eix = 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.