

Justifications that e^{i }
= cos() + i sin()
e^{i x} = COs( x ) + i sin( x )
Justification #1: from the derivative
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 separation of
variables
(1/g) dg = i dx
(1/g) dg =
i dx
ln g  = i x + C
 g  = e^{i x + C} = e^{C} e^{i x}
 g  = C_{2} e^{i x}
g = C_{3} e^{i x}
So we need to determine what value (if any) of the constant C_{3}
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) = C_{3} e^{i 0} = C_{3}
These functions are equal when C_{3} = 1.
Therefore,
COs( x ) + i sin( x ) = e^{i x}
Justification #2: the series method
(This is the usual justification given in textbooks.)
By use of Taylor's Theorem, we can show the following to be true for all
real numbers:
sin x = x  x^{3}/3! + x^{5}/5! 
x^{7}/7! + x^{9}/9!  x^{11}/11! + ...
COs x = 1  x^{2}/2! + x^{4}/4!  x^{6}/6!
+ x^{8}/8!  x^{10}/10! + ...
e^{x} = 1 + x + x^{2}/2! + x^{3}/3! +
x^{4}/4! + x^{5}/5! + x^{6}/6! + x^{7}/7!
+ x^{8}/8! + x^{9}/9! + x^{10}/10! + x^{11}/11!
+ ...
Knowing that, we have a mechanism to determine the value of e^{i},
because we can express it in terms of the above series:
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:
i^{1} = i
i^{2} = 1 terms repeat every four
i^{3} = i
i^{4} = 1
i^{5} = i
i^{6} = 1
etc...
We can see that it repeats every four terms. Knowing this, we can simplify
the above expansion:
e^(i) = 1 +
i  ^{2}/2!
 i^{3}/3! + ^{4}/4!
+ i^{5}/5!  ^{6}/6!
 i^{7}/7! + ^{8}/8!
+ i^{9}/9!  ^{10}/10!
 i^{11}/11! +
...
It just so happens that this power series can be broken up into two very
convenient series:
e^(i) =
[1  ^{2}/2!
+ ^{4}/4!  ^{6}/6!
+ ^{8}/8!  ^{10}/10!
+ ...]
+
[i  i^{3}/3!
+ i^{5}/5!  i^{7}/7!
+ i^{9}/9!  i^{11}/11!
+ ...]
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
e^(i) =
COs() + i sin()
An interesting case is when we set
= , since the above equation
becomes
e^(
i) = 1 + 0i = 1.
which can be rewritten as
e^(
i) + 1 = 0. special case
which 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 e^{x} series converge as indicated for imaginary
numbers.


