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

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 $\cos \theta$ and $\sin \theta$ 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 $z = i \cdot \theta$ 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 ${e}^{i \theta} = \cos \theta + i \sin \theta$ is known as Euler's formula

Mar 13, 2016

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 $\frac{d}{\mathrm{dt}} \cos t = - \sin t$ and $\frac{d}{\mathrm{dt}} \sin t = \cos t$

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 $f \left(t\right) = \cos t$ we have:

${f}^{\left(0\right)} \left(0\right) = \cos \left(0\right) = 1$

${f}^{\left(1\right)} \left(0\right) = - \sin \left(0\right) = 0$

${f}^{\left(2\right)} \left(0\right) = - \cos \left(0\right) = - 1$

${f}^{\left(3\right)} \left(0\right) = \sin \left(0\right) = 0$

${f}^{\left(4\right)} \left(0\right) = \cos \left(0\right) = 1$

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.