# Finding the conjugate of a complex number

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))

• 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. – Peter Nov 3 '19 at 11:29
• 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. – Gerry Myerson Nov 3 '19 at 12:10

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.
If you have an expression of the form $$A + Bj,$$ where $$A$$ and $$B$$ are both real, then its conjugate is $$A + B(-j) = A - Bj.$$
• 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. – Gerry Myerson Nov 3 '19 at 21:05