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I'm having (another) thick moment (complex numbers) 
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#1
Nov1906, 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 


#2
Nov1906, 09:23 AM

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P: 15,067

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]. 


#3
Nov1906, 09:24 AM

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#4
Nov1906, 09:25 AM

P: 230

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



#5
Nov1906, 09:26 AM

P: 230




#6
Nov1906, 09:39 AM

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P: 15,067

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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