Defining $e^{i \theta}$

We want to define $$f(z)=e^z$$ in such a way that it satisfies the following properties: $$ f'(z) = f(z) , \quad f(x+ 0i) = e^x$$ In other words we want it to be its own derivative and we would like it to reduce to the regular exponential function when the exponent is purely real. Lets explore what properties such a function would have to have.

To make things easier write $f(z) = f(x+iy) = u + iv$ with $ u = u(x,y) , v = v(x,y)$.

By the Cauchy-Riemann equations we know that $$f'(x+iy) = u_x + i v_x = v_y + i (-u_y) = u + iv.$$ where the rightmost equality comes from the fact that$ f' = f$. Equating real/imaginary parts we see that $$u(x,y) = u_x(x,y)$$ $$ v(x,y) = v_x(x,y)$$ for all x,y. The general solutions to these equations are $$u(x,y) = a(y)e^x$$ $$ v(x,y)= b(y)e^x $$ Looking at the other constraint we have $$ e^{x} + 0i = e^ x = f(x+0i) = u(x,0) + i v(x,0) = a(0)e^x + i b(0)e^x$$ This gives our "initial conditions" $$ a(0) = 1 ,\ b(0) = 0.$$

Going back to the C-R equations we have $$ u_x = v_y \implies a(y)e^x = b'(y)e^x$$ $$v_x = - u_y \implies a'(y)e^x = -b(y)e^x$$ Giving the system $$a = b'$$ $$- b = a'$$ Which we can cleverly write as $$ \vec{x}' = \begin{bmatrix} a'(y) \\ b' (y) \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a(y) \\ b(y) \end{bmatrix} = A \vec{x} $$ It turns out the solution to a linear system like this is given by the matrix exponential

$$ \vec{x}(y) = e^{Ay} \vec{x}_0 $$

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where $ \vec{x}_0 = \begin{bmatrix} a(0) \\ b(0) \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $ e^{Ay} = \displaystyle\sum_{k=0}^{\infty} \frac{A^ky^k}{k!} $ and$A^k$ denotes matrix exponentiation.

Note that $ A^2 = -I$ so that $A^3 = - A$, $A^4 = I$, $A^5 = A $ etc.

This gives $$ e^{Ay} = \displaystyle\sum_{k=0}^{\infty} \frac{A^ky^k}{k!} = \displaystyle\sum_{\text{even } k} \frac{A^ky^k}{k!} + \displaystyle\sum_{\text{odd } k} \frac{A^ky^k}{k!}$$

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$$ = \displaystyle\sum_{k=0}^{\infty} \frac{(-1)^k Iy^{(2k)}}{2k!} + \displaystyle\sum_{k=0}^{\infty} \frac{(-1)^kAy^{(2k+1)}}{(2k+1)!}$$ $= \begin{bmatrix} \sum_{k=0}^{\infty} \frac{(-1)^k y^{(2k)}}{2k!} & 0 \\ 0 & \sum_{k=0}^{\infty} \frac{(-1)^k y^{(2k)}}{2k!}\end{bmatrix} + \begin{bmatrix} 0 & - \sum_{k=0}^{\infty} \frac{(-1)^ky^{(2k+1)}}{(2k+1)!} \\ \sum_{k=0}^{\infty} \frac{(-1)^ky^{(2k+1)}}{(2k+1)!} & 0 \end{bmatrix} $

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$= \begin{bmatrix} \sum_{k=0}^{\infty} \frac{(-1)^k y^{(2k)}}{2k!} & - \sum_{k=0}^{\infty} \frac{(-1)^ky^{(2k+1)}}{(2k+1)!} \\ \sum_{k=0}^{\infty} \frac{(-1)^ky^{(2k+1)}}{(2k+1)!} & \sum_{k=0}^{\infty} \frac{(-1)^k y^{(2k)}}{2k!} \end{bmatrix}= \begin{bmatrix} \cos(y) & - \sin(y) \\ \sin(y) & \cos(y) \end{bmatrix}$

As mentioned above, multiplying by the initial conditions vector gives us our solution : $$ \begin{bmatrix} a(y) \\ b(y) \end{bmatrix} = \begin{bmatrix} \cos(y) & - \sin(y) \\ \sin(y) & \cos(y) \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} cos(y) \\ sin(y) \end{bmatrix}$$ We finally arrive at $$f(x+iy) = u(x,y) + i v(x,y) = e^x\cos(y) + i e^x\sin(y).$$ In other words, if we want the complex exponential to naturally generalize the real exponential then

we must DEFINE it as $ e^{x+iy} = e^x(\cos(y) + i \sin(y))$.

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.