e^i(theta)=cis(theta) (idea)
This has been reffered to the most beautiful relation in mathematics. For those who have trouble reading it is e to the power of i times theta(where theta is a variable for the angle of the argument (angle)of a complex number) is equal to cos(theta)+isin(theta) or cis(theta). And cis(theta) is a very convenient way of expressing a complex number.
Now onto the proof: A long While back some irish mathematician named something MacLauren started analysing a function equal to a series A0 + A1x + A2x2 + A3x3 + .... He started finding the derivatives of such a function.
f'(x)= A1 + 2A2x + 3A3x2 + 4A3x3 + ...
f''(x)= 2!A2 + 3•2A3x + 4•3A4x2 + 5•4A5x3 + ...
f'''(x)= 3!A3 + 4•3•2A4x + 5•4•3A5x2 + 6•5•4A6x3 + ...
f''''(x)= 4!A4 + 5•4•3•2A5x + 6•5•4•3A6x2 + 7•6•5•4A7x3 + ...
...
Then he observed that if he knew f(0) and f'(0) and f''(0) and so on, he could solve the equation for f(x)
f(0)=A0 + A1x + A2x2 + A3x3 + ....
f'(0)= A1 + 2A2(0) + 3A3(0)2 + 4A3(0)3 + ...
f''(0)= 2!A2 + 3•2A3(0) + 4•3A4(0)2 + 5•4A5(0)3 + ...
f'''(0)= 3!A3 + 4•3•2A4(0) + 5•4•3A5(0)2 + 6•5•4A6(0)3 + ...
f''''(0)= 4!A4 + 5•4•3•2A5(0) + 6•5•4•3A6(0)2 + 7•6•5•4A7(0)3 + ...
f(0)=A0
f'(0)=1!A1
f''(0)=2!A2
f'''(0)=3!A3
f''''(0)=4!A4
Then he figured that A1=f'(0)/1! , A2=f''(0)/2!, A(3)=f'''/3! and so on, that An=f(0)n/n! where ! denotes the factorial notation. sp...
f(x)= f(0) + f'(0)x + f''(0)x2/2! + f'''(0)x3/3! + ... + f n(0)xn/n! + ...
This is function let's us figure out any function for which we know f(0) and all the derivatives of zero. Three functions for which we know this are sinx, cosx and e^x.
Briefly about the calculus of trigonometric functions
The derivative of f(x)=sinx is equal to cosx that is f'(x)=cosx, and the derivative of cosx is -sinx so f''(x)=-sinx and then the derivative of this; f'''(x)=-cosx and f''''(x)=sinx and then it repeats for further derivatives. (There is a rather convenient proof for this but it requires the aid of graphical representation to accurately convey the idea, and is difficult to prove in HTML.) So now sin 0=0 and cos 0=1 now inserting this into our above identity sinx= 0 + 1*x + 0*x^2/2! + (-1)x^3/3! + (-0)x^4/4! + ...now this can be rewritten as sinx= x - x^3/3! + x^5/5! - x^7/7! + x^9/9! + ... and cos x can be written as 1-x^2/2! + x^4/4! - x^6/6! + x^8/8! + ...
Now assuming that e is a number such that the derivative of e^x=e^x then we can find and any number to the power of zero equals one, then we can safely say that all derivatives of e^x for x=0 are 1. Again taking MacLauren's series.
e^x = 1 + x + x^2/2! + x^3/3! + x^4/4! + ...
now taking a number i such that i = ?-1 and putting it into the equation for e^x we get:
e^i = 1 + i +i^2/2! + i^3/3! + i^4/4! + i^5/5! + ...
well first off we know that for all even values of n in i^n/n! that i^n/n! will be real(will not utilize i). And seeing as i^2=-1 then i^4=1 and i^6=-1 we can conclude that all values of n that are only divisible by 2 and not 4 will be negative, and all values of n divisible by four will be positive. so looking at only even values of n in e^i, e^i= 1 + i^2/2! + i^4/4! + i^6/6! + i^8/8! + ...
and simplifying with our two theories e^i = 1- 1/2! + 1/4! - 1/6! + 1/8! + ... which looks startlingly familiar to our equation of cosx and indeed if we had ^ix then the real part would be equal to cosx.
As for the nonreal part; for all values of n that aren't divisible by two we know that the i's won't multiply out, and that this part is imaginary, and agaain using theories similar to those above we can state that for all values of n-1 divisble by 2 and not four, i^n/n! is negative, and for all values of n-1 divisible by 4, i^n/n! is positive.
e^i = 1 + i - 1/2! - i/3! + 1/4! + i/5! - 1/6! - i/7! + 1/8! + i/9! - ...
rearranged e^i = 1 - 1/2! + 1/4!- 1/6! +1/8! -... + i - i/3! + i/5! - i/7! + i/9! - ...
which looks suspiciously like cos(1) + isin(1) and indeed we could substitute x or any variable(such as theta) in and it will come out to cos(x) + isin(x) or cis(x).