I want to prove the following thing, $$c_j \to 0, \, a_j \to \infty, \text{ and } a_j c_j \to \lambda, \text{ then } (1 + c_j)^{a_j} \to e^\lambda$$ where $c_j \in \mathbb{C}$ and $a_j \in \mathbb{R}$.
It is easy if $c_j \in \mathbb{R}$ since we can take logarithm for $(1 + c_j)^{a_j}$ and then continue. However, if $c_j \in \mathbb{C}$, there might need a different technique. Any help would be appreciated!
Complex powers can be tricky but here we use the following two unambiguous definitions:
$e^{\lambda}=\exp \lambda=\sum_{n \ge 0}{\frac{\lambda^n}{n!}}$ for any complex $\lambda$
$\log (1+c)=\sum_{n \ge 1}{(-1)^{n-1}\frac{c^n}{n}}$ for any complex $|c| <1$,
so then we define $(1+c)^a=\exp(a\log (1+c))$ for any $|c| <1, a$ complex
With the above $\exp$ is an entire function (hence continuous) so if we have $|c_k| <1, a_k$ arbitrary, and $a_k\log(1+c_k) \to A$, then $(1+c_k)^{a_k} \to \exp(A)=e^A$.
In our case we definitely have $|c_k| <\frac{1}{2}<1, k \ge k_0$ since $c_k \to 0$, so we need to show $a_kc_k \to \lambda$ implies $a_k\log(1+c_k) \to \lambda$.
But $a_kc_k \to \lambda$ means $|a_kc_k| \le 2(|\lambda|+1), k \ge k_1$, hence taking $k \ge k_0+k_1$ so we also have $|c_k| <\frac{1}{2}$ we then get:
$|a_k\log(1+c_k) -\lambda| \le |a_kc_k-\lambda|+ |a_kc_k||c_k|\sum_{n \ge 2}{\frac {|c_k|^{n-2}}{n}} \le |a_kc_k-\lambda|+4|c_k|(|\lambda|+1)$ by the inequality above for $|a_kc_k|$ and estimating the sum by the geometric one $\sum_{n \ge 2}{|c_k|^{n-2}} \le 2$.
Since $a_kc_k \to \lambda, c_k \to 0$ we are done!