by $-i$ doesn't change $F(x)$ and so it follows $$F(x) = y_1 - iy_2$$ $$2F(x) = F(x) + F(x) = y_1 + iy_2 + y_1 - iy_2 = 2y_1$$ and so $$F(x) = y_1.$$

For your $H$ function I'm assuming that the domain is a subset of $\mathbb{R}$ so that $\omega$ is a real number, otherwise you need to replace $\omega$ by its conjugate $\bar{\omega}$ when you replace $i$ by $-i$ (in which case the function $H$ would not be equal to it's own conjugate so it wouldn't be real valued).

Also note that I've taken a particular interpretation of what you mean by "replacing i by −i". To be specific, I interpreted it as to apply the complex conjugate map $\psi : \mathbb{C} \to \mathbb{C}, \psi(x) = \bar{x}$ to $F(x)$ so $\psi(F(x)) = F(x)$. If you meant something different my answer may not be correct.