((sin(π*z))/(z*(z^(2)+1)^(2))) why z≠0?

In the expression 

\[
f(z) = \frac{\sin(\pi z)}{z \left( z^2 + 1 \right)^2},
\]

the variable \(z\) cannot be equal to 0 for the following reasons:

1. **Singularity at \(z = 0\)**: The term \( \frac{1}{z} \) in the denominator will become undefined when \( z = 0 \). Division by zero is undefined in mathematics, making the expression singular at this point.
  
   Specifically, the denominator \( z \left( z^2 + 1 \right)^2 \) becomes zero when \( z = 0 \) because the \( z \)-term directly multiplies the whole denominator, causing the whole denominator to equal zero.

2. **Removable Singularity**: Even though \( \sin(\pi z) \) is finite and continuous at \( z = 0 \) (since \( \sin(0) = 0 \)), this does not resolve the issue of dividing by zero. The singularity is not automatically removable unless the entire numerator approaches zero fast enough to cancel the \( z \) in the denominator.

Therefore, \( z = 0 \) is a singularity where the function is not defined, so we must restrict \( z \neq 0 \).

Would you like further details or have any questions?

### Related Questions:
1. What is the type of singularity at \( z = 0 \) for this function?
2. How do we classify singularities in complex functions?
3. What happens to the function's behavior as \( z \to \infty \)?
4. How can we check if a singularity is removable or not?
5. What would be the limit of \( f(z) \) as \( z \to 0 \) if we could remove the singularity?

### Tip:
Always check for terms in the denominator when determining where a function is undefined.

Explore why the function (sin(πz))/(z(z²+1)²) has a singularity at z=0, making it undefined. This problem dives into key complex analysis concepts, including singularities, removable singularities, and the behavior of trigonometric functions around undefined points. Suitable for undergraduate students.

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