A simpler demo of Euler’s formula

Following on from yesterday I found the following at:

Differentiate \displaystyle\frac{\cos\theta + i\sin\theta}{e^{i\theta}} to get zero. Then put \theta=0 to show that the constant value of \displaystyle\frac{\cos\theta + i\sin\theta}{e^{i\theta}} must be 1.

That’s it!



A simple derivation of Euler’s formula

The usual way of proving Euler’s formula, \exp(i\theta) = \cos\theta + i\sin\theta, is to use the Maclaurin series expansions of \exp(x), \sin(x) and \cos(x). But suppose you want to introduce this formula to a class that has a good grasp of complex arithmetic and calculus, but hasn’t yet met Mclaurin’s series. The following is a non-rigorous demonstration of this result using only basic tools, and could form an extended homework.

Three basic results first need to be established. The first two were taken from the wikipedia page on Euler’s formula (1 August 2013).

First the student should show that

\displaystyle \int\frac{dx}{1+ax} = \frac{1}{a}\ln(1+ax)

(the arbitrary constant isn’t important here). Second, that

\displaystyle\frac{2}{1+x^2} = \frac{1}{1-ix} + \frac{1}{1+ix}

noting that this can be shown by direct manipulation and does not require a knowledge of partial fractions. Thirdly, by differentiating both sides for example, that

\displaystyle \int\frac{dx}{1+x^2} = \tan^{-1}(x)

Then, step-by-step, show that:

  1. \displaystyle 2\tan^{-1}(x) = \int\frac{dx}{1-ix} + \int\frac{dx}{1+ix}.
  2. \displaystyle 2i\tan^{-1}(x) = \ln\left(\frac{1+ix}{1-ix}\right).
  3. \displaystyle \frac{1+ix}{1-ix} = \frac{1-x^2 + 2ix}{1+x^2}.
  4. (by putting x=\tan(\alpha)) \displaystyle \frac{1+ix}{1-ix} = \cos(2\alpha)+i\sin(2\alpha).
  5. \displaystyle 2i\tan^{-1}(x) = 2i\alpha.
  6. 2i\alpha = \ln(\cos(2\alpha)+i\sin(2\alpha)).

Writing \theta=2\alpha and taking antilogs then results in Euler’s formula:

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

If you can find a simpler `easy’ demonstration of this result I would appreciate hearing about it.