Find all $z \in \Bbb{C}$ such that $\sum_{n=1}^{\infty}\sin(n)z^n$ converges 
Find all $z \in \Bbb{C}$ such that $\sum_{n=1}^{\infty}\sin(n)z^n$ converges

I started by trying to compute the radius of convergence of the series, however Im absolutely stuck in finding $\limsup\sqrt[n]{|\sin(n)|}$. 
I know that $\sin(n) \neq 1$ for every $n \in \Bbb{N}$ but I don't know how close can it get to $1$. Any hint?
 A: As (at the time I was typing) no-one else had posted a complete answer, I'm reinstating mine, with corrections due to insightful comments from A.S.
Because $\pi$ is irrational, the additive subgroup of $\Bbb{R}$ generated by $\Bbb{Z}$ and $\pi/2$ is dense in $\Bbb{R}$. This means that for any $N \in \Bbb{N}$ and $\epsilon > 0$ there is $n > N$ such that $|\sin(n) - 1| < \epsilon$. So $\sin(n)$ and hence $|\sin(n)z^n|$ does not tend to $0$ as $n$ tends to $\infty$ if $|z| \ge 1$, so $\sum_{n=1}^{\infty}\sin(n)z^n$ cannot converge. If $|z|< 1$, then $\sum_{n=1}^{\infty}\sin(n)z^n$  converges by the ratio test or comparison with a geometric series.
A: EDITED $-1 \le \sin(n) \le 1$, so the series converges for $|z| < 1$.
Irrationality of $\pi$ implies $\sin(n)$ takes values arbitrarily close to $\pm 1$, so it diverges for $|z| \ge 1$.  
A: Hint: Write as
$$\Im \left[ \sum_{n=1}^\infty (e^{i}z)^n \right].$$
A: Look at my question and resolution, this might help you with your exercise:
Is series $\displaystyle\sum^{\infty}_{n=1}\frac{\cos(nx)}{n^\alpha}$, for $\alpha>0$, convergent?
It's almost the same situation, apply it to your problem.
A: There is a more general result here, and it has nothing to do with the irrationality of $\pi.$ Claim: $\sum \sin (nx)z^n$ diverges for all $z,|z|=1$ and for all $x\in \mathbb {R}\setminus \pi\mathbb {Z}.$ I'll treat only the case $0< x \le 1.$ (See if you can fill in the details for the other values of $x.$) The proof is simple: As we view $e^{inx}$ marching around the unit circle infinitely many times, we see it has to land in the arc $A=\{e^{it}: \pi/4 <t < 3\pi/4\}$ infinitely many times. Why? Because the steps are of arc-length less than $1,$ the arc-length of $A$ is $\pi/2 > 1,$ and therefore we land in $A$ at least once every orbit. (Just like your momma said: You can't step over a puddle longer than your step.) So $\sin (nx)\ge 1/\sqrt 2$ for infinitely many $n,$ giving the result. There's surely a result here for any periodic function.
