IIT JEE , AIEEE Forum - Mathematics , Algebra

elastiboysai

And as for the proof of the Eulers formula here u go

using series expansions

$egin{align} i^0 &{}= 1, quad & i^1 &{}= i, quad & i^2 &{}= -1, quad & i^3 &{}= -i, i^4 &={} 1, quad & i^5 &={} i, quad & i^6 &{}= -1, quad & i^7 &{}= -i, end{align}$

and so on. The functions e^x, cos(x) and sin(x) (assuming x 2 b real ofc) can be expressed using their Taylor expansions around zero:

$egin{align} e^x &{}= 1 + x + rac{x^2}{2!} + rac{x^3}{3!} + cdots cos x &{}= 1 - rac{x^2}{2!} + rac{x^4}{4!} - rac{x^6}{6!} + cdots sin x &{}= x - rac{x^3}{3!} + rac{x^5}{5!} - rac{x^7}{7!} + cdots end{align}$

For complex z we define each of these functions by the above series, replacing x with z. This is possible because the radius of convergence of each series is infinite. We then find that

$egin{align} e^{iz} &{}= 1 + iz + rac{(iz)^2}{2!} + rac{(iz)^3}{3!} + rac{(iz)^4}{4!} + rac{(iz)^5}{5!} + rac{(iz)^6}{6!} + rac{(iz)^7}{7!} + rac{(iz)^8}{8!} + cdots &{}= 1 + iz - rac{z^2}{2!} - rac{iz^3}{3!} + rac{z^4}{4!} + rac{iz^5}{5!} - rac{z^6}{6!} - rac{iz^7}{7!} + rac{z^8}{8!} + cdots &{}= left( 1 - rac{z^2}{2!} + rac{z^4}{4!} - rac{z^6}{6!} + rac{z^8}{8!} - cdots ight) + ileft( z - rac{z^3}{3!} + rac{z^5}{5!} - rac{z^7}{7!} + cdots ight) &{}= cos (z) + isin (z) end{align}$

The rearrangement of terms is justified because each series is convergent. Taking z = x to be a real number gives the original identity as Euler discovered it.

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