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

• RJLiberator
In summary, the conversation discusses a simple problem of showing 1/e^(ix) = e^(-ix) and how to express 1/(a+ib) as A+iB. The solution involves multiplying the numerator and denominator by the complex conjugate and using trigonometric identities. The conversation also addresses the question of whether e^(-ix) is equal to e^(ix)*, and the need to prove this. The solution shows that e^(-ix) is the multiplicative inverse of e^(ix).
RJLiberator
Gold Member

## 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?

Last edited by a moderator:
RJLiberator said:

## 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.

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.

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

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.

Anama Skout
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 $e^{ix}$.

RJLiberator

## 1. What is the definition of "Simple equality help e^(-ix)"?

The simple equality help e^(-ix) refers to the mathematical expression that represents the equality between two complex numbers, where one is the exponential function e raised to the power of negative ix.

## 2. How is "Simple equality help e^(-ix)" used in scientific research?

Simple equality help e^(-ix) is commonly used in quantum mechanics and signal processing to represent the conversion between time and frequency domains. It is also used in Fourier analysis, where it helps to simplify complex integrals.

## 3. What is the significance of the letter "i" in "Simple equality help e^(-ix)"?

The letter "i" in "Simple equality help e^(-ix)" represents the imaginary unit, which is the square root of -1. It is used to denote the imaginary part of a complex number and is essential in representing complex numbers in mathematical equations.

## 4. Can "Simple equality help e^(-ix)" be simplified further?

No, "Simple equality help e^(-ix)" is already in its simplest form as the exponential function e^(-ix) cannot be further simplified. It is a fundamental mathematical expression and cannot be reduced to a simpler form.

## 5. What are the real and imaginary components of "Simple equality help e^(-ix)"?

The real component of "Simple equality help e^(-ix)" is cos(x), while the imaginary component is -isin(x). This means that the complex number represented by e^(-ix) has a real part of cos(x) and an imaginary part of -isin(x).

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