# Complex conjugate of a complex function [closed]

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?

• I think you are asking if $f(\bar z) = \overline{f(z)}$ for all functions $f$. That's false. – jjagmath Sep 26 at 16:10
• 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? – Xander Henderson Sep 26 at 16:44
• 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". – John Hughes Sep 26 at 18:58

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.

• 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. – Sai Krishna Garlapati Sep 26 at 16:27
• 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. – John Hughes Sep 26 at 16:38
• 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? – Sai Krishna Garlapati Sep 26 at 16:41
• 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. – John Hughes Sep 26 at 18:55
• 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. – John Hughes Sep 26 at 19:38

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 $$z.$$

• 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. – Sai Krishna Garlapati Sep 26 at 16:29