Question

Can soneone send a simplified proog of
e^iπ+1=0 best answer will receive the branliest answer

Answer 1

Since, e^(iθ) = cos(θ) + i sin(θ)

so we get, e^iπ = cosπ + i sinπ

and sin(nπ) = 0 where n is any whole number

therefore, sin(π) = 0

and cos((2n+1)π)= -1 => cos(π)= -1

thus we get, 
e^iπ+1=0

Answer 2

I have quite a few good proofs of Euler's identity but I will give you the easiest one.

For this we will first prove Euler's Formula(I would use the Calculus approach, you can use Taylor series expansions as well.)

First let there be a complex function 'z' such that:

$z = \cos(x) + i \sin(x)$

The function here depends on x and 'i' is the imaginary unit.

Differentiating the function,
$dz = - \sin(x) + i \cos(x) \: dx \\ dz = {i}^{2} \sin(x) + i \cos(x) \: dx \\ dz = i(i \sin(x) + \cos(x) ) \: dx \\ dz = iz \: dx \\ \frac{dz}{z} = i \: dx$

Now integrate both sides,

$\int \frac{dz}{z} = \int i dx$

$\log(z) = ix$

${e}^{ix} = z \\ {e}^{ix} = \cos(x) + i \sin(x)$
This is Euler's formula.

Now just replace
$x = \pi$
Then,

${e}^{i\pi} = \cos(\pi) + i \sin(\pi) \\ {e}^{i\pi} = - 1 + 0 \\ {e}^{i\pi} + 1 = 0$
Well this is precisely we wanted to prove!!