Are the transfer functions resulting from $\frac{Y(s)}{X(s)}$ and $\frac{Y^*(s)}{X^*(s)}$ equal?

Question

If $\frac{Y(s)}{X(s)} = G(s)$ in the Laplace domain, does $\frac{Y^*(s)}{X^*(s)}$ also equal $G(s)$. Where $Y^*(s)$ and $X^*(s)$ are the complex conjugates of $Y(s)$ and $X(s)$.

Answer 1

No, it isn't totally equal. Their absolute values are equal, but their phases are inverted. Proof: $Y(s)$ and $X(s)$ are just complex numbers. Let us assume that $ Y(s) = a + ib $ and $ X(s) = c + id $ which implies that $ Y^*(s) = a - ib $ and $ X^*(s) = c - id $. Let us now find the transfer functions in terms of polar coordinates: $$ G_1(s) = \frac{Y(s)}{X(s)} = \frac{a+ib}{c+id} = \frac{\sqrt{a^2+b^2}e^{i\arctan(b/a)}}{\sqrt{c^2+d^2}e^{i\arctan(d/c)}} = \frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}} \cdot e^{i(\arctan(b/a)-\arctan(d/c))}$$ $$ G_2(s) = \frac{Y^*(s)}{X^*(s)} = \frac{a-ib}{c-id} = \frac{\sqrt{a^2+b^2}e^{-i\arctan(b/a)}}{\sqrt{c^2+d^2}e^{-i\arctan(d/c)}} = \frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}} \cdot e^{-i(\arctan(b/a)-\arctan(d/c))} $$ We see that their absolute values are the same whereas the phases are inverted. I hope that the proof will help you.