Discussion
The residue of the function \(\frac{z^{2}}{z^{2} + a^{2}}\) at \(z = ia\) is :
Answer: A
The residue at \(z = ia\) is \(\frac{ia}{2}\).
The denominator factorises as \((z - ia)(z + ia)\), so \(z = ia\) is a simple pole. For a simple pole of \(p(z)/q(z)\), the residue is \(p(z_{0})/q'(z_{0})\):
\[ \begin{aligned} \text{Res} &= \frac{z^{2}}{2z}\bigg|_{z = ia} \\ &= \frac{ia}{2} \end{aligned} \]
The same follows from cancelling the factor \(z - ia\) and putting \(z = ia\) into \(\frac{z^{2}}{z + ia}\), which gives \(\frac{(ia)^{2}}{2ia} = \frac{ia}{2}\). The value \(-a^{2}\) is just the numerator at the pole, and infinity is the value of the function there, not its residue.
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