logaritmic function in complex plane

Question

$$ \lim_{n\to \infty} log (-1+\frac{1}{n}i) = \pi i $$ and $$ \lim_{n\to \infty} log (-1-\frac{1}{n}i) =- \pi i $$

From $$ log (z)= log(\mid z \mid) + \theta i $$

$$ log (-1+\frac{1}{n}i)= log(1+\frac{1}{n^2}) +arctan(....)i $$