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We know that the complex conjugate of a complex number is the number with an equal real part and an imaginary part equal in magnitude but opposite in sign. 

So, given a certain complex number, it is possibile to find its conjugate by writing it as:

Z = Re {Z} + j Im {Z}

and by considering:

Z* = Re {Z} - j Im {Z}

But in many applications (ex: signal theory etc) I saw people apply this rule: you have to replace "j" with "-j". Of course in case Z is written as shown before, it works. But in general?

For instance:

Z = (exp(4j)+sqrt(17j))/(exp(6j))

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  • $\begingroup$ Just verify whether it is the case for the expression in the bottom. I guess it is not. If it is not, you have found a counterexample and have shown that the rule does not hold in general. $\endgroup$
    – Peter
    Nov 3, 2019 at 11:29
  • $\begingroup$ Note that $e^{ix}=\cos x+i\sin x$ and $e^{-ix}=\cos(-x)+i\sin(-x)=\cos x-i\sin x$, so "replace $i$ with $-i$" works for such exponentials. The trouble with $\sqrt i$ is that first you have to decide which of the two solutions of $x^2=i$ you mean by that notation. $\endgroup$
    – Gerry Myerson
    Nov 3, 2019 at 12:10

2 Answers 2

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There are functions such that $(f(z))^*\neq f(z^*)$, for example, $$ f(z) = \mathrm{Re}z + \mathrm{Im}z.$$ However, if $f$ is analytic then the trick always works, i.e., $(f(z))^* = f(z^*)$. In your specific example the outcome will depend on how you choose to interpret the square root.

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If you have an expression of the form $$ A + Bj, $$ where $A$ and $B$ are both real, then its conjugate is $$ A + B(-j) = A - Bj. $$

If your expression has any other form -- involving quotients, for instance, then the rule above does not apply, and may lead to an incorrect answer, as your example shows. (It may also lead to a correct answer, but it's not guaranteed to do so, and guarantees are what you need when doing mathematics.)

It's important, when you learn a trick like "just negate the j", that you also learn the context and assumptions of the trick. That's why theorems in mathematics all have hypotheses --- the things you have to check before applying the conclusion.

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  • $\begingroup$ The conjugate of $2^{17i}$ is $2^{-17i}$, even though $2^{17i}$ isn't written in the form $A+Bi$. Of course, it can be rewritten in the form $A+Bi$, but that just means that the real question is, which expressions can be rewritten in that form. $\endgroup$
    – Gerry Myerson
    Nov 3, 2019 at 21:05
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    $\begingroup$ I carefully did not say that the rule only worked when the number was written in that form, but looking back at what I wrote, I realize that one could easily read it that way, so I've edited slightly. Thanks for pointing that out, Gerry. $\endgroup$
    – John Hughes
    Nov 3, 2019 at 21:21

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