Let f(z) = exp\(\rm\left(z+\frac{1}{z}\right)\), z ∈ ℂ\{0}. The residue of f at z = 0 is

Concept:

Residue of f(z)  at z = 0 is the coefficient of \(\frac1z\) in the Maclaurin series expansion of f(z)

Explanation:

f(z) = exp\(\rm\left(z+\frac{1}{z}\right)\)

     = \(e^z.e^{\frac1z}\)

    = \((1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+...)\)\(.(1+\frac1zz+\frac{1}{z^22!}+\frac{1}{z^33!}+...)\)

Hence the coefficient of \(\frac1z\) in the above expression

 = \(\frac11+\frac1{2!.1!}+\frac1{3!.2!}+\frac1{4!.3!}+...\)

 = \(\sum_{l=0}^{\infty} \frac{1}{l !(l+1) !}\)

Hence option (3) is correct