Key to "How can you prove eπi+1=0?"

osnomu3

Answered question

2022-01-17

How can you prove

e^{π i} + 1 = 0 ?

Ella Williams

Beginner2022-01-18Added 28 answers

e^{x} = \sum_{n = 0}^{\infty} \frac{x^{n}}{n!}

e^{x} = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!} + \dots

e^{i x} = 1 + i x + \frac{{(i x)}^{2}}{2} + \frac{{(i x)}^{3}}{3!} + \frac{{(i x)}^{4}}{4!} + \frac{{(i x)}^{5}}{5!} + \frac{{(i x)}^{6}}{6!} \dots

e^{i x} = (1 + \frac{{(i x)}^{2}}{2} + \frac{{(i x)}^{4}}{4!} + \frac{{(i x)}^{6}}{6!} + \dots) + (i x + \frac{{(i x)}^{3}}{3!} + \frac{{(i x)}^{5}}{5!} + \frac{{(i x)}^{7}}{7!} + \dots)

e^{i x} = (1 - \frac{x^{2}}{2} + \frac{x^{4}}{4!} - \frac{x^{2}}{6!} + \dots) + i (x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + \dots)

e^{i x} = \cos x + i \sin x

e^{i π} = \cos π + i \sin π

e^{i π} = - 1 + 0

e^{i π} + 1 = 0

Andrew Reyes

Beginner2022-01-19Added 24 answers

Taking the derivative of the function

f (x) = \cos x + i \sin x

we obtain

f^{'} (x) = - \sin x + i \cos x = if (x)

and we have

f (0) = 1

. Solving the differential equation gives

f (x) = e^{i x}

.
And

\cos x + i \sin x = e^{i x}

immediately gives

e^{i π} + 1 = 0

alenahelenash

Expert2022-01-24Added 556 answers

Well known Euler's formula is writen as

e^{i θ} = \cos θ + i \sin θ, w h e n θ = π

, we can write

e^{i π} = \cos π + i \sin π

, we know that

\cos π = - 1 and \sin π = 0

, hence

e^{i π} = \cos π + i \sin π \Rightarrow e^{i π} = - 1 + i \cdot 0

\Rightarrow e^{i π} \Rightarrow e^{π i} = - 1 + 0

\Rightarrow e^{π i} + 1 = 0

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