It is given that the residue of the complex function \(\dfrac{e^{1/z}}{z^{n}}\) at an isolated singular point \(z = 0\) is \(\dfrac{1}{9!}\). The value of \(n\) must be equal to:
\(n = -8\)
Use the Laurent series \(e^{1/z} = \sum_{k=0}^{\infty} \dfrac{1}{k!\,z^{k}}\), so \(\dfrac{e^{1/z}}{z^{n}} = \sum_{k=0}^{\infty} \dfrac{1}{k!\,z^{k+n}}\).
The residue at \(z = 0\) is the coefficient of \(\dfrac{1}{z}\), i.e. the term with \(k + n = 1\), giving residue \(\dfrac{1}{(1-n)!}\).
Setting \(\dfrac{1}{(1-n)!} = \dfrac{1}{9!}\) yields \(1 - n = 9\), so \(n = -8\).
Hence, \(n = -8\).
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