I'm having (another) thick moment (complex numbers)


by Brewer
Tags: complex, moment, numbers
Brewer
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#1
Nov19-06, 08:50 AM
P: 230
This is a question thats stumping both myself, and my friends who are on maths degrees!

So...

cos(x) can be written as [tex]\frac{1}{2}(e^{ix}+e^{-ix})[/tex] correct?

so does that make its conjugate [tex]\frac{1}{2}(e^{-ix}+e^{ix})[/tex], i.e. cos(x) again? or does the switching of the sign go in front of the e? Its been a long time since I used complex numbers, so I (and my friends) are a little rusty! Any help would be appreciated.

Thanks

Brewer
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D H
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#2
Nov19-06, 09:23 AM
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You are assuming that [tex]e^{iz}[/tex] and [tex]e^{-iz}[/tex] are conjugates of each other. This is only true if z is pure real.

If [tex]z=x+iy, (x,y)\in\mathbb R\times \mathbb R)[/tex], then [tex]e^{iz}=e^{-y}e^{ix}[/tex] and [tex]e^{-iz}=e^{y}e^{-ix}[/tex]. Taking the conjugates, [tex]e^{iz^\ast}=e^{-y}e^{-ix} \ne e^{-iz}[/tex] and [tex]e^{-iz^\ast}=e^{y}e^{ix} \ne e^{iz}[/tex].
HallsofIvy
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#3
Nov19-06, 09:24 AM
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Quote Quote by Brewer
This is a question thats stumping both myself, and my friends who are on maths degrees!

So...

cos(x) can be written as [tex]\frac{1}{2}(e^{ix}+e^{-ix})[/tex] correct?

so does that make its conjugate [tex]\frac{1}{2}(e^{-ix}+e^{ix})[/tex], i.e. cos(x) again? or does the switching of the sign go in front of the e? Its been a long time since I used complex numbers, so I (and my friends) are a little rusty! Any help would be appreciated.

Thanks

Brewer
In order that your equation [tex]\frac{1}{2}(e^{ix}+e^{-ix})[/tex] be correct, x must be a real number and then cos(x) is a real number. Is the complex conjugate of cos(x) equal to cos(x)? Of course it is: the complex conjugate of any real number is itself!

Brewer
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#4
Nov19-06, 09:25 AM
P: 230

I'm having (another) thick moment (complex numbers)


You are assuming that [math]e^{ix}[/math] and [math]e^{-ix}[/math] are conjugates of each other. This is only true if [math]x[/math] is pure real.
Well in the example I'm doing this is the case.
Brewer
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#5
Nov19-06, 09:26 AM
P: 230
Quote Quote by HallsofIvy
In order that your equation [tex]\frac{1}{2}(e^{ix}+e^{-ix})[/tex] be correct, x must be a real number and then cos(x) is a real number. Is the complex conjugate of cos(x) equal to cos(x)? Of course it is: the complex conjugate of any real number is itself!
Good that makes me feel better, as thats the reasoning I came up with, and the other thought was conceived by 2 maths students!
D H
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#6
Nov19-06, 09:39 AM
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Quote Quote by HallsofIvy
In order that your equation [tex]\frac{1}{2}(e^{ix}+e^{-ix})[/tex] be correct, x must be a real number ...
Halls, you should know better!

The equation [tex]\cos(x) = \frac{1}{2}(e^{ix}+e^{-ix})[/tex] follows directly from Euler's formula, [tex]e^{ix} = \cos(x) + i\sin(x)[/tex], which is valid for all real and complex x. Thus the given expression for [tex]\cos(x)[/tex] is valid for all real and complex x.


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