e^(ix) = cos(x) + i*sin(x) (idea)

e^ix = cos(x) + isin(x)

Here is a proof of a trigonometric identity commonly used in representing complex numbers; it provides a simple relationship between polar and exponential form. So here goes...

We know the following Taylor Series:

              x2     x3     x4   
ex = 1 + x + ---- + ---- + ---- + ... ...
              2!     3!     4!

              x3     x5
sin(x) = x - ---- + ---- - ... ...
              3!     5!

              x4     x6
cos(x) = 1 - ---- + ---- - ... ...
              4!     6!

And we know that the imaginary number i represents the square root of -1, so that i² = -1.

So if we raise e to the power ix:

                (ix)2     (ix)3     (ix)4 
eix = 1 + ix + ------- + ------- + ------- + ... ...
                 2!        3!        4!

Seperating out the powers gives:

                i2x2     i3x3      i4x4 
eix = 1 + ix + ------ + ------ + ------ + ... ...
                 2!       3!       4!

Evaluating the powers of i simplifies things a bit:

                x2     ix3     x4   
eix = 1 + ix - ---- - ----- + ---- + ... ...
                2!      3!     4!

And finally, seperate the odd- and even- powered terms, and factorize out the i, to give:

      /      x2     x4            \        /      x3     x5         \
eix = | 1 - ---- + ---- - ... ... |  +  i | x - ---- + ---- ... ... |
      \      2!     4!           /        \      3!     5!         /

Which, given the Taylor Identities given above, is equal to:

eix = cos(x) + isin(x)

This has been an essay in the craft of the <PRE> tag.