Question

e^iπ +1=0. pLs explain​

Answer 1

it is proved from complex no.

according to eüler's formula:-

$\cos( \alpha ) + i \sin( \alpha ) = {e}^{i \alpha }$

proof of above formula:

$z = cos \alpha + isin \alpha \\ iz = icos \alpha - sin \alpha \\ \frac{dz}{d \alpha } = icos \alpha - sin \alpha$

equating both:

$\frac{dz}{z} = id \alpha$

integrating both the sides we get

$ln(z) = i \alpha$

hence we get

${e}^{i \alpha } = cos \alpha + isin \alpha$

putting

$\alpha = \pi$

${e}^{ i\pi} = - 1$

${e}^{i\pi } + 1 = 0$

plz mark brainliest if anyone is able to do that. ......xD