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

1. Nov 19, 2006

Brewer

This is a question thats 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

2. Nov 19, 2006

D H

Staff Emeritus
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 $$e^{-iz^\ast}=e^{y}e^{ix} \ne e^{iz}$$.

Last edited: Nov 19, 2006
3. Nov 19, 2006

HallsofIvy

Staff Emeritus
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!

4. Nov 19, 2006

Brewer

Well in the example I'm doing this is the case.

5. Nov 19, 2006

Brewer

Good that makes me feel better, as thats the reasoning I came up with, and the other thought was conceived by 2 maths students!

6. Nov 19, 2006

D H

Staff Emeritus
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.