power series representation in terms of another Hi how to do the following:
Given 
$f(z) = \sum c_n z^n$
How to express $\sum c_{3n} z^{3n}$ in terms of $f(z)$
Thanks a lot!
 A: Hint: We give an approach using complex numbers. Let $\omega_1=\frac{-1+\sqrt{-3}}{2}$ and $\omega_2=\frac{-1-\sqrt{-3}}{2}$. If $k$ is divisible by $3$, then $1^k+\omega_1^k+\omega_2^k=3$. In all other cases, $1^k+\omega_1^k+\omega_2^k=0$. This is because if $k\equiv 1\pmod{3}$, then $\omega_1^k=\omega_1$ and $\omega_2^k=\omega_2$, while if $k\equiv 2\pmod{3}$, then  $\omega_1^k=\omega_2$ and $\omega_2^k=\omega_1$.
Now consider 
$$\frac{f(z)+f(\omega_1 z)+f(\omega_2 z)}{3}.$$
Substitute in the power series, and add. The terms involving powers not divisible by $3$ will vanish. 
Remark: A similar approach, using fourth roots of unity instead of cube roots, will give an expression for $\sum c_{4n} z^{4n}$. At a simpler level, using the square roots of unity, we obtain an expression for $\sum c_{2n}z^{2n}$. 
A: If you allow the use of complex numbers (numbers of the form $a + bi$ where $i=\sqrt{-1}$ and $a$ and $b$ are ordinary real numbers) then there are three cube roots of 1:
$$
\begin{array}{l}
1 \\
\frac{-1+i\sqrt{3}}{2} \equiv \omega \\
\frac{-1-i\sqrt{3}}{2} = \omega^2 \\
\end{array}
$$ 
You can easily verify that 
$$
\omega^2 = \left(\frac{-1+i\sqrt{3}}{2}\right)^2 = 
\frac{1 -2i\sqrt{3} - 3}{4} = \frac{-1-i\sqrt{3}}{2}
$$ 
and that 
$$
\omega^3 = \frac{-1+i\sqrt{3}}{2} \frac{-1-i\sqrt{3}}{2} = \frac{1+3}{4} = 1
$$
Now let's look at what $f(z\omega)$ would be: When the power is a multiple of 3, the term will remain $c_{3n}z^{3n}$ because $\omega^{3n} = 1$. But when the power is $3n+1$ the term is $\omega c_{3n+1}z^{3n+1}$ and when the power is $3n+2$ the term is 
$\omega^2 c_{3n+2}z^{3n+2}$.
$$
f(\omega z) = \sum c_{3n} z^{3n} + \omega \sum c_{3n+1}z^{3n+1}  + \omega^2 \sum c_{3n+2}z^{3n+2}  
$$
Similarly, 
$$
f(\omega^2 z) = \sum c_{3n} z^{3n} + \omega^2 \sum c_{3n+1}z^{3n+1}  + \omega \sum c_{3n+2}z^{3n+2}  
$$ because $\omega^4 = \omega$.
Now here is the cute step: Notice that $\omega + \omega^2 = -1$. So 
$$
f(\omega z) + f(\omega^2 z)= 2 \sum c_{3n} z^{3n} -\sum c_{3n+1}z^{3n+1}  -\sum c_{3n+2}z^{3n+2}
$$
and the answer to your question is obtained by adding $f(z)$ to get rid of all the $3n+1$ and $3n+2$ powers:
$$
\sum c_{3n} z^{3n} = \frac{f(z)+f(\omega z) + f(\omega^2 z)}{3}
$$
