Consider the function on the right hand side (RHS)
f(x) = cos( x ) + i sin( x )
Differentiate this function
f ' (x) = -sin( x ) + i cos( x) = i f(x)
So, this function has the property that its derivative is i times the original function.
What other type of function has this property?
A function g(x) will have this property if
dg / dx = i g
This is a differential equation that can be solved with seperation of variables
(1/g) dg = i dx
(1/g) dg = i dx
ln| g | = i x + C
| g | = ei x + C = eC ei x
| g | = C2 ei x
g = C3 ei x
So we need to determine what value (if any) of the constant C3 makes g(x) = f(x).
If we set x=0 and evaluate f(x) and g(x), we get
f(x) = cos( 0 ) + i sin( 0 ) = 1
g(x) = C3 ei 0 = C3
These functions are equal when C3 = 1.
cos( x ) + i sin( x ) = ei x
(This is the usual justification given in textbooks.)
By use of Taylors Theorem, we can show the following to be true for all real numbers:sin x = x - x3/3! + x5/5! - x7/7! + x9/9! - x11/11! + ...Knowing that, we have a mechanism to determine the value of ei, because we can express it in terms of the above series:
cos x = 1 - x2/2! + x4/4! - x6/6! + x8/8! - x10/10! + ...
ex = 1 + x + x2/2! + x3/3! + x4/4! + x5/5! + x6/6! + x7/7! + x8/8! + x9/9! + x10/10! + x11/11! + ...e^(i) = 1 + (i) + (i)2/2! + (i)3/3! + (i)4/4! + (i)5/5! + (i)6/6! + (i)7/7! + (i)8/8! + (i)9/9! + (i)10/10! + (i)11/11! + ...We know how to evaluate an imaginary number raised to an integer power, which is done as such:i1 = iWe can see that it repeats every four terms. Knowing this, we can simpliy the above expansion:
i2 = -1 terms repeat every four
i3 = -i
i4 = 1
i5 = i
i6 = -1
etc...e^(i) = 1 + i - 2/2! - i3/3! + 4/4! + i5/5! - 6/6! - i7/7! + 8/8! + i9/9! - 10/10! - i11/11! + ...It just so happens that this power series can be broken up into two very convenient series:Now, look at the series expansions for sine and cosine. The above above equation happens to include those two series. The above equation can therefore be simplified to
[1 - 2/2! + 4/4! - 6/6! + 8/8! - 10/10! + ...]
[i - i3/3! + i5/5! - i7/7! + i9/9! - i11/11! + ...]e^(i) = cos() + i sin()An interesting case is when we set = , since the above equation becomese^( i) = -1 + 0i = -1.which can be rewritten ase^( i) + 1 = 0. special casewhich remarkably links five very fundamental constants of mathematics into one small equation.
Again, this is not necessarily a proof since we have not shown that the sin(x), cos(x), and ex series converge as indicated for imaginary numbers.