By using Hadamard's factorization theorem show that Sin(πz) = πz ∏(1 - z²/n²) for n = 1, Find the nature and location of the zeros.

Steps to Solve

1. Understanding Hadamard's Factorization

Hadamard's factorization expresses an entire function as a product involving its roots. For the sine function, we can represent it as: $$ \sin(\pi z) = \pi z \prod_{n=1}^{\infty} \left(1 - \frac{z^2}{n^2}\right) $$

2. Identifying Zeros

The sine function has zeros where its value is zero. From the factorization, we see that: $$ \sin(\pi z) = 0 \quad \text{when} \quad z = n \quad \text{for } n \in \mathbb{Z} $$

3. Analyzing the Product Term

The infinite product $\prod_{n=1}^{\infty} \left(1 - \frac{z^2}{n^2}\right)$ also contributes to the zeros. Each factor $(1 - \frac{z^2}{n^2}) = 0$ implies: $$ z^2 = n^2 \implies z = \pm n $$

4. Finding Locations and Nature of Zeros

Combining the information above:

• The zeros of $\sin(\pi z)$ occur at all integer values: $$ z = n, , n \in \mathbb{Z} $$
• Each zero is of a simple (first) order.

5. Conclusion

The sine function has simple zeros located at every integer. The nature of these zeros is simple, meaning that near each zero, the function behaves linearly.