Proving 1/e^(ix) = e^(-ix) using Complex Conjugate Property

[RJLiberator]

Homework Statement

Very simple problem I'm having:

The problem is: Show 1/e^(ix) = e^(-ix)

Homework Equations

* is the complex conjugate.
e^(ix) = cos(x)+isin(x)

The Attempt at a Solution

RHS = $e^{-ix} = e^{ix^*} = (cos(x)+isin(x))^* = cos(x)-isin(x)$

LHS = $\frac{1}{e^{ix}} = \frac {1}{cos(x)+isin(x)} = cos(x)-isin(x)$

My question:
1) How does the LHS go from 1/(cos(x)+isin(x)) to cos(x)-isin(x)? I assume there is some trig property involved here?
2) Is it safe to state that e^(-ix)=e^(ix)* where * is the complex conjugate? Do I need to really prove anything here?

 

[Ray Vickson]

Homework Statement

Very simple problem I'm having:

The problem is: Show 1/e^(ix) = e^(-ix)

Homework Equations

* is the complex conjugate.
e^(ix) = cos(x)+isin(x)

The Attempt at a Solution

RHS = $e^{-ix} = e^{ix}^* = (cos(x)+isin(x))^* = cos(x)-isin(x)$

LHS = $\frac{1}{e^{ix}} = \frac {1}{cos(x)+isin(x)} = cos(x)-isin(x)$

My question:
1) How does the LHS go from 1/(cos(x)+isin(x)) to cos(x)-isin(x)? I assume there is some trig property involved here?
2) Is it safe to state that e^(-ix)=e^(ix)* where * is the complex conjugate? Do I need to really prove anything here?

For real $a$ and $b$, do you know how to express $1/(a + ib)$ as $A + iB$, where $A,B$ are also real?

 

[RJLiberator]

BOOM.

There it is. The connection.

We multiply numerator and denominator by the complex conjugate to see
cos(x)-isin(x) over cos^2(x)+sin^2(x) which we know by trig is just the numerator.

That iss Q1 cleared.

 

[Anama Skout]

1. That's because $$\frac{1}{\cos x+i\sin x}=\frac{\cos x-i\sin x}{(\cos x +i\sin x)(\cos x-i\sin x)}=\ldots$$ Then use the Pythagorean identity.
2. I would say: it isn't safe.

 

[pbuk]

$$ e^{-ix} = e^{(ix)(-1)} = (e^{ix})^{-1} = \frac 1{e^{ix}} $$

 

[RJLiberator]

Ah, Q2) isn't too hard to prove as well.

e^(-ix) = cos(-x)+isin(-x) = cos(x)-isin(x).

This is complete now.

Kind regards for your words of motivation.

 

[HallsofIvy]

I don't see any reason to change to the trig form $\left(e^{-ix}\right)\left(e^{ix}\right)= e^{I(x- x)}= e^0= 1$ is sufficient to prove that $e^{ix}$ is the multiplicative inverse of $e^{ix}$.