Let $f_1(x) = (x-2)^2$, $f_2(x) = ((x-2)^2 - 2)^2$, $f_3(x) = \left((x-2)^2 - 2)^2 - 2\right)^2$, and so on; so that $f_k(x) = \left(...\left((x-2)^2 - 2\right)^2 - ... - 2\right)^2 = A_k + B_k x + C_k x^2 + D_k x^3 + ...$. Then $C_3$ is equal to:
Step-by-Step Solution
Key Concept: When a product of complex ratios is positive real, the arguments sum to zero; this concyclic condition is equivalent to equal inscribed angles.
Compute $\arg\left(\frac{3-z_1}{2-z_1}\right) + \arg\left(\frac{2-z_2}{3-z_2}\right) = \arg\left(\frac{3-z_1}{2-z_1} \cdot \frac{2-z_2}{3-z_2}\right)$. If the product $\frac{(3-z_1)(2-z_2)}{(2-z_1)(3-z_2)}$ is a positive real number, the argument is zero. This makes the angles $\theta_1$ and $\theta_2$ equal in magnitude but opposite in sign, so chord $DC$ subtends equal angles at points $A$ and $B$. Therefore, the four points are concyclic for $K > 0$.
Correct Answer: I need to find C₃, the coefficient of x² in f₃(x).
Let me work through this systematically.
**Finding f₁(x):**
f₁(x) = (x-2)² = x² - 4x + 4
So: A₁ = 4, B₁ = -