Euler's Formula explained with exponential function

Question

Why is $$ e^x = \lim_{n \to \infty} \left( 1+ \frac{x}{n} \right)^n $$ also true for an imaginary $x$? Per definition, $$ e = \lim_{n \to \infty} \left( 1+ \frac{1}{n} \right)^n $$ so $$ e^x = \lim_{n \to \infty} \left( 1+ \frac{1}{n} \right)^{nx} $$ and $$ e^x = \lim_{n \to \infty} \left( 1+ \frac{x}{nx} \right)^{nx} $$ I understand now that we can substitute $n$ for $n*x$ as both approach $\infty$, but why does this also work for $x=i$, where $\lim_{n \to \infty} \left(n*i\right) $ would approach $\infty*i$, which is all the way to the top of the complex plane instead of to the right? $$ e^i = \lim_{n \to \infty} \left( 1+ i × \frac{1}{ni} \right)^{ni} $$

(This question came to me at 7:47 in Mathologer's video on $e^{\pi*i}$: https://youtu.be/-dhHrg-KbJ0?t=7m47s)

Answer 1

When one is dealing with the famous limit $$\lim_{n\to\infty} \left(1+\frac{z}{n}\right)^n\tag{1}$$ it is implicitly assumed that $n$ is a positive integer. It can be proved with some effort that the limit above exists for all $z\in\mathbb{C} $ and thus defines a function $f(z) $. Further by definition $e=f(1)$ and again it can be proved using elementary arguments that $f(z_1+z_2)=f(z_1)f(z_2)$ for all $z_1,z_2\in \mathbb {C} $. Now by using algebra it is seen that $f(x) =\{f(1)\}^x=e^x$ for $x\in\mathbb{Q} $. And it therefore makes sense to define the symbol $e^x$ as $f(x) $ even when $x\notin\mathbb{Q} $. Your approach somehow assumes that $e^i$ (or imaginary exponents in general) has a well defined meaning. This assumption needs to be substantiated by proper definitions of imaginary or complex exponents before one can deal with your question.

Also the transition from $e=\lim_{n\to\infty} \left(1+\dfrac{1}{n}\right)^n$ to $e^x=\lim_{n\to\infty} \left(1+\dfrac{1}{n}\right)^{nx}$ involves the concept of raising to power $x$ and the continuity of $g(t) =t^x, t\in\mathbb{R} $. These ideas are non-trivial if $x\notin\mathbb{Q} $ and it is best to work using proper definitions involving the limit $(1)$ as described earlier (see my comments to your question also).

The youtube video you have linked also uses the definition based on limit $(1)$ and the exponent $m$ in video is a positive integer and the guy in video takes $z=i\pi$ and evaluates the expression $(1+i\pi/m)^m$ for $m=1,2,\dots 99$. You can see that imaginary exponents do not come into picture at all.

Answer 2

Yes we can show that in general for $z=x+iy$ we have that $$\lim\limits_{n\to\infty}\left(1+\frac{z}{n}\right)^n=e^z=e^x(\cos{y}+i\sin{y})$$

but not by that limit.

For a proof look here Suppose $z=x+iy$, prove that $\lim\limits_{n\to\infty}(1+\frac{z}{n})^n=e^x(\cos{y}+i\sin{y})$? and to the related OP.

Answer 3

Use the binomial theorem to expand the nth power. As $n\to\infty$, the coefficient of $x^k$ approaches $1/k!$.

Answer 4

It is valid in a simple generality.. where we are decomposing in binomial series to the log of complex arguments and then re-composing it back.

$$ z= e^{\log \,z} = \lim_{n \to \infty} \left( 1+ \frac{\log \,z} {n} \right)^n $$