- #1

RJLiberator

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## 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 = [itex] e^{-ix} = e^{ix^*} = (cos(x)+isin(x))^* = cos(x)-isin(x)[/itex]

LHS = [itex] \frac{1}{e^{ix}} = \frac {1}{cos(x)+isin(x)} = cos(x)-isin(x)[/itex]

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?

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