Does swapping $i \to -i$ in $f(x)$ always give $\overline{f(x)}$?

Question

I am a physics student. Often we wish to find the complex conjugate of a function , for example

$$ f: \mathbb{R}\to \mathbb{C}, f(x)=e^{ix},$$

To do this, we just replace all $i \to -i$. In the special case $f(x)=a(x)+b(x)i$ it is very clear that this is equivalent to conjugation. And for a particular function, like the one above, it is also easy to prove, so that is not my question. But is there a way to see that for arbitrary $f(x)$ or even $f(z)$, making the replacement $i \to -i$ always gives the complex conjugate?

Most valuable to me would be a "proof sketch" wherein I understand the reasoning, but maybe not every epsilon.

Answer 1

It does not work for $f(z)= iz$ where changing $i$ t0 $-i$ results in $-iz$ which is not the conjugate of $iz$

Another counter example is $f(z)=e^{iz}$ where changing $i$ to $-i$ results in $e^{-iz}$ which is not the conjugate of $e^{iz}$

Answer 2

Well, if you mean $f(\bar z)=\overline{f(z)}$, then, yes (at least for most well-behaved functions).

To start off, define $h(z)$ (where $z\in\mathbb C$) as being $=\bar z$ (for all $z$).

Now, suppose $u$ and $v$ are any two complex numbers with $u=a+bi$ and $v=c+di$.

Then:

$h(u+v)=h((a+bi)+(c+di))=h(a+c+(b+d)i)=a+c-(b+d)i=(a-bi)+(c-di)=h(a+bi)+h(c+di)=h(u)+h(v)$

And:

$h(uv)=h((a+bi)(c+di))=h(ac+adi+bci+bdi^2)=h(ac-bd+(ad+bc)i)=\\ ac-bd-(ad+bc)i=ac-bd-adi-bci=a(c-di)-b(d+ci)=a(c-di)-bi(c-di)=(a-bi)(c-di)=h(a+bi)h(c+di)=h(u)h(v)$

So we have, for any $u,v \in\mathbb C$: $\quad h(u+v)=h(u)+h(v)\quad$ & $\quad h(uv)=h(u)h(v)$

i.e. $\overline{u+v}=\bar u +\bar v\quad$ & $\quad\overline{uv}=\bar u\bar v$

So, suppose we have some polynomial $p(z)=a_0+a_1z+a_2z^2+...+a_nz^n$ where $a_0,a_1,...,a_n$ are all real (so that $\overline {a_i}=a_i$ for all i).

Based on what we have shown,

$p(\bar z)=a_0+a_1\bar z+a_2\bar z^2+...+a_n\bar z^n=a_0+a_1\bar z+a_2\overline {z^2}+...+a_n\overline {z^n}=\overline {a_0}+\overline {a_1z}+\overline {a_2z^2}+...+\overline{a_nz^n}=\overline {a_0+a_1z+a_2z^2+...+a_nz^n}=\overline {p(z)}$

Now, no part of this argument excluded $f$ from having an 'infinite' degree, so to speak. So you could apply this to any function with a Taylor Series expansion (at least, over the domain where the series was valid).

Edit: $f$ restricted to $\mathbb R$ must map back to $\mathbb R$ so that the coefficients of its Taylor series are real