On Euler's Identity $e^{\ i\theta}$

Three interesting numbers in Maths are $\pi$, the ratio of a circle's circumference to its diameter, Euler's number $e$, approx 2.7181828459, and i, $\sqrt{-1}$. Amazingly, $e^{i\pi} = -1$.

This comes from his identity, $e^{i\theta} = cos(\theta) + i sin(\theta)$. Here are some remarks on it and a circular plot on the Argand plane.

You can also use his identity for expressions for $cos(\theta)$ and $sin(\theta)$ in terms of $e^{i\theta}$, as is shown here.

See also, my pages explaining a use of $e^\pi$, which start here.

Leonhard Euler is credited with finding the identity (and many other concepts in mathematics) published in his 1748 book Introductio im analysin infinitorum.

In fact, the British Mathematician Roger Coates, who worked with Isaac Newton, had a geometric demonstration in 1714 that $i\theta = ln (cos(\theta) + i sin(\theta))$. This becomes Euler's formula if both sides are raised to the power of $e$!

Whilst Euler was the first to use the letter e, the number was discovered by Jacob Bernoulli in 1683, who was trying to determine how wealth grew if compound interest was applied many times a year!

Often the identity is demonstrated using power series of exp, sin and cos. Here is a shorter method using differentiation.

Deriving the Identity

It is known that the differential of $e^{n\theta}$ is $ne^{n\theta}$, the differential of $sin(n\theta)$ is $n cos(n\theta)$ and that of $cos(n\theta)$ is $-n sin(\theta)$.

Let us consider $z = cos(n\theta) + i sin(n\theta)$ and its differential with respect to $\theta$.

$\frac{dz}{d\theta}=\frac{d}{d\theta}(cos(n\theta) + i sin(n\theta)) =-n sin(n\theta) + i n cos(n\theta)$

But $i n z = i n cos(n\theta) + i^2 n sin(n\theta) = i n cos(n\theta) -n sin(n\theta)$

Hence $\frac{dz}{d\theta} = i n z$

Given the differential of $e^{n\theta}$ this means $z = e^{in\theta}$

So we have demonstrated Euler's identity $e^{i\theta} = cos(\theta) + i sin(\theta)$.

On $e^{\ i\theta}$

If $n\theta = \pi$, $cos(n\theta) = -1$ and $sin(n\theta) = 0$, so $e^{i\pi}=-1$

Cos and sin in terms of $e^{\ i\theta}$

$e^{-i\theta} = cos(-\theta) + i sin(-\theta) = cos(\theta) - i sin(\theta)$

$e^{i\theta} + e^{-i\theta} = 2 cos(\theta)$ and $e^{i\theta} - e^{-i\theta} = 2i sin(\theta)$

Hence $cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}$ and $sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}$

Euler on Argand Plane

A complex number $a + ib$ can be plotted on the Argand Plane, horizontal value a, vertical b. We can thus plot $z = cos(\theta) + i sin(\theta)$ on the plane. Its distance from the origin is $cos^2(\theta) + sin^2(\theta)$ = 1. Hence as $\theta$ varies from 0 to 360 degrees, $z$ forms a unit circle, as shown below.