# CVR6

Find the residues of $f(z)\,$ at all its isolated singular points and at infinity (if infinity is not a limit point of singular points), where $f(z)\,$ is given by

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

This has poles at $z=0\,$ of multiplicity $k=2\,$ and at $z=\pm 3i\,$ of multiplicity $k=1\,$.

For the simple roots, the residue is the numerator divided by the first derivative of the denominator when both are evaluated at the root.

$\mathrm{Res}_{z=-3i} f(z) = \frac{e^z}{4z^3+18z} \Big|_{z=-3i} = \frac{e^{-3i}}{4(27i)+18(-3i)} = \frac{e^{-3i}}{54i}\,$

$\mathrm{Res}_{z=3i} f(z) = \frac{e^z}{4z^3+18z} \Big|_{z=3i} = \frac{e^{3i}}{ 4(27(-i))+18(3i)} = \frac{-e^{3i}}{54i}\,$

For 0, use the formula $\mathrm{Res}_{z=z_0} f(z) = \frac{1}{(k-1)!}\lim_{z\to z_0} \frac{d^{k-1}}{{dz}^{k-1}}\left[(z-z_0)^k f(z)\right]\,$

$\mathrm{Res}_{z=0} =\lim_{z\to 0} \frac{d}{dz} z^2 f(z) = \lim_{z\to 0} \frac{d}{dz} \frac{e^z}{z^2+9}\,$

$=\lim_{z\to 0} \frac{(z^2+9)e^z-e^z 2z}{(z^2+9)^2} = \frac{9}{9^2} = \frac{1}{9}\,$

If $f(z)\,$ is analytic except at isolated singular points, then the sum of all the residues of $f(z)\,$ equals 0.

So $\mathrm{Res}_{z=-3i} +\mathrm{Res}_{z=3i} +\mathrm{Res}_{z=0} +\mathrm{Res}_{z=\infty} = \frac{e^{-3i}}{54i} - \frac{e^{3i}}{54i} + \frac{1}{9} + \mathrm{Res}_{z=\infty}=0\,$

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