Complex Conjugates: Replacing i & Taking Alpha's Conjugate?

[dyn]

Hi. If I have a complex number αe ^iα where α is complex what is the conjugate ? Usually I just replace i with -i but do i also take the conjugate of all α's ?
Thanks

 

[mfb]

There are rules how to take the complex conjugate of a product of complex numbers, for example. If you don't know the rules you can look them up or derive them yourself (e.g. by writing $\alpha = \beta+i \gamma$ with real $\beta$ and $\gamma$).

 

[dyn]

I get the complex conjugate of αe ^{i α} as α^*e^{- i α*}
Is that right ?

 

[andrewkirk]

Is that right ?

Not quite. Replacing $e^{i\alpha}$ by $e^{-i\alpha}$ only gives the conjugate if $\alpha$ is real. Try writing $\alpha=b+ic$ and then work out the conjugate of $e^{b+ic}$.

 

[mfb]

Try writing $\alpha=b+ic$ and then work out the conjugate of $e^{b+ic}$.

Or directly find the complex conjugate of $e^{i(b+ic)}$

 

[dyn]

I want to emphasize this is not homework. I self-study at home. I do not get homework.
I don't think my previous answer was very clear. So I will try again. I think the complex conjugate of ze^iz is z*e^-iz*.
Is that correct ?
Thanks

 

[Mark44]

I want to emphasize this is not homework. I self-study at home. I do not get homework.

Even if you are not in a formal class, if the problem is from a textbook or is a textbook-style problem, it should be posted in one of the Homework & Coursework sections, not here in the technical math section.

 

[dyn]

I want to emphasize this is not homework. I self-study at home. I do not get homework.
I don't think my previous answer was very clear. So I will try again. I think the complex conjugate of ze^iz is z*e^-iz*.
Is that correct ?
Thanks

Could somebody please tell me if my answer is correct for the complex conjugate ?

 

[mfb]

andrewkirk did in post 4. It is not correct.

 

[dyn]

Not quite. Replacing $e^{i\alpha}$ by $e^{-i\alpha}$ only gives the conjugate if $\alpha$ is real. Try writing $\alpha=b+ic$ and then work out the conjugate of $e^{b+ic}$.

I have replaced iz in the exponential by - i times the conjugate of z. If that is not right can someone please tell what the right answer is ?

 

[mfb]

What is the complex conjugate of $e^{a+ib}$ with real a and b?
Can you bring $i \alpha$ in this shape?

 

[dyn]

conjugate of e^a+ib is e^a-ib
conjugate of e^i(a+ib) is e^-i(a-ib)
So the conjugate of e^iz is e^-iz* ?
So all i's are replaced by -i and and all complex numbers by their complex congates ?

 

[Mark44]

conjugate of e^a+ib is e^a-ib
conjugate of e^i(a+ib) is e^-i(a-ib)

What is the simplified form of the last expression above?

So the conjugate of e^iz is e^-iz* ?
So all i's are replaced by -i and and all complex numbers by their complex congates ?

 

[mfb]

conjugate of e^i(a+ib) is e^-i(a-ib)

It is not. Consider a=0, b=1 for example.

Edit: See below.

 

[dyn]

It is not. Consider a=0, b=1 for example.

If a=0 and b=1 then e^i(a+ib) is a real number e^-1 so the number and its complex conjugate are the same. I have no idea why I keep being told I am wrong. Can someone please tell me what the complex conjugate of ze^iz is ?

 

[Mark44]

One of the properties of the complex conjugate (alluded to in post #2) is that $\overline{zw} = \bar z \cdot \bar w$.
So, $\overline{ze^{iz}} = \bar z \cdot \overline{e^{iz}}$

Although $\overline{e^{ia}} = \overline{e^{-ia}}$ if a is real, this formula isn't applicable if a is a complex number. As already suggested, replace z in the expression $e^{iz}$ by a + bi, and find the conjugate of $e^{iz}$.

 

[FactChecker]

I think that your answer is correct. Setting z = a +ib where a & b are real, I get:
(e^i(a+ib))^* = e^-be^-ia
and
e^-i(a+ib)* = e^-be^-ia

So (e^iz)^* = e^-iz*
Using this in your problem gives your result.

 

[dyn]

I got an answer from a lecturer at a top university who told me just replace i with -i and z with z* which is what i had been saying all along. I'm not sure why some people disagreed with this

 

[mfb]

Sorry, I missed a minus sign in your previous post.
Your answer is right.

 

[dyn]

Sorry, I missed a minus sign in your previous post.
Your answer is right.

Thank you. I was beginning to think I was going mad !

 

[FactChecker]

I got an answer from a lecturer at a top university who told me just replace i with -i and z with z* which is what i had been saying all along. I'm not sure why some people disagreed with this

My initial reaction was that there could not be an identity that simple that was not well known. I was surprised by it. That is probably why most people, including myself, assumed that it was not true.