sequences and series - Show $(1+frac{z}{n})^{n} underset{n to +infty}{longrightarrow }e^{z}$

To show that $$z_n=\left(1+\dfrac{z}{n} \right)^{n} \underset{n \longmapsto +\infty}{\longrightarrow \exp(z)}$$
the author of textbook use the following method but there is some steps that i'm not sure if i got it right so would someone elaborate it

Let $x = \Re(z)$ and $y=\Im(z)$

• ${\displaystyle r_n=\sqrt{\left(1+\dfrac{x}{n} \right)^{2}+\dfrac{y^{2}}{n^2}}=\sqrt{1+\dfrac{2x}{n}+o\left(\dfrac{1}{n}\right)}=1+\dfrac{x}{n}+o\left(\dfrac{x}{n}\right)}$


• why if $r_n=1+\dfrac{x}{n}+o\left(\dfrac{x}{n} \right)$ then $\ln(r_n)\sim \dfrac{x}{n}$


• how we can get the expression of $\tan(\alpha_n)$


• why if $\tan\left( \alpha_n \right)\sim \dfrac{y}{n}$ then $\alpha_{n}\sim \dfrac{y}{n}$

My thoughts:

first i think there is typo in $z_n=r_n^{n}e^{n\alpha_n}$ should we write $z_n=r_n^{n}e^{i n\alpha_n}$ instead.

• ${\displaystyle r_n=\sqrt{\left(1+\dfrac{x}{n}
  \right)^{2}+\dfrac{y^{2}}{n^2}}=\sqrt{1+\dfrac{2x}{n}+o\left(\dfrac{1}{n}\right)}=1+\dfrac{x}{n}+o\left(\dfrac{x}{n}\right)}$

note that

$$\left(1+x \right)^{\alpha}\underset{x\to 0}{=}1+\alpha x +o\left( x\right) $$

$$\begin{cases} \left(1+\dfrac{x}{n} \right)^{2}=1+2\dfrac{x}{n}+o\left( \dfrac{1}{n}\right) \\ \dfrac{y^{2}}{n^{2}}=o\left(\dfrac{1}{n}\right)\end{cases}\implies 1+\dfrac{2x}{n}+o\left(\dfrac{1}{n}\right)$$

on the other hand
\begin{aligned}\sqrt{1+\dfrac{2x}{n}+o\left(\dfrac{1}{n}\right)}=\left(1+\dfrac{2x}{n}+o\left(\dfrac{1}{n}\right) \right)^{\frac{1}{2}}&=1+\dfrac{1}{2}\left( \dfrac{2x}{n}+o\left( \dfrac{1}{n}\right)\right)+o\left(o\left(\dfrac{1}{n}\right)\right) \\
&=1+\dfrac{x}{n}+o\left( \dfrac{1}{n}\right)+o\left(\dfrac{1}{n}\right)\\
&=1+\dfrac{x}{n}+o\left( \dfrac{1}{n}\right)

\end{aligned}
then $$\fbox{$r_{n}=1+\dfrac{x}{n}+o\left( \dfrac{1}{n}\right)$}$$

• why if $r_n=1+\dfrac{x}{n}+o\left(\dfrac{x}{n} \right)$ then $\ln(r_n)\sim \dfrac{x}{n}$

note that :

if $u_n\sim v_n $ and $v_n \sim w_n $ then $u_n \sim w_n$
$u_n\sim v_n \iff u_n=v_n+o(v_n)$

i can't show this

• how we can get the expression of $\tan(\alpha_n)$

If $a+ib=\rho e^{i\theta}$ with $a>0$ then $\tan(\theta)=\frac b a$

since $\left( 1+\dfrac{z}{n}\right)=\left( 1+\dfrac{x}{n}\right)+i\dfrac{y}{n}=r_{n}e^{i\alpha_n} $ then
$$\tan(\alpha_n)=\dfrac{\dfrac{y}{n}}{1+\dfrac{x}{n}}=\dfrac{y}{x+n}$$
$$\fbox{$\tan(\alpha_n)=\dfrac{y}{x+n} $} $$

• why if $\tan\left( \alpha_n \right)\sim \dfrac{y}{n}$ then $\alpha_{n}\sim \dfrac{y}{n}$

if $u_n\sim v_n $ and $v_n \sim w_n $ then $u_n \sim w_n$

So should show that $\alpha_n \underset{n \to +\infty}{\overset{}{\longrightarrow}}0$ to be able to say that $\tan(\alpha_n)\sim r_n$

we've $\tan(\alpha_n)=\dfrac{y}{x+n}$ then $\alpha_n=\arctan\left(\dfrac{y}{x+n}\right) $
$$\lim_{n\to +\infty}\alpha_n=\lim_{n\to +\infty} \arctan\left(\dfrac{y}{x+n}\right)=\arctan\left(\dfrac{y}{x+\lim_{n\to +\infty} n}\right)=\arctan(0)=0 $$

then $$\begin{cases}\tan(\alpha_n) \sim \alpha_n \\ \tan(\alpha_n)\sim \dfrac{y}{n} \end{cases} \implies \alpha_n\sim \dfrac{y}{n}$$

• If my proof wrong would you elaborate the steps