# Find the limit of complex number exponentiation

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?

• I would use $\operatorname{cis}$ representation for the complex number and then use De Moivre's formula. – Don Thousand Nov 15 '18 at 19:23
• Hint: do you recognize which well-known function is equal to $\lim_{n \to \infty}(1+\frac{x}{n})^n$? – Cuspy Code Nov 15 '18 at 19:27
• To add to @CuspyCode's comment, you can find a proof here. – MSDG Nov 15 '18 at 19:46
• What is your definition of the complex exponential function? Let's start with that. – Mark Viola Nov 15 '18 at 20:18

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 $$Z_n\to f(a) (\cos b+i\sin b)$$