Think A Little Bit

Since $z_1 = \frac{3-\sqrt{5}}{8}$ , it is a root of $4x^2 -3x + \frac{1}{4} = 0$ . So that $4z_1^2 = 3z_1 - \frac{1}{4}$ .

Now, we have:

$\begin{aligned} z_2 & = 4z_1(1-z_1) \\ & = 4z_1 - 4z_1^2 \\ & = 4z_1 - 3z_1 + \frac{1}{4} \\ \Rightarrow z_2 & = z_1 + \frac{1}{4} \\ z_3 & = 4z_2 (1-z_2) \\ & = 4 \left( z_1+\frac{1}{4} \right) \left(1 - z_1-\frac{1}{4} \right) \\ & = (4z_1+1) \left(1 - z_1-\frac{1}{4} \right) \\ & = 4z_1 \left(1 - z_1 \right) - z_1 + 1 - z_1-\frac{1}{4} \\ & = z_2 - 2z_1 + \frac{3}{4} \\ & = z_1 + \frac{1}{4} - 2z_1 + \frac{3}{4} \\ \Rightarrow z_3 & = 1 - z_1 \\ z_4 & = 4z_3(1-z_3) \\ & = 4 (1-z_1)z_1 \\ \Rightarrow z_4 & = z_2 \\ z_5 & = 1 - z_1 \\ z_6 & = z_2 \\ ... & \quad ... \end{aligned}$

$\Rightarrow z_n = \begin{cases} 1-z_1 & \text{if } n > 1 \text{ is odd.} \\ z_2 & \text{if } n \text{ is even.} \end{cases}$

Therefore,

$\begin{aligned} z_5z_{10} & = (1-z_1)z_2 \\ & = (1-z_1) \left( z_1+\frac{1}{4} \right) \\ & = \frac{1}{4} + \frac{3z_1}{4} - z_1^2 \\ & = \frac{1}{4} + \frac{3z_1}{4} - \frac{3z_1}{4} + \frac{1}{16} \\ & = \frac{5}{16} \end{aligned}$

$\Rightarrow a + b = 5 + 16 = \boxed{21}$

$Z_1=\dfrac{3-\sqrt5} 8=Cos^272.\\ \therefore~Z_2=4*Cos^272*(1-Cos^272)=Sin^2(2*72)=Sin^2(36)=\dfrac{5 - \sqrt5}8.\\ \therefore~Z_3=4*Sin^2(36)*Cos^2(36)=Sin^2(72)=\dfrac{5 + \sqrt5}8.\\ \therefore~Z_4=4*Sin^2(72)*Cos^2(72))=Sin^2(144)=Sin^2(36)\\ \text{So for odd n>1, }Z_n=\dfrac{5 + \sqrt5}8\\ \text{And for even n, }Z_n=\dfrac{5 - \sqrt5}8\\ \therefore~~Z_5*Z_{10}=\dfrac{5 + \sqrt5}8*\dfrac{5 - \sqrt5}8=\dfrac{25 - 5}{8^2}=\dfrac 5 {16}\\ \implies~~a+b=5+16=21.$