1. The problem asks for the residue of the function $$f(z) = \frac{1}{(z-2)(z+2)(z-3)}$$ at the point $$z_0 = 3$$. 2. The residue at a simple pole $$z_0$$ of a function $$f(z) = \frac{g(z)}{h(z)}$$ where $$h(z_0) = 0$$ and $$h'(z_0) \neq 0$$ is given by: $$\text{Res}(f, z_0) = \frac{g(z_0)}{h'(z_0)}$$ 3. Here, rewrite $$f(z)$$ as: $$f(z) = \frac{1}{(z-3)(z-2)(z+2)}$$ 4. Define: $$g(z) = 1$$ $$h(z) = (z-3)(z-2)(z+2)$$ 5. Compute the derivative $$h'(z)$$ using the product rule: $$h'(z) = \frac{d}{dz}[(z-3)(z-2)(z+2)]$$ 6. First, expand partially or use product rule stepwise: $$h'(z) = (z-2)(z+2) + (z-3)(z+2) + (z-3)(z-2)$$ 7. Evaluate each term at $$z=3$$: - $$(3-2)(3+2) = 1 \times 5 = 5$$ - $$(3-3)(3+2) = 0 \times 5 = 0$$ - $$(3-3)(3-2) = 0 \times 1 = 0$$ 8. Sum these values: $$h'(3) = 5 + 0 + 0 = 5$$ 9. Calculate the residue: $$\text{Res}(f, 3) = \frac{g(3)}{h'(3)} = \frac{1}{5}$$ 10. Now, identify which function among the options matches the original function: - Option a: $$\frac{1}{z^2 + 4}$$ - Option b: $$\frac{1}{z^2 + 4z - 6}$$ - Option c: $$\frac{1}{z^2 - 4}$$ - Option d: $$\frac{1}{z^2 + 4z + 6}$$ 11. The original denominator is: $$(z-2)(z+2)(z-3) = (z^2 - 4)(z-3) = z^3 - 3z^2 - 4z + 12$$ 12. None of the options match the full denominator, but the question likely asks which function corresponds to the factorization related to the poles. 13. The factor $$z^2 - 4$$ corresponds to $$(z-2)(z+2)$$, which is option c. 14. Therefore, the function $$f(z) = \frac{1}{(z-2)(z+2)(z-3)}$$ contains the factor $$\frac{1}{z^2 - 4}$$ as part of its denominator. Final answer: Option c: $$f(z) = \frac{1}{z^2 - 4}$$