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I was curious in showing that $\left(1+\frac{ix}{n}\right)^n$ goes to $e^{ix}$ as n goes to infinity. I believe this is right as included it is included numerous times in here: Proof of Euler's formula that doesn't use differentiation?. I have:

$e^{ix}=\left(\left(1+\frac{1}{n}\right)^n\right)^{ix}=\left(1+\frac{1}{n}\right)^{ixn}$ then name ixn as k so we have: $\left(1+\frac{ix}{ixn}\right)^{ixn}=\left(1+\frac{ix}{k}\right)^k$. As per the definition of e, n approaches infinity, and so k (which equals ixn) approaches $i\infty$.

My problem is that k is going to $i\infty$, not just $\infty$ as in the original. Did I make an error or am I misunderstanding something?

Thanks and regards!

Edit: My question is where did I make a mistake? I am aware there may not be one and that it may be salvageable, but I don't know how.

Pineapple Fish
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  • Try taking $ix/n =1/k$ – Koro May 15 '20 at 01:39
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    @Koro if you do that k approaches $-i\infty$, which is not the definition of e. This seems so simple I feel like I am making a huge mistake. – Pineapple Fish May 15 '20 at 01:45
  • Then, expand using binomial theorem and then take limit. – Koro May 15 '20 at 01:52
  • The issue is more basic. One needs a definition of $w^z$ for complex values of $w$ and $z$. The definition of $e^{z}$ is given as $e^x \cos(y)+ie^x \sin(y)$ where $x$ and $y$ are the real and imaginary parts of $z$, respectively. – Mark Viola May 15 '20 at 02:28

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$$a_n=\left(1+\frac{ix}{n}\right)^n\implies \log(a_n)=n\log\left(1+\frac{ix}{n}\right)$$ Now, by Taylor $$\log(a_n)=n\left(\frac{i x}{n}+\frac{x^2}{2 n^2}+O\left(\frac{1}{n^3}\right)\right)=i x+\frac{x^2}{2 n}+O\left(\frac{1}{n^2}\right)$$ $$a_n=e^{\log(a_n)}=e^{ix}\left(1+\frac{x^2}{2 n}\right)+O\left(\frac{1}{n^2}\right)$$

Claude Leibovici
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