The Modulus of Sequences of Complex Numbers Find the limit of each sequence that converges. If the sequence diverges, explain why.

(a) $z_n=\frac{(1-i)^{n+1}}{2^{n/2}}$
(b) $z_n=n(1-\cos(\frac{\pi}{n})-i\sin(\frac{\pi}{n}))$

For part(a), my reasoning is to rewrite the sequence as:
$z_n=(1-i)\left({\frac{(1-i)}{\sqrt{2}}}\right)^n$. Then, observe that $|\frac{1-i}{\sqrt{2}}|$=1 and so the sequence $w_n=\left({\frac{(1-i)}{\sqrt{2}}}\right)^n$ diverges because the terms of $w_n$ go around a circle of radius 1 counterclockwise forever and never approach a single value. Because $w_n$ diverges, would this imply that $z_n$ diverges? This was true in calculus, that any constant multiple of a divergent sequence was also divergent.
For part(b), my reasoning is that $|z_n|=n\sqrt{1-2\cos(\frac{\pi}{n}})$ and so since $|z_n|$ converges to $\infty$, it follows that the limit of our original sequence doesn't exist.
If someone could analyze my arguments and see if they are acceptable and correct I would greatly appreciate it.
 A: The essence of the first part is quite ok but the reasoning should be more concise. The sequence in the second part converges. So, your arguments for this part are not correct.
In case of sequences where rotation matters it is often convenient to transform the complex number in polar coordinates.
\begin{align*}
  z=x+iy=|z|e^{i\varphi}
  \end{align*}

Part (a): $z_n=2^{-\frac{n}{2}}\left(1-i\right)^{n+1}$
Using polar coordinates $1-i=\sqrt{2}e^{i\frac{\pi}{4}}$ we obtain
  \begin{align*}
  z_n&=2^{-\frac{n}{2}}\left(1-i\right)^{n+1}
  =2^{-\frac{n}{2}}\left(\sqrt{2}e^{-\frac{i\pi}{4}}\right)^{n+1}
=2^{-\frac{n}{2}}2^{\frac{n+1}{2}}e^{-\frac{i(n+1)\pi}{4}}\\
    &=\sqrt{2}e^{-\frac{i(n+1)\pi}{4}} \tag{1}
  \end{align*}

In (1) we see the part $e^{-\frac{i(n+1)\pi}{4}}$ is relevant for rotation. The formulation goes around a circle of radius $1$ counterclockwise forever could be made somewhat more concise. One aspect of this rotation forever   is that different values are infinitely often attained by the sequence $\left(z_n\right)$.
Now recall that following is valid:
If $(z_n)$ is a sequence convergent with limit $z$ then every subsequence $(z_{n_k})$ of $z_n$ converges to $z$.

We select two subsequences of $(z_n)$ which converge to different values proving that $(z_n)$ is not convergent. We observe
  \begin{align*}
z_{8n-1}&=\sqrt{2}e^{-\frac{\pi8n}{4}}=\sqrt{2}e^{-i2n\pi}=\sqrt{2}\\
z_{8n+3}&=\sqrt{2}e^{-\frac{\pi(8n+4)}{4}}=\sqrt{2}e^{-i(2n+1)\pi}=\sqrt{2}e^{-i\pi}=-\sqrt{2}\\
\end{align*}
We conclude: Since the subsequences  $(z_{8n-1})$ and $(z_{8n+3})$ converge to different values $\sqrt{2}$ resp. $-\sqrt{2}$ the sequence $(z_n)$ is divergent.

