I am reading Complex Variables and Applications 8th ed by James Brown and Churchill. On Pg $183$ Example $2$, it says
Let $z_n=-2+\iota\frac{(-1)^n}{n^2}\;\; n\in N$
Using, $\lim_{n\to \infty} z_n =\lim_{n\to \infty} (x_n+\iota y_n) = \lim_{n\to \infty}x_n + \iota \lim_{n\to \infty} y_n$
We get $\lim_{n\to \infty} z_n = \lim_{n\to \infty} -2 +\iota \lim_{n\to \infty}\frac{(-1)^n}{n^2} = -2 $
But the book further says that,
If we use polar coordinates, then
$r_n = |z_n|$ and $\theta_n = Arg(z_n)$ where $Arg(z_n)$ denotes the principal argument ($-\pi<\theta<\pi$), we find that,
$\lim_{n\to \infty} r_n = \lim_{n\to \infty} \sqrt{(4+ \frac1{n^4})} = 2$
but that
$\lim_{n\to \infty}\theta_{2n}= \pi$ ---($1$) and
$\lim_{n\to \infty} \theta_{2n-1}=-\pi$ ---($2$)
Evidently, $\theta_n$ limit does not exists.
First of all , I want to know why the author is trying to use polar coordinates?
Secondly, I don't understand how he got $1$ and $2$?
Because $\theta_n = \arctan(\frac{\frac{(-1)^n}{n^2}}{-2})\to \arctan(0) = 0$
Also is it true that if $z_n = r_n e^{\iota \theta_n}$ and $z=re^{\iota \theta}$
then $z_n \to z$ $\iff$ $r_n \to r$ and $\theta_n \to \theta$ -- ($3$)
Is $3$ true?
First of all, it is certainly true that $z_n \to -2$ as you have observed. What the book is trying to say is that the principal logarithm of $z_n$ does not tend to principal logarithm of $-2$. The principal logarithm is defined using the principal argument which is always assumed to be a number in $(-\pi, \pi]$. For a point like $-2+it$ with $t$ positive and close to $0$ the argument is less than and close to $\pi$. On the other hand, for a point like $-2-it$ with $t$ positive and close to $0$ the argument is gretater than and close to $-\pi$. Hence the limit of $arg(z)$ as $z \to -2$ does not exist.