Proof of Convergence of Exponential When Defined as the Limit of a Sequence of Functions I'm trying to prove that the sequence $(1 + \frac{z}{n})^n$ converges, for any complex $z$, without using the equivalent series definition of $exp$ or the properties of the complex logarithm. All I need is pointwise convergence. I've tried expanding the sequence by the binomial theorem, but the fact that the number of summands in the expansion depends on the position in the sequence seems to prevent me from moving forward. 
Any hints would be greatly appreciated. Thank you.
 A: Lets assume that $(1+\frac{x}{n})^n$ converges for every $x∈\Bbb R$. Then 
$$\left( 1+a+ib\right)^n = \left( 1+a\right)^n\left(1 + \frac{ib}{1+a}\right)^n$$
using $a=x/n,b=y/n$, we reduce by product rule and continuity to needing to show the result for purely imaginary $z=0+iy/n$, since
$$\left(1 + \frac{iy/n}{1+x/n}\right)^n = \Big(1 + \frac{iy}{n} - \underbrace{\frac{ixy}{n(x+n)}}_{=O(n^{-2})}\Big)^n$$
For $(1+\frac{iy}{n})^n$, we avoid the complex logarithm by using the geometrically defined(say) $\arg$ function, which satisfies the log-like rule
$$ \arg \left(1+i\frac{y}{n}\right)^n =  n \arg \left(1+i\frac{y}{n}\right) = n \arctan \frac{y}{n}$$
(up to a multiple of $2π$, but lets choose that multiple to be 0.)
Since 
$$ n\arctan \frac{y}{n} = y \frac{\arctan(0+y/n)-\arctan(0)}{y/n} \xrightarrow[n→∞]{} y$$
But note also that $$\left|\left(1+\frac{iy}{n}\right)^n\right| = \left|1+\frac{iy}{n}\right|^n = \left(1+\frac{y^2}{n^2}\right)^{n/2} → 1$$ By properties of the real exponential which I'm assuming. Hence by the polar representation of complex numbers we have that $(1+iy/n)^n → e^{iy}$.
A: Hint :
Let $z_n = 1+\frac{z}{n}$ and $u_n=(z_n)^n$.
We can write $u_n = r_n \exp(i \theta_n)$. To show that $(u_n)$ converges, it is sufficient to show that $r_n$ and $\theta_n$ converge.

Let $z = a + ib$.

$r_n = |u_n| = |z_n|^n = (1+ 2\frac{a}{n} + \frac{a^2 + b^2}{n^2})^{\frac{n}{2}}$. Then you can show  that $r_n$ converges to $e^{a}$.

$\theta_n = \arg(u_n) = n \arg(z_n) = n \arctan(\frac{\frac{b}{n}}{1+\frac{a}{n}}) = n \arctan(\frac{b}{n+a})$  Similarly you can show that this converges to $b$.