$$ $$

Part (b): $z_n=n(1-\cos \frac{\pi}{n}-i\sin \frac{\pi}{n})$
Here we have two competitive aspects. On the one hand the factor $n$ tends (linearly) to $\infty$. But observe, that both other parts
  \begin{align*}
\lim_{n\rightarrow\infty}\left(1-\cos\frac{\pi}{n}\right)=0\qquad\text{and}
\qquad\lim_{n\rightarrow\infty}\sin\frac{\pi}{n}=0
\end{align*}
  tend to $0$.
Per se it is not obvious which tendency is dominant, if any. So, we have to analyse it and we could try to apply L'Hospital's Rule.
We note that a sequence $(z_n)$ converges if and only if both sequences, $(\operatorname{Re} z_n)$ and
  $(\operatorname{Im} z_n)$ converge. We obtain
  \begin{align*}
\lim_{n\rightarrow\infty}&\frac{1-\cos \frac{\pi}{n}}{\frac{1}{n}}
=\lim_{n\rightarrow\infty}\frac{-\frac{\pi}{n^2}\sin \frac{\pi}{n}}{-\frac{1}{n^2}}
=\lim_{n\rightarrow\infty}\left(\pi\sin\frac{\pi}{n}\right)=0\tag{2}\\
\lim_{n\rightarrow\infty}&\frac{\sin \frac{\pi}{n}}{\frac{1}{n}}
=\lim_{n\rightarrow\infty}\frac{-\frac{\pi}{n^2}\cos \frac{\pi}{n}}{-\frac{1}{n^2}}
=\lim_{n\rightarrow\infty}\left(\pi\cos\frac{\pi}{n}\right)=\pi\tag{3}\\
\end{align*}

We can now apply L'Hospital's Rule, since with $f(n)=1-\cos \frac{\pi}{n}$  and $g(n)=n$, we have that $\lim_{n\rightarrow\infty}f(n)=0, \lim_{n\rightarrow\infty}g(n)=0$ and the limit 
$
\frac{f^\prime}{g^\prime}
$ exists and is finite. The same holds with $f(n)=\sin\frac{\pi}{n}$ and $g(n)=n$.

We conclude from (2) and (3) the sequence $(z_n)$ converges to $-i\pi$:
  \begin{align*}
\lim_{n\rightarrow \infty}z_n&=\lim_{n\rightarrow\infty}n\left(1-\cos \frac{\pi}{n}-i\sin \frac{\pi}{n}\right)\\
&=\lim_{n\rightarrow\infty}n\left(1-\cos \frac{\pi}{n}\right)-i\lim_{n\rightarrow\infty}\sin \frac{\pi}{n}\\
&=-i\pi
\end{align*}

A: The sequence in (a) doesn't converge, the one in (b) does. Unfortunately, both your arguments are wrong: the first one is not conclusive, because you don't give any real reason for nonconvergence; the second one is wrong because the sequence does converge.
For (a) observe that $(1-i)^2=1-2i-1=-2i$, so $(1-i)^3=-2i(1-i)=-2(1+i)$ and $(1-i)^8=(-2i)^4=2^4$; thus
$$
z_{8m}=\frac{(1-i)^{8m+1}}{2^{4m}}=
\frac{(1-i)^{8m}(1-i)}{2^{4m}}=1-i
$$
and
$$
z_{8m+2}=\frac{(1-i)^{8m+3}}{2^{4m+1}}=
\frac{(1-i)^{8m}(1-i)^3}{2^{4m+1}}=
\frac{-2^{4m+1}(1+i)}{2^{4m+1}}=-(1+i)
$$
Since the sequence has two subsequences that converge to different limits, it doesn't converge.
For (b) recall the basic limit
$$
\lim_{x\to0}\frac{\sin x}{x}=1
$$
from which it follows that
$$
\lim_{x\to0}\frac{1-\cos x}{x}=
\lim_{x\to0}\frac{1-\cos^2x}{x(1+\cos x)}=
\lim_{x\to0}\frac{\sin x}{x}\,\frac{\sin x}{1+\cos x}=0
$$
The basic limit has also the consequence that
$$
\lim_{n\to\infty}\frac{\sin(\pi/n)}{\pi/n}=1
$$
and therefore
$$
\lim_{n\to\infty}n\sin\frac{\pi}{n}=\pi
$$
On the other hand,
$$
\lim_{n\to\infty}\frac{1-\cos(\pi/n)}{\pi/n}=0
$$
and so, recalling that convergence of a sequence in $\mathbb{C}$ is equivalent to convergence of the real and the imaginary part,
$$
\lim_{n\to\infty}n\left(1-\cos\frac{\pi}{n}-i\sin\frac{\pi}{n}\right)=
-i\pi
$$
A: I think your answer in part (b) is sufficient. In part (a), though, you can actually compute $z^n$ for every $n$. Sure, it spins around in a circle, but how?
