I am trying to prove that $\lim_{n \to \infty} (1 + \frac{z}{n} + o(\frac{1}{n}))^n = e^z$ for complex $z$. There are many proofs for the case where there is no $o(\frac{1}{n})$ term (this is the Landau small o). One proof I like is found in the first answer here, and this proof for real values of $z$ is easily modified to include the complex case. However, I have not been able to extend it to include the $o(\frac{1}{n})$ term. How can the proof linked be extended? Or is there a different proof (from scratch) that can prove the result I state in the first sentence?
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By triangle inequality, if $a,b$ are complex numbers and $n\geq 1$, then $|(a+b)^n-a^n| \leq (|a|+|b|)^n-|a|^n \leq n|b|(|a|+|b|)^{n-1}$.
Now take $a=1+z/n$, $b=x_n/n$, where $x_n \rightarrow 0$, then $|(a+b)^n-a^n| \leq |x_n|(|a|+|b|)^{n-1}$.
Thus the result follows as long as $R_n=(1+|z|/n+|x_n|/n)^{n-1}$ is a bounded sequence. But there is a $M>0$ such that $|x_n|+|z| < M$ for any $n$, so that $R_n \leq (1+M/n)^n\leq e^{M}$. QED.
Aphelli
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Excellent, thanks! Just a small point: The last equality should really be $\leq$, right? – Harmenszoon Mar 23 '22 at 17:57
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Oops, that’s correct, thanks for pointing it out! – Aphelli Mar 23 '22 at 19:05