Why $e^{-n\pi i} + (-1)^{n}i$ is divergent?

Question

This is not a homework question.

The series in question is: $$z_{n} = e^{-n\pi i} + (-1)^n i$$

Simplification would results into: $$ z_{n} = \cos(n\pi)+ (-1)^n i$$

and I think neither of these limits, $lim_{n \to \infty} \cos(n\pi)$ and $ \lim_{n \to \infty} (-1)^n$, exists. Correct?

Answer 1

Hint:

If $a_n,b_n\in\mathbb R$ then $(a_n+ib_n)_n$ will converge if and only if $(a_n)_n$ and $(b_n)_n$ both converge.

Answer 2

Starting from the beginning where we take $n\in\Bbb Z$ with

$$z_{n} = e^{-n\pi i} + (-1)^n i\tag 1$$

we have $e^{-n\pi i}=\cos(n\pi)-i\sin(n\pi)=\cos(n\pi)-i\cdot 0$, so we can rewrite $(1)$ as

$$z_n=\cos(n\pi)+(-1)^ni\tag 2$$

Now we have two sequences, $a_n=\cos(n\pi)=1,-1,1,-1,\dots$ and $b_n=(-1)^ni=i,-i,i,-i,\dots$ Individually, neither sequence converges. Summed together, the result (i.e., $z_n$) cannot converge, since we have

$$z_n=\cos(n\pi)+(-1)^ni=1+i,-1-i,1+i,-1-i,\dots$$