## Trigonometry (11th Edition) Clone

We can see that the value is approaching $w_3 = -\frac{1}{2}-\frac{\sqrt{3}}{2}i$. Therefore, the point $~~z_1 = -1-i~~$ should be colored yellow.
$f(z) = \frac{2z^3+1}{3z^2}$ $z_1 = -1-i$ We can evaluate the function for $z_1$: $z_2 = f(z_1)$ $z_2 = \frac{2(-1-i)^3+1}{3(-1-i)^2}$ $z_2 = \frac{2(2-2i)+1}{3(2i)}$ $z_2 = (\frac{5-4i}{6i})(\frac{i}{i})$ $z_2 = \frac{4+5i}{-6}$ $z_2 = -0.667 - 0.833i$ We can evaluate the function for $z_2$: $z_3 = f(z_2)$ $z_3 = \frac{2(-0.667 - 0.833i)^3+1}{3(-0.667 - 0.833i)^2}$ $z_3 = \frac{2(1.092-0.534i)+1}{3(-0.249+1.111i)}$ $z_3 = (\frac{3.184-1.068i}{-0.747+3.333i})(\frac{-0.747-3.333i}{-0.747-3.333i})$ $z_3 = \frac{-5.938-9.814i}{11.667}$ $z_3 = -0.509-0.841i$ We can see that the value is approaching $w_3 = -\frac{1}{2}-\frac{\sqrt{3}}{2}i$. Therefore, the point $~~z_1 = -1-i~~$ should be colored yellow.