Given a power series $f(z) = \sum\limits_{n=0}^\infty a_nz^{n}$ where $z\in \mathbb{C}$ and radius of convergence $R.$ Then my goal is to find, $$\sum_{n=0}^\infty a_{kn}z^{kn}$$ for $|z|<R$ and $k\in \mathbb{N}.$
So I tried two examples and I think there is a connection with roots of unity. \begin{align*} \sum_{n=0}^\infty a_{2n}z^{2n} &=\frac{1}{2}(f(z)+f(i^2z))\\ \sum_{n=0}^\infty a_{3n}z^{3n} &=\frac{1}{2}(f(z)+f(iz) + f(i^2z)) \end{align*} My guess is that $$\sum_{n=0}^\infty a_{kn}z^{kn} =\frac{1}{2}(f(z)+f(iz) + f(i^2z)+ \cdots + f(i^{k-1}z))$$ if $k$ is odd. For $k$ even I am not sure. Any generalizations of this fact or proof ideas will be much appreciated.
Let $\omega \ne 1$ be a $k$-th root of unity: $\omega^k = 1$. Then $$ \sum_{j=0}^{k-1} \omega^{nj} = \begin{cases} k & \text{ if $n$ is a multiple of $k$,} \\ 0 & \text{ otherwise.} \end{cases} $$ Consequently, $$ \frac 1k \sum_{j=0}^{k-1} f(\omega^j z ) = \sum_{n=0}^\infty a_{kn}z^{kn} $$
For $k=2$ this gives your result $$ \sum_{n=0}^\infty a_{2n}z^{2n} =\frac{1}{2}(f(z)+f(-z))\\ $$ For $k=3$ the correct result is $$ \sum_{n=0}^\infty a_{3n}z^{3n} =\frac{1}{3}(f(z)+f(\omega z) + f(\omega^2 z)) $$ with $\omega = - \frac 12 \pm \frac{\sqrt 3}{2}$.