Question

Prove that :-

$\quad \qquad { \bigstar { \underline { \boxed { \tt { {e}^{iπ} + 1 = 0 } } } } }$ ​

Answer 1

I hope it's helpful to you

Answer 2

Prove :

$\sf \longrightarrow e ^{x} = 1 + x + \frac{ {x}^{2} }{2!} + \frac{ {x}^{3} }{3!} + \frac{ {x}^{4} }{4!} + ...$

$\sf \longrightarrow sin(x) = x - \frac{ {x}^{3} }{3!} + \frac{x ^{5} }{5!} - \frac{ {x}^{7} }{7!} + ...$

$\sf \longrightarrow cos(x) = 1 - \frac{ {x}^{2} }{2!} + \frac{ {x}^{4} }{4!} + \frac{ {x}^{6} }{6!} + ...$

Now, add sin(x) + cos(x)

• $\sf sin(x) + cos(x) = 1 + x - \frac{ {x}^{2} }{2!} - \frac{ {x}^{3} }{3!} + \frac{ {x}^{4} }{4!} + \frac{ {x}^{5} }{5!} - \frac{ {x}^{6} }{6!} - ...$

• $\sf e ^{ix} = 1 + ix + \frac{(ix )^{2} }{2!} + \frac{(ix)^{3} }{3!} + \frac{(ix) ^{4} }{4!} + ...$

$\sf \bigg ( i¹ = i \: \: , \: \: i ² = -1 \: \: , \: \: i ³ = - i \: \: , \: \: i ⁴ = 1 \: \: , \: \: i ⁵ = i \: \: , \: \: i ⁶ = - 1 \: \: \bigg)$ ____(1)

Simplify :

• $\sf e ^{ix} = 1 + ix + \frac{i ^{2}x ^{2} }{2!} + \frac{i ^{3} {x}^{3} }{3!} + \frac{i ^{4} {x}^{4} }{4!} + \frac{ {i}^{5} {x}^{5} }{5!} + \frac{i ^{6} {x}^{6} }{6!} + ...$

Now, According to (1)

• $\sf e ^{ix} = 1 + ix - \frac{ {x}^{2} }{2!} - \frac{ {ix}^{3} }{3!} + \frac{ {x}^{4} }{4!} + \frac{ix ^{5} }{5!} - \frac{ {x}^{6} }{6!} + ...$

• $\sf \sf {e}^{ix} = \bigg(1 - \frac{ {x}^{2} }{2!} + \frac{ {x}^{4} }{4!} - \frac{ {x}^{6} }{6!} + ... \bigg) + i \bigg( x - \frac{x ^{3} }{3!} + \frac{ {x}^{5} }{5!} + ... \bigg)$

• $\sf \bigg(1 - \frac{ {x}^{2} }{2!} + \frac{ {x}^{4} }{4!} - \frac{x^{6} }{6!} + ... \bigg) \longrightarrow$ Real Coverage With cos(x)

• $\sf \bigg(x - \frac{ {x}^{3} }{3!} + \frac{ {x}^{5} }{5!} + ... \bigg) \longrightarrow$ Imaginary Coverage With sin(x)

• $\sf e ^{ix} = \cos(x) + i \sin(x) : \longrightarrow \red{\underline{\underline{Euler's \: \: Formula}}}$

Now, put X = π

• $\sf {e}^{i\pi} = \cos(\pi) + i \sin(\pi)$

$\longrightarrow \sf \cos(\pi) = - 1$

$\longrightarrow \sf \sin(\pi) = 0$

• $\sf {e}^{i\pi} = \cos(\pi) + i \sin(\pi)$

• $\sf {e}^{i\pi} = - 1 + i(0)$

• $\sf e ^{i\pi} = - 1$

• $\sf e^{i\pi} + 1 = 0 : \longrightarrow \green{\underline{\underline{Euler's \: \: Identity}}}$

Hence, it proven.

I hope it helps you ❤️✔️