Complex conjugate of a complex function

Question

Does just replacing the $i$ ( $=\sqrt{-1}$ ) by $-i$ everywhere give the complex conjugate of any complex number of a function? Will that be the same as changing the sign of imaginary part of the finally computed complex value?

Answer 1

Let $f(z) = i z$. You're proposing writing $$ g(z) = (-i) z $$ and hoping that's the complex conjugate of $f$. Let's see it in parts. We have $$ f(a + bi) = i(a + bi) = ai -b = -b + ai\\ g(a + bi) = -ai + b = b - ai $$ But $\overline{-b + ai}$ is not $b - ai$, but is actually $-b - ai$.

So no, your proposed approach does not work, even for this very simple function.

Answer 2

I'm guessing what you mean is: Is $f(\overline z) = \overline {f(z)}$ for every function $f$ and every complex number $z.$ That is false. For example, suppose \begin{align} f(z)=i|z|. \\[8pt] \text{Then } f(i)= i|i| = i\cdot 1 & = i \\[4pt] \text{and } f(-i) = i\left|-i\right| = i\cdot1 & =i \ne -i. \end{align}

However if $f$ is differentiable on its domain and its domain is an open set in $\mathbb C$ and $f(z)$ is real whenever $f$ is real, then $f(\overline z) = \overline {f(z)}$ for every complex $z.$