Verified answer

OK, Take a deep breath:

One proves in elementary calculus (it's an easy

application of MacLaurin's formula) that for all real x

e^x = 1 + x/1! + x²/2! + x³/3! + x⁴/4! +...

cosx = 1 - x²/2! + x⁴/4! - x⁶/6! +...

sinx = x - x³/3! + x⁵/5! -... .

(I suppose you know that n!, read "n factorial", stands for the

product of all integers from 1 to n. Thus 4! = 1·2·3·4 = 24)

Formally DEFINING e^x, sinx and cosx for imaginary x by those formulas

(there are, after all, no imaginary angles to take their sine and cosine!)

we get the famous formula e^(it) = cost + i·sint

(just substitute it for x in e^x and separate real from purely imaginary terms,

remembering that even powers of i are real and odd powers imaginary).

If t=π, we get e^(iπ) = cosπ+isiniπ = -1+ι0 = -1, an amazing result indeed!

One of the easiest ways of explaining this is by using Taylor expansions of e^x, sin(x) and cos(x). Those are ways of approximating these functions by polynomials. It can be shown that 1 + x + x^2/2 + x^3/6 + ... + x^n / (n!) + ... will converge to e^x. The function sin(x) can be represented as x - x^3/6 + x^5/5! - x^7/7! + ... The function cos(x) can be represented by 1 - x^2/2 + x^4/4! - x^6/6! + ...

Now, what happens if we take e(ix) ? We get 1 + ix - x^2/2 - ix^3/6 + x^4/4! ...

This is exactly equal to cos(x) + i sin(x). Fill in x = π, and you get cos(π) + i sin(π) = -1 + i*0 = -1.

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In forth answer angle is in radians. Using first and second wonder limits and ading i as common number one can prove this very well.

e^(iπ) = 1/2(e^(iπ) + e^(-iπ)) + 1/2(e^(iπ) - e^(-iπ)) = cos π + i sin π

= -1

e is the limiting value as n tends to infinity of (1 + 1/n)^n which is approx. 2.71828182846

Recall Euler's identity;

e^(iθ) = cos(θ) + isin(θ)

e^(i*π) = cos(π) + isin(π) = -1