Find the accumulation points in overlinemathbb C=mathbb…

Find the accumulation points in $\overline{\mathbb C}=\mathbb C\cup\{\infty\}$ for: $\ 1^\circ\ \left[(-1)^n+i^n\right]\sqrt n;\quad 2^\circ\ \dfrac1n+e^{in\alpha}\ (\alpha\in\mathbb R)$.

Part $1^\circ$: $z_n=\left[(-1)^n+i^n\right]\sqrt n$

Since $(-1)^n$ and $i^n$ both have period dividing $4$, so does their sum:

$$ (-1)^n+i^n=\begin{cases}-1+i & n\equiv1\!\!\pmod4\\ 0 & n\equiv2\!\!\pmod4\\ -1-i & n\equiv3\!\!\pmod4\\ 2 & n\equiv0\!\!\pmod4\end{cases} $$

Along each residue class:

$$ n\equiv2\!\!\pmod4:\ z_n=0\to0; \qquad \text{all other classes: } z_n=(\text{nonzero constant})\cdot\sqrt n\to\infty $$

The three unbounded subsequences head off in three different directions ($\arg=\frac{3\pi}4,-\frac{3\pi}4,0$ respectively), but $\overline{\mathbb C}$ has only a single point at infinity, so all three collapse to the same accumulation point there.

$$ \boxed{\text{Accumulation points: } \{0,\ \infty\}} $$

Part $2^\circ$: $z_n=\dfrac1n+e^{in\alpha}$

Since $\frac1n\to0$, the limiting behavior is governed entirely by $e^{in\alpha}$.

Case A · $\alpha=2\pi\dfrac pq$ a rational multiple of $2\pi$ ($\gcd(p,q)=1$)

Then $e^{in\alpha}=e^{i2\pi np/q}$ is periodic in $n$ with period $q$, and because $\gcd(p,q)=1$, as $n$ ranges over residues $0,\ldots,q-1\pmod q$ the values $np\bmod q$ sweep out all of $0,\ldots,q-1$ too — so the $q$ subsequence limits are exactly the $q$-th roots of unity:

$$ \boxed{\text{Accumulation points: } \left\{e^{2\pi ij/q}: j=0,1,\ldots,q-1\right\}} $$

(Special case $\alpha=0$, i.e. $q=1$: a single accumulation point, $z_n\to1$.)

Case B · $\alpha/2\pi$ irrational

By Weyl's equidistribution theorem, $\{e^{in\alpha}\}$ is dense on the unit circle (in fact equidistributed). Since $\frac1n\to0$ only perturbs each term toward the circle without changing the set of limit points:

$$ \boxed{\text{Accumulation points: the full unit circle } \{z:|z|=1\}} $$