# How do you show #e^(it) = cos t + i sin t# ?

##### 2 Answers

Refer to explanation

#### Explanation:

The exponential of a real number x, written as

Also

The exponential of a complex number z is written

Let set

Thus the proof concluded.

The identity **Euler's formula**

Supplementary to Konstantinos's proof:

How do we show this from scratch, without knowing the series expansions of

#### Explanation:

Consider a point on the Complex plane at

Next suppose the point moves anticlockwise around the unit circle at a rate of

The rate of change of the position (i.e. velocity) at time

Looking at the

We can then use Taylor's theorem to derive the Maclaurin series for

#f(t) = sum_(k=0)^oo f^((k))(0)/(k!) t^n#

For example, in the case of

#f^((0))(0) = cos(0) = 1#

#f^((1))(0) = -sin(0) = 0#

#f^((2))(0) = -cos(0) = -1#

#f^((3))(0) = sin(0) = 0#

#f^((4))(0) = cos(0) = 1#

etc.

Hence:

#cos t = sum_(k = 0)^oo (-1)^k/((2k)!) t^(2k) = 1 - t^2/(2!) + t^4/(4!) -...#

Similarly:

#sin t = sum_(k = 0)^oo (-1)^k/((2k+1)!) t^(2k+1) = t - t^3/(3!) + t^5/(5!) -...#

If we then define:

#e^z = sum_(k=0)^oo 1/(k!) z^n = 1 + z/(1!) + z^2/(2!) + z^3/(3!) +...#

we find:

#e^(i t) = cos t + i sin t#

as Konstantinos showed.