1
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ I think you are asking if $f(\bar z) = \overline{f(z)}$ for all functions $f$. That's false. $\endgroup$ – jjagmath Sep 26 at 16:10
  • $\begingroup$ It is not clear to me what you are asking. Can you please clarify your question? Can you give an example of what you are looking for? $\endgroup$ – Xander Henderson Sep 26 at 16:44
  • 1
    $\begingroup$ I think the hidden question might be this: "Suppose $f(a + bi)$, where $a, b \in \Bbb R$, is written in the form $f(a + bi) = g(a,b) + h(a, b)i$, where $g$ and $h$ are real-valued functions of two real arguments. If we write $u(a + bi) = g(a, b) - h(a, b)i$, will $u$ be the complex conjugate of the function $f$?" Of course, once it's written this clearly, the answer is evidently "yes". $\endgroup$ – John Hughes Sep 26 at 18:58
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ You did it wrong. According to your definition of g(a+bi), it should be -ai-b only by replacing i by -i. So, it works here. $\endgroup$ – Sai Krishna Garlapati Sep 26 at 16:27
  • 1
    $\begingroup$ My definition of $f$ is $f(z) = iz$. There's exactly one $i$ in that expression. I negated it to get my definition of $g$. If that's not what "just replace the $i$ by $-i$ everywhere" means to you, then our problem is with English rather than mathematics. But you seem to already know the answer you want, so I'll just tune out here. $\endgroup$ – John Hughes Sep 26 at 16:38
  • $\begingroup$ I expect the definition of g(a+bi) to be -i * zbar since I mentioned in the question that 'replacing i by -i everywhere'. So, if we define it like that does that work? $\endgroup$ – Sai Krishna Garlapati Sep 26 at 16:41
  • 1
    $\begingroup$ Ah...so your operation on $f$ changes depending on the way in which $f$ is written, even though the different forms define exactly the same function. That's ill-defined, so ... you need to think more about expressing your question clearly and unambiguously. $\endgroup$ – John Hughes Sep 26 at 18:55
  • 1
    $\begingroup$ BTW: saying "You did it wrong" isn't a very good way to get folks to want to help you; I say that as someone who's provided more than 3000 answers at this point. $\endgroup$ – John Hughes Sep 26 at 19:38
1
$\begingroup$

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 $z.$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ No, my question was not that. I have asked whether replacing i by -i 'everywhere', which does not mean only zbar. The function f may have 'i's at some other places as well. $\endgroup$ – Sai Krishna Garlapati Sep 26 at 16:29

Not the answer you're looking for? Browse other questions tagged or ask your own question.