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

by Brewer
Tags: complex, moment, numbers
 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 $$\frac{1}{2}(e^{ix}+e^{-ix})$$ 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
 Mentor P: 14,243 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 $$e^{-iz^\ast}=e^{y}e^{ix} \ne e^{iz}$$.
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P: 38,706
 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 $$\frac{1}{2}(e^{ix}+e^{-ix})$$ 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!

P: 230

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

 You are assuming that $e^{ix}$ and $e^{-ix}$ are conjugates of each other. This is only true if $x$ is pure real.
Well in the example I'm doing this is the case.
P: 230
 Quote by HallsofIvy 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 thats the reasoning I came up with, and the other thought was conceived by 2 maths students!
Mentor
P: 14,243
 Quote by HallsofIvy 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.

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