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 \(\displaystyle e^{ix}\) and \(\displaystyle e^{-ix}\) are conjugates of each other. This is only true if \(\displaystyle x\) 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.