Step-by-Step Solution
Key Concept: When solving equations with inverse trigonometric functions, convert to algebraic equations and find common solutions.
Since $z^2 - 5z + 5 \geq 0$, we have $0 \leq \sin^{-1}(\sqrt{z^2 - 5z + 5}) \leq \frac{\pi}{2}$. Since $4z - z^2 - 3 \geq 0$, we have $0 \leq \cos^{-1}(\sqrt{4z - z^2 - 3}) \leq \frac{\pi}{2}$. For LHS = RHS, we need $\sin^{-1}(\sqrt{z^2 - 5z + 5}) = \frac{\pi}{5}$ and $\cos^{-1}(\sqrt{4z - z^2 - 3}) = \frac{\pi}{5}$. Solving: $z^2 - 5z + 5 = 1 \& 4z - z^2 - 3 = 0$, giving $z^2 - 5z + 4 = 0$ and $z^2 - 4z + 3 = 0$. The common solution is $z = 1$.
Correct Answer: 1