Question

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

Answer 1

Refer to explanation

Explanation:

The exponential of a real number x, written as `e^x` , is defined by an sum of infinite series, as follows

`e^x = ∑_(k=0) ^ ∞ (x^k/(k!)) = 1 + x + (x^ 2/(2!)) + (x^3 /(3!)) + ...`

Also `costheta` and `sintheta` can be expressed as sum of infinite series as follows

`cos(θ) = 1 - (θ^2/(2!)) + (θ^4 /(4!)) + ...`

`sin(θ) = θ - (θ^3/(3!)) + (θ^5/(5!)) + ...`

The exponential of a complex number z is written `e^z` , and is defined in the same way as the exponential of a real number

`e^z = ∑_(k=0) ^ ∞ (z^k/(k!)) = 1 + z + (z^ 2/(2!)) + (z^3 /(3!)) + ...`

Let set `z=i*theta` in the previous relation to get

#e^(i*theta) = ∑_(k=0) ^ ∞ ((i*theta)^k/(k!)) = 1 + (i*theta) + ((i*theta)^ 2/(2!)) + ((i*theta)^3 /(3!)) + ...= [1 - (θ^2/(2!)) + (θ^4/(4!)) - ...] + i[θ - (θ^3/(3!)) + (θ^5/(5!)) + ...]= =cos(θ) + i sin(θ)#

Thus the proof concluded.

The identity `e^(itheta)=costheta+isintheta` is known as Euler's formula

Answer 2

Supplementary to Konstantinos's proof:

How do we show this from scratch, without knowing the series expansions of `cos theta` and `sin theta`?

Explanation:

Consider a point on the Complex plane at `cos t + i sin t`. This will lie on the unit circle for any Real value of `t`.

Next suppose the point moves anticlockwise around the unit circle at a rate of `1` radian per second. That is `t` increases at a rate of `1` per second.

The rate of change of the position (i.e. velocity) at time `t` will be a tangential vector, perpendicular to the radius and will have length `1`.

Looking at the `x` and `y` (or Real and imaginary) components we see that `d/(dt) cos t = - sin t` and `d/(dt) sin t = cos t`

We can then use Taylor's theorem to derive the Maclaurin series for `cos t` and `sin t`.

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

For example, in the case of `f(t) = cos t` we have:

`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.