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$$f(z)=\frac{e^{z}}{z^{4}+9z^{2}}$$

For a simple pole alternate formula : $$\operatorname{Res}(𝑓(𝑧),𝑎) = \frac{𝑃(𝑎)}{𝑄′(𝑎)}$$ While the general formula for finding residue still holds.

The answer is coming very different for simple pole $z=3i$ from the general residue formula and the alternate formula.

I got e^3i / -36+54i and e^3i / 54i respectively.

Same issue with many problems.

Are there any limitations to the alternate formula? if so please state that.

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    $\begingroup$ What did you find by using the two formulas? $\endgroup$ – Robert Z Sep 15 at 12:49
  • $\begingroup$ e^3i / -36+54i and e^3i / 54i respectively. $\endgroup$ – Prakhar Shrivastava Sep 15 at 17:45
  • $\begingroup$ Did you read my answer below? Finally, have you understood what you did wrong? $\endgroup$ – Robert Z Sep 16 at 21:10
  • $\begingroup$ ya thanks, calculation error. $\endgroup$ – Prakhar Shrivastava Sep 16 at 21:18
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The second value is the correct one.

General formula: we should evaluate $$\lim_{z\to 3i}\frac{e^{z}(z-3i)}{z^{4}+9z^{2}}=\lim_{z\to 3i}\frac{e^{z}}{z^2(z+3i)}=\frac{e^{3i}}{(3i)^{3}+(3i)^{3}}=\frac{e^{3i}}{-54i}=\frac{ie^{3i}}{54}$$ where $z^{4}+9z^{2}=z^2(z+3i)(z-3i)$.

Second formula with $P(z)=e^z$ and $Q(z)=z^{4}+9z^{2}$: $$\operatorname{Res}(𝑓(𝑧),3i) = \frac{P(3i)}{Q'(3i)}=\frac{e^{3i}}{4(3i)^{3}+18(3i)}=\frac{e^{3i}}{-54i}=\frac{ie^{3i}}{54}$$

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