171k views
0 votes
Prove Euler's identity using Euler's formula.
e^ix = cos x + i sin x

User CHANist
by
4.5k points

1 Answer

5 votes

First list all the terms out.

e^ix = 1 + ix/1! + (ix)^2/2! + (ix)^3/3! ...

Then, we can expand them.

e^ix = 1 + ix/1! + i^2x^2/2! + i^3x^3/3!...

Then, we can use the rules of raising i to a power.

e^ix = 1 + ix - x^2/2! - ix^3/3!...

Then, we can sort all the real and imaginary terms.

e^ix = (1 - x^2/2!...) + i(x - x^3/3!...)

We can simplify this.

e^ix = cos x + i sin x

This is Euler's Formula.

What happens if we put in pi?

x = pi

e^i*pi = cos(pi) + i sin(pi)

cos(pi) = -1

i sin(pi) = 0

e^i*pi = -1 OR e^i*pi + 1 = 0

That is Euler's identity.

User Ahmetunal
by
5.9k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.