Justifying complex number exponential notation i.e. Euler’s Formula without Calculus

Background

I’m assuming that you’re already familiar with polar coordinates and cis notation for complex numbers. If not look here: Link

So we start with any complex number being represented in polar coordinate on the complex plane as:  $ r ( \cos(\theta) + i \sin(\theta)) $

By directly evaluating the expressions we can start off by showing the complex numbers have the all same properties as an exponential function

 

- $ f(0) = 1 $
- $ f(x + y) = f(x) \cdot f(y) $
- $ f(x - y) = \dfrac{f(x)}{f(y)} $
- $ f(-x) = \dfrac{1}{f(x)} $
- $ f(x)^n = f(n\cdot x) $
  
  

  
That's very suggestive but what exponential expression is equivalent in this
case?  In fact Euler's Formula  says that    $ r ( \cos(\theta) + i \sin(\theta))  =
r \cdot e^{i \theta} $
  
  

  
Usually that's justified via Calculus and Taylor Series but its possible to
mostly ground with simpler techniques.
  
  

  
Lets start by looking at how e is defined:  $  \displaystyle{e
=  \lim_{n \to -\infty} \left(1 + \dfrac{1}{n}\right)^n} $ 
 or conceptually the result of infinitely compounding a value. 
 That's usually been introduced before in the context of interest or
other proportional growth.  
  
  

  

  
We'd also like to have an expression for $e^x $ so  we  raise
both sides to the xth power. Then on the right hand side via the binomial
theorem we get:
  
  

  
$ ( 1 + {x \choose 1} \cdot \frac{1}{n} + {x \choose 2} \cdot \frac{1}{n^2}
... )^n $
  
  

  
But as x goes to infinity all of the smaller powers of n drop off and we're
only left with the first two terms  giving us: 
  
  

  

  $  \displaystyle{e^x =  \lim_{n \to -\infty} \left(1 +
  \dfrac{x}{n}\right)^n} $



  
  
So how to connect these two seemingly unrelated concepts?  We're going
to start with De Moivre's Formula
  
  

  

  
$ (\cos(nx) + i \sin(nx)) = ((\cos(x) + i \sin(x))^n $
  
  

  

  
We'd like this to look similar in structure to the definition of e and the
normal cis notation so first step is to scale everything down:
  
  


  
  
$ (\cos(x) + i \sin(x)) = \left((\cos(\dfrac{x}{n}) + i
  \sin(\dfrac{x}{n})\right)^n $ 
  
  


  
  

  
Now what happens as n heads towards infinity?
  
  


  
  
 For the cosine value $\frac{x}{n}$  heads towards so 0 so in
  this case $ \cos(\frac{x}{n}) \approx \cos(0) = 1 $.   Likewise as n
  goes to infinity  $\frac{x}{n}$ is getting closer and closer to 0 and so is the sine
  value:   $ \sin(\frac{x}{n}) \approx\ \sin(0) \approx \frac{x}{n}  $
  
  


  
  
So combining things we get as 
 $ \displaystyle{ \lim_{n \to -\infty}} (\cos(x) + i \sin(x)) \approx
\left(1 + i \cdot \frac{x}{n} \right)^n $   
  
  


  
  
We then look back and realize this is exactly the same as the
  definition for:   $ e^{ix} $ !