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How do I find the limit of $$ Z_n=\left(1+{\frac{a+bi}{n}}\right)^n $$

Should I have real and imaginary parts? Like this $$ \lim Z_n = \lim \left(\left(1+{\frac{a}{n}}\right)+{\frac{b}{n}}\right)^n $$ What's the next step then?

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  • $\begingroup$ I would use $\operatorname{cis}$ representation for the complex number and then use De Moivre's formula. $\endgroup$ – Don Thousand Nov 15 '18 at 19:23
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    $\begingroup$ Hint: do you recognize which well-known function is equal to $\lim_{n \to \infty}(1+\frac{x}{n})^n$? $\endgroup$ – Cuspy Code Nov 15 '18 at 19:27
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    $\begingroup$ To add to @CuspyCode's comment, you can find a proof here. $\endgroup$ – MSDG Nov 15 '18 at 19:46
  • $\begingroup$ What is your definition of the complex exponential function? Let's start with that. $\endgroup$ – Mark Viola Nov 15 '18 at 20:18
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In case you know the following limit $$\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n=e^x,\forall x\in\mathbb {R} $$ then it can be shown without much effort that $Z_n\to e^a(\cos b+i\sin b) $.

Let's then assume that

For all real $x$ the limit $$\lim_{n\to \infty} \left(1+\frac{x}{n}\right) ^n$$ exists and is positive and thus defines a function, say $f$ from $\mathbb {R} $ to $\mathbb {R} ^{+} $ and thus the above limit may be denoted by $f(x) $.

Another key fact to be used is the lemma of Thomas Andrews :

Lemma: If $\{a_n\} $ is a sequence of real or complex terms such that $n(a_n-1)\to 0$ then $a_n^n\to 1$.

Using this we prove that if $z, w$ are complex then $$\lim_{n\to\infty} \left(1+\frac{z+w}{n}\right)^n= \lim_{n\to\infty} \left(1+\frac{z}{n}\right)^n\cdot \lim_{n\to\infty} \left(1+\frac{w}{n}\right)^n\tag{ 1}$$ provided both the limits on right side exist and are non-zero.

Consider the sequence $$a_n=\dfrac{1+\dfrac{z+w}{n}} {\left(1+\dfrac{z}{n}\right)\left(1+\dfrac{w}{n}\right)} $$ It can be proved easily that $n(a_n-1)\to 0$ and therefore by the lemma stated above $a_n^n\to 1$ which proves equation $1$.

Next consider the sequence $$b_n=\dfrac{1+\dfrac{ib}{n}}{\cos(b/n)+i\sin(b/n)},b\in\mathbb {R} $$ and again it can be easily proved that $n(b_n-1)\to 0$ so that $b_n^n\to 1$. It follows that $$\lim_{n\to\infty} \left(1+\frac{ib}{n}\right)^n=\cos b+i\sin b,\forall b\in\mathbb {R} \tag{2}$$ Also note that the above limit is non-zero. By assumption $(1+(a/n))^n\to f(a) $ for all real $a$ and $f(a) >0$. Using formula $(1)$ with $z=a, w=ib$ it follows that $$Z_n\to f(a) (\cos b+i\sin b) $$

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