The convergence radius of a complex series its the same of his real and imaginary series?

Question

I'm trying to calculate the radius of convergence of $\displaystyle\sum (1-1/n)^n\sin (n \alpha) z^{n}$ and my idea was to use the Hadamard formula, but I have no idea what to do about the $\sin$ and then I noticed that $\sin(n\alpha) = -i Im(e^{in\alpha})$ and then I thought that if the ratio would be the same as $\displaystyle\sum (1-1/n)^ne^{in\alpha} z^{n}$.

Answer 1

For a fixed real value of $\alpha$ :

(1) .For brevity let $A_n=(1-1/n)^n\sin n \alpha. $ We have $|A_n|^{1/n}|\leq 1-1/n, $ so the radius $r$ of the convergence is at least $1.$

().If $\alpha /\pi \in \mathbb Z$ then every term in the series is $0$ for all $z,$ the sum is always $0,$ and the radius of convergence is $\infty.$

(3).When $\alpha /\pi \not \in \mathbb Z :$

(3a). If $\alpha /\pi \in \mathbb Q$ \ $\mathbb Z ,$ let $S=\{n \in \mathbb N: \sin n \alpha=\sin \alpha\}$. Then $S$ is infinite. And $\sin \alpha \ne 0.$ So $\lim_{n\to \infty, n\in S}|A_n|^{1/n}=$ $\lim_{n\to \infty,n\in S}(1-1/n)|\sin \alpha |^{1/n}=1.$ So the radius $ r $ of convergence is at most $1.$ From (1) therefore $r=1.$

(3b). If $\alpha /\pi$ is irrational, we quote the following theorem: For real $x$ let $[x]$ denote the largest integer not exceeding $x.$ If $x \in \mathbb R$ \ $\mathbb Q$ then $\{nx-[nx]: n\in \mathbb N\}$ is dense in $[0,1].$ Corollary. If $\alpha \in \mathbb R$ and $\alpha /\pi \not \in \mathbb Q$ then the set $T=\{n\in \mathbb N: \sin n\alpha \in (1/2,1)\}$ is infinite.

So $\lim_{n\to \infty,n\in T}|A_n|^{1/n}\geq$ $ \lim_{n\to \infty , n\in T}(1-1/n)(1/2)^{1/n}=1.$ So $r\leq 1$ and by (1) we have $r=1.$

Answer 2

It's enough to consider $0<\alpha < \pi.$ Let $A = \{e^{it}: (\pi - \alpha)/2\le t \le (\pi + \alpha)/2\}.$ Then as $e^{in\alpha}$ marches around the unit circle as $n\to \infty$ through integer values, it must land in $A$ infinitely many times. Why? The arc length of $A$ is precisely $\alpha.$ It's impossible to step over this arc if you are going around the circle in steps of arc-length $\alpha.$ So let $n_1,n_2,\dots $ be the subsequence of $\mathbb N$ for which this happens. Then $\limsup |\sin(n_k\alpha)|^{1/n_k} \ge \limsup [\sin((\pi - \alpha)/2)]^{1/n_k}.$ Since $\sin((\pi - \alpha)/2)>0,$ that $\limsup $ is $1.$ It follows that the radius of convergence in question is $1.$