Limit of $-1+i\frac{(-1)^n}{n^2}$ in Polar Coordinates

Question

Consider the sequence $z_n=-1+i\frac{(-1)^n}{n^2}$m which converges to $-1$. However, in polar coordinates, $$r_n=|z_n|=\sqrt{1+\frac{1}{n^4}} ,\\\Theta_n=\text{Arg}(z_n).$$ We have $r_n\to 1$, but $\lim \Theta_n$ does not exist since $\Theta_{2n}\to \pi$ and $\Theta_{2n-1}\to -\pi.$

If the limit is $-1=e^{\pi i}$, why doesn't $\Theta_n$ converge to $\pi$?

Answer 1

Since we are studying a sequence that converges to $-1\in{\mathbb C}$ we better use $$\phi(z):={\rm the}\bigl({\rm arg}(z)\,\cap\>]0,2\pi[\>\bigr)$$ as real representant of the ${\rm arg}$ function. (Note that ${\rm arg}$ takes values in ${\mathbb R}/(2\pi{\mathbb Z})$.)

We then have $$\phi(z_n)=\pi+\alpha_n\>,\qquad |\alpha_n|<{\pi\over2}\ ,$$ whereby $|\tan\alpha_n|={1\over n^2}\to0$ $\>(n\to\infty)$. It follows that $\lim_{n\to\infty}\phi(z_n)=\pi$.