TITLE: Defining $e^{i \theta}$
QUESTION [6 upvotes]: I want to define $e^{i z}$ without using power series. My approach was as follows:
Start with 
$$\lim_{n \to \infty} \left(1+\frac{i}{n}\right)^{nz}$$
Then make the substitution
$$\frac{1}{m} = \frac{i}{n},\ \text{ so } n=mi.$$
And we have
$$\lim_{n \to \infty} \left(1+\frac{1}{m}\right)^{miz}$$
This almost works, but for $n \to \infty$ we have $m \to i \infty.$ So at this point we can't say
$$\left[\lim_{m \to \infty} \left(1+\frac{1}{m}\right)^{m} \right]^{iz}=e^{iz}.$$
Because we don't have $m \to \infty.$ 
So my next idea was to try
$$\lim_{x \to 0} \left(1+ ix\right)^{z/x}$$
Then make the substitution $y=ix.$ Then for $x \to 0$ we have $y \to 0.$ So we can say
$$\lim_{y \to 0} (1+y)^{iz/y}$$
It is tempting now to write
$$\left[\lim_{y \to 0} (1+y)^{1/y}\right]^{iz} = e^{iz}.$$
But we don't know that 
$$e = \lim_{y \to 0} (1+y)^{1/y}.$$
We do know that
$$e = \lim_{y \to 0+} (1+y)^{1/y}.$$
For $y \in \mathbb{R}.$ In other words, we know that the right limit of $(1+y)^{1/y}$ is $e$, but I wan't to be able to show that its true for all $y \in \mathbb{C}$ and for all "paths".

REPLY [2 votes]: Let's simply define $\ \exp(z)\ $ instead of $\ \exp(i\cdot z).\ $ Here:
$$ \exp(z)\,\ :=\,\ \lim_{n\rightarrow\infty}\ \left(1 +\frac zn\right)^n $$
Then, as requested in another Answer, $\ \exp(x+i\cdot 0) = \exp(x)\ $ (of course), and derivative $\ exp\,'\ =\ \exp.$

REMARK.  A clean definition (like the one here) should not use irrational exponents nor even any non-integer exponent.