To find the poles of the function

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

we need to determine the values of $z$ where the function becomes undefined, i.e., where the denominator is zero. Poles occur when a function blows up (becomes infinite) due to a zero in the denominator while the numerator remains finite or nonzero.

Denominator:

The denominator is:

$z \left( z^2 + 1 \right)^2$

We set the denominator equal to zero to find the poles:

$z \left( z^2 + 1 \right)^2 = 0$

Step 1: Solve $z = 0$

The factor $z$ gives a solution $z = 0$. This is a pole.

Step 2: Solve $z^2 + 1 = 0$

The term $z^2 + 1 = 0$ implies:

$z^2 = -1$

$z = \pm i$

So, $z = i$ and $z = -i$ are additional points where the denominator is zero.

Nature of the Poles:

• At $z = 0$: The term $z$ in the denominator suggests that $z = 0$ is a simple pole.
• At $z = \pm i$: Since $(z^2 + 1)$ appears as $(z^2 + 1)^2$, the poles at $z = i$ and $z = -i$ are poles of order 2 (since the power of the factor is 2).

Conclusion:

The function has:

• A simple pole at $z = 0$,
• Poles of order 2 at $z = i$ and $z = -i$.

Would you like further details on how to classify these poles, or any other related concept?

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Here are 5 related questions to expand on this topic:

1. How do you determine the residue at a pole of a meromorphic function?
2. What is the significance of higher-order poles in complex analysis?
3. Can you find the Laurent series of this function around $z = 0$?
4. How would the behavior of the function change near $z = i$ or $z = -i$?
5. How do poles contribute to contour integrals in the residue theorem?

Tip: The nature of a pole (simple, double, etc.) is determined by how many times the function becomes infinite as you approach the pole.