# Simple equality help e^(-ix)

1. Sep 20, 2015

### RJLiberator

1. The problem statement, all variables and given/known data
Very simple problem I'm having:

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

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

3. 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?

Last edited by a moderator: Sep 21, 2015
2. Sep 20, 2015

### Ray Vickson

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

3. Sep 20, 2015

### 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.

4. Sep 20, 2015

### 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.

5. Sep 20, 2015

### MrAnchovy

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

6. Sep 20, 2015

### 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.

7. Sep 21, 2015

### 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 $e^{ix}$.