Find all entire functions that satisfy $f(2z) = (1-2z)f(z)$ This is for homework, and I could use a little help.  The question asks

Find all entire functions that satisfy $f(2z) = (1-2z)f(z)$.

Here is what I have done so far.  Since $f$ is entire, I wrote
$$ f(z) = \sum_{n=0}^{\infty} a_n z^n = a_0 + a_1z + a_2z^2 + a_3z^3 + a_4z^4 + \dotsb $$
for some $z \in \mathbb{C}$.  Then
$$ f(2z) = a_0 + 2a_1z + 4a_2z^2 + 8a_3z^3 + 16a_4z^4 + \dotsb $$
and
$$ (1-2z)f(z) = a_0 + (a_1-2a_0)z + (a_2-2a_1)z^2+(a_3-2a_2)z^3 + (a_4-2a_3)z^4 + \dotsb. $$
Comparing coefficients, I find that
\begin{align*}
a_0 &= a_0 \\
a_1 &= -2a_0 \\
a_2 &= \frac{2^2}{3}a_0 \\
a_3 &= -\frac{2^3}{7 \cdot 3}a_0 \\
a_4 &= \frac{2^4}{15 \cdot 7 \cdot 3}a_0 \\
a_5 &= \frac{2^5}{31 \cdot 15 \cdot 7 \cdot 3} a_0 \\
&\vdots
\end{align*}
Now $f$ looks like
$$ f(z) = a_0 \left( 1 - 2z + \frac{2^2}{3}z^2 - \frac{2^3}{7 \cdot 3}z^3 + \frac{2^4}{15 \cdot 7 \cdot 3}z^4 - \frac{2^5}{31 \cdot 15 \cdot 7 \cdot 3}z^5 + \dotsb \right). $$
Does the series in parenthesis represent any elementary function?  Besides the denominators, it looks like the Taylor expansion of $e^{-2z}$.

Your function has zero at $z=1/2$. I think that your formulas should be easier to work with if you expand at that point.
