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

[Brewer]

This is a question that's stumping both myself, and my friends who are on maths degrees!

So...

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

so does that make its conjugate \frac{1}{2}(e^{-ix}+e^{ix}), 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

 

[D H]

You are assuming that e^{iz} and e^{-iz} are conjugates of each other. This is only true if z is pure real.

If z=x+iy, (x,y)\in\mathbb R\times \mathbb R), then e^{iz}=e^{-y}e^{ix} and e^{-iz}=e^{y}e^{-ix}. Taking the conjugates, e^{iz^\ast}=e^{-y}e^{-ix} \ne e^{-iz} and e^{-iz^\ast}=e^{y}e^{ix} \ne e^{iz}.

 

[HallsofIvy]

This is a question that's stumping both myself, and my friends who are on maths degrees!

So...

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

so does that make its conjugate \frac{1}{2}(e^{-ix}+e^{ix}), 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 \frac{1}{2}(e^{ix}+e^{-ix}) 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]

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]

In order that your equation \frac{1}{2}(e^{ix}+e^{-ix}) 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 that's the reasoning I came up with, and the other thought was conceived by 2 maths students!

 

[D H]

In order that your equation \frac{1}{2}(e^{ix}+e^{-ix}) be correct, x must be a real number ...

Halls, you should know better!

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