This identity can be obtained or demonstrated by a method of developing functions in series of powers:
The function of: Sen(x)=x-x^3/3!+x^5/5!-x^7/7!+⋯(-1)^(n-1) x^(2n-1)/((2n-1) )
The function of: Cos(x)=1-x^2/2!+x^4/4!-x^6/6!+⋯(-1)^(n-1) x^(2n-2)/((2n-2) )
And what is exponential: e^x=1+x/1!+x^2/2!+x^3/3!+⋯+x^(n-1)/(n-1)!+⋯,-∞<x<∞
We see that the function "e" is elevated to a variable x, if x equals the number iπ it would be x=iπ
Then we have "e^iπ"
And the terms of the exponential series are as follows
e^xπ=1+iπ+(i^2 π^2)/2!+(i^3 π^3)/3!+(i^4 π^4)/4!+(i^5 π^5)/5!
Complex units that have an exponent greater than one can be replaced
e^xπ=1+iπ+(i^2 π^2)/2!+(i^3 π^3)/3!+(i^4 π^4)/4!+(i^5 π^5)/5!
Since i^2=-1 we can replace
e^xπ=1+iπ-π^2/2!-(iπ^3)/3!+π^4/4!+(iπ^5)/5!…
After this we can reorder the formula
e^xπ=1-π^2/2!+π^4/4!+iπ-(iπ^3)/3!+π^5/5!-…
In addition, we can group terms
e^xπ=(1-π^2/2!+π^4/4!-…)+i(π-(iπ^3)/3!+π^5/5!-…)
We note that we have factored i
Looking like: e^xπ=(1-π^2/2!+π^4/4!-...)+i(π-(iπ^3)/3!+π^5/5!-...)
To simplify, we can do this:
e^iπ=cos(π)+i sen (π)
Furthermore, we know that sen (π)=0 and cos(π)= -1, so
e^iπ=-1+0
Finally ordering is
e^iπ+1=0
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