$$f(z)=\frac{e^{z}}{z^{4}+9z^{2}}$$
For a simple pole alternate formula : $$\operatorname{Res}((),) = \frac{()}{′()}$$ 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.
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 $Q(z)=z^{4}+9z^{2}$: $$\operatorname{Res}((),3i) = \frac{P(3i)}{Q'(3i)}=\frac{e^{3i}}{4(3i)^{3}+18(3i)}=\frac{e^{3i}}{-54i}=\frac{ie^{3i}}{54}$$