New user? Sign up

Existing user? Sign in

\[ e^{i \pi } + 1 = 0 \]

is the great equation given by “EULER”.

Note by Hemanth Koundinya 2 years, 2 months ago

Easy Math Editor

*italics*

_italics_

**bold**

__bold__

- bulleted- list

1. numbered2. list

paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)

> This is a quote

This is a quote

# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

\boxed{123}

Sort by:

Complex number can be exponentially expressed as e^ix =cosx+isinx

Therefore e^ipi=cos(pi)+i sin(pi)

Since sin(pi)=0 Cos(pi)=-1

the above equation

e^i(pi)+1. => -1+1. => 0

Log in to reply

Can you prove it?

You can see the brilliant wiki on this.

Please provide a link .

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestComplex number can be exponentially expressed as e^ix =cosx+isinx

Therefore

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

Since sin(pi)=0 Cos(pi)=-1

the above equation

e^i(pi)+1. => -1+1. => 0

Log in to reply

Can you prove it?

Log in to reply

You can see the brilliant wiki on this.

Log in to reply

Please provide a link .

Log in to reply