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Part BCSIR NET June 2023residue-by-series-product

Residue by series product

Let $f (z) = exp (z + 1/ z), z \in C$ ∖{0}. The residue of f at z = 0 is

A. $\sum_{l = 0}^{\infty}$ 1/(l+1)!
B. $\sum_{l = 0}^{\infty}$ 1/(l!(l+1))
C. $\sum_{l = 0}^{\infty}$ 1/(l!(l+1)!)✓
D. $\sum_{l = 0}^{\infty}$ $1/ ((l^{2} + l)!)$

Solution

Multiply the series eᶻ $= \sum z^{m} / m$ ! and $e^{1/ z}$ $= \sum$ $z^{- k} / k$ !; the coefficient of $z^{- 1}$ comes from $k = m + 1 : \sum_{m} 1/ (m! (m + 1)!)$ .

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Related counterexample: |f| bounded near an isolated singularity ⇒ pole

From Singularities and Residues › Laurent series, classification of singularities, Casorati–Weierstrass

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