Let $f(z)$ be the function given by the power series $\sum_{n=0}^{\infty}a_nz^n$ with radius of convergence $R$. Put $$g(z)=\frac{1}{3}(f(z)+f(w\cdot z)+f(w^2\cdot z)),$$where $w=\cos(\frac{2\pi}{3})+i\sin(\frac{2\pi}{3})$. Find the power series for $g(z)$ and estimate the radius of convergence.
My attempt:
So first we notice that $w^3=1$ and $1+w+w^2=0$. Then we find \begin{align*} g(z)&=\frac{1}{3}(\sum_{n=0}^{\infty}a_nz^n+\sum_{n=0}^{\infty}a_n(wz)^n+\sum_{n=0}^{\infty}a_n(w^2z)^n)\\ &=\sum_{n=0}^{\infty}(\frac{1+w^n+(w^n)^2}{3})a_nz^n=\sum_{n=0}^{\infty}b_nz^n \end{align*} When $n$ is a multiplum of $3$, so $n=3k,k\in \mathbb{N}_0$, then $1+w^n+(w^n)^2=1+(w^3)^k+(w^3)^{nk}=1+1+1=3$ so $b_n=a_n$ when $n=3k, k\in \mathbb{N}_0$.
After having tried some of the others like $b_1,b_2,b_4,b_5$ (when $n$ is not a multiplum of $3$) it's easy enough to see that those should be equal to $0$ because $w^3=1$ and $1+w+w^2=0$, however I'm not quite sure how to put together a strigent argument for those. Is there a way to do this without just testing a lot of them and then noticing the pattern? Also if anyone has a hint how to find the radius of convergence for $g(z)$ I would really appreciate it. Thank you for your time!
If $n$ is a multiple of $3$ then $\omega^n = 1$, so that $1+w^n+w^{2n} = 3$.
Otherwise $\omega^n \ne 1$, and the formula for (finite) geometric sums gives $$ 1+w^n+w^{2n} = \frac{w^{3n}-1}{w^n-1} = \frac{(w^3)^n-1}{w^n-1}= 0\, . $$
So we have $$ b_n = \begin{cases} a_n & \text{ if } n = 3k, k \in \Bbb Z \, .\\ 0 & \text{ otherwise.} \end{cases} $$ and therefore $g(z) = \sum_{k=0}^\infty a_{3k}z^{3k}$.
$g$ is holomorphic for $|z| < R$ so that he radius of convergence of that power series it at least $R$.
The power series for $g$ can have a larger radius of convergence. Example: $$ f(z) = z + z^4 + z^7 + \ldots = \frac{z}{1-z^3} $$ is convergent for $|z| < 1$, but $$ g(z)=\frac{1}{3}\bigl(f(z)+f(w z)+f(w^2 z) \bigr) = 0 = \sum_{k=0}^\infty 0 z^n $$ is convergent for all $z \in \Bbb C$.