I'm looking for the proof of the following statement:$$\mathrm{arg}(1+\frac{z}{n})=\mathrm{Im}(\frac{z}{n})+o(\frac{1}{n}).$$
I found it on page 108 of Arnold's『Ordinary Differential Equations(1973)』, though it is not apparently related to differential equations.
Here is my attempt at a proof.
Since $n\cdot \rm{arg(\it{w}\rm)}=\rm{arg(\it{w}^n\rm)}\pmod{2\pi}$ and $\rm{arg(\it e^z\rm)}=\rm{Im}(\it z\rm)\pmod{2\pi},$
$$n[\mathrm{arg}(1+\frac{z}{n})-\mathrm{Im}(\frac{z}{n})]$$ $$=\mathrm{arg}\{{(1+\frac{z}{n})}^n\}-\mathrm{Im}(z)\pmod{2\pi}$$ $$=\mathrm{arg}\{{(1+\frac{z}{n})}^n\}-\mathrm{arg}(e^z\rm)\pmod{2\pi}$$ $$=\mathrm{arg}\{(1+\frac{z}{n})^n/e^z\}\pmod{2\pi}.$$ But $(1+\frac{z}{n})^n/e^z \to 1$ as $n\to\infty$, so $n[\mathrm{arg}(1+\frac{z}{n})-\mathrm{Im}(\frac{z}{n})]\to2k\pi$ as $n \to \infty$, for some integer k.
Therefore $$\mathrm{arg}(1+\frac{z}{n})=\mathrm{Im}(\frac{z}{n})+\frac{2k\pi}{n}+o(\frac{1}{n}).$$
How can I verify, if my argument is correct, that $k=0$?
This is not a problem of coping with "modulo $2\pi$", but a limit problem.
Assume that $z=x+iy$ is given and fixed. Then $${\rm Re}\left(1+{z\over n}\right)\geq1-{|z|\over n}>0$$ as soon as $n>|z|$, which we shall assume from now on. Then $${\rm arg}\left(1+{z\over n}\right)={\rm arg}\left(1+{x+iy\over n}\right)=\arctan{{y\over n}\over 1+{x\over n}}$$ and therefore $${\rm arg}\left(1+{z\over n}\right)-{y\over n}=\arctan{y\over n+x}-{y\over n}=:a_n\ .$$ It remains to show that $a_n=\,o\bigl({1\over n}\bigr)$ when $n\to\infty$. Now $\arctan t=t+o(t)$ when $t\to0$. It follows that $$a_n={y\over n+x}+o\left({y\over n+x}\right)-{y\over n}={-xy\over n(n+x)}+o\left({y\over n+x}\right)=o\left({1\over n}\right)\qquad(n\to\infty)\ .$$