find the maximum of the $C$ such $|z_{1}+z_{2}+\cdots+z_{m}|\ge C,\forall m\in N^{+}$ let the complex sequence $\{z_{n}\}$ such 
$$4z^2_{n+1}+2z_{n}z_{n+1}+z^2_{n}=0,\forall n\in N^{+}$$
where $|z_{1}|=1$,find the maximum of the $C$ such
$$|z_{1}+z_{2}+\cdots+z_{m}|\ge C,\forall m\in N^{+}$$
I try
$$4(z^2_{n+1}+\dfrac{1}{2}z_{n+1}z_{n}+\dfrac{1}{16}z^2_{n})=-\dfrac{3}{4}z^2_{n}$$
so we have
$$(z_{n+1}+\dfrac{1}{2}z_{n})^2=\left(-\dfrac{\sqrt{3}}{2}z_{n}i\right)^2$$
so we have
$$z_{n+1}=-\dfrac{1}{2}z_{n}\pm \dfrac{\sqrt{3}}{2}z_{n}i=-z_{n}e^{ix}$$
wher $x=\pm\dfrac{\pi}{3}$
then I want use this geomtric sum to get it,But I meet a problem,there are some cese,so How to solve this?
 A: @achille ’s comment shows that $C_{\max}\leq\frac1{\sqrt3}$.
On the other hand, wlog we assume that $z_1=1$ and $z_2=\frac{\omega}2$.
Suppose $z_n=\frac{\omega^{a_n}}{2^{n-1}}$, where $a_1=0,a_2=1,a_n-a_{n-1}=\pm 1(n=2,3,\cdots)$.
Let
$$A_i=\{1\leq n\leq m\mid a_n\equiv i\mod 3\}, i=0,1,2$$
and assume $A_2=\{b_1,b_2,\cdots,b_t\}$, where $b_1<b_2<\cdots<b_t$. Then $b_1\geq3$ and $b_j-b_{j-1}\geq2(j=2,3,\cdots,t)$. Hence $b_j\geq 2j+1,j=1,\cdots,t$.
Therefore
\begin{align*}
|z_1+z_2+\cdots+z_m|&=\left|\frac{\sqrt3-i}2z_1+\cdots+\frac{\sqrt3-i}2z_m\right|\\
&\geq \Re\left(\frac{\sqrt3-i}2z_1+\cdots+\frac{\sqrt3-i}2z_m\right)\\
&=\sum_{n\in A_0}\Re\frac{\sqrt3-i}{2^n}+\sum_{n\in A_1}\Re\frac{\sqrt3-i}{2^n}\frac{\sqrt3i-1}{2}+\sum_{n\in A_2}\Re\frac{\sqrt3-i}{2^n}\frac{-1-\sqrt3i}2\\
&=\sum_{n\in A_0}\frac{\sqrt3}{2^n}+\sum_{n\in A_1}0+\sum_{n\in A_2}\frac{-\sqrt3}{2^n}\\
&\geq \frac{\sqrt3}2-\sum_{n\in A_2}\frac{\sqrt3}{2^n}\\
&\geq\frac{\sqrt3}2\left(1-\frac1{2^2}-\frac1{2^4}-\cdots\right)\\
&=\frac1{\sqrt3}.
\end{align*}
This shows $C_{\max}\geq\frac1{\sqrt3}$. Hence $C_{\max}=\frac1{\sqrt3}$.
