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- 2082 Chaitra (new course) · 2 marks
Residues of a function
Short answer
At simple pole z=3: Res = (2·3+1)/(3+5)=7/8. At z=−5: Res = (2·(−5)+1)/(−5−3)=9/8.
Find the residue of f(z) = (2z + 1)/((z − 3)(z + 5)) at each of the singularities.
f(z) = (2z + 1)/((z − 3)(z + 5)) has two simple poles, z = 3 and z = −5. For a simple pole at z = a, Res = lim (z − a) f(z).
At z = 3: Res = (2z + 1)/(z + 5) at z = 3 = 7/8
At z = −5: Res = (2z + 1)/(z − 3) at z = −5 = (−9)/(−8) = 9/8
Answer: residue = 7/8 at z = 3 and 9/8 at z = −5. (Check: the sum of the residues, 7/8 + 9/8 = 2, equals the limit of z·f(z) as z → ∞, which is 2, as it must.)
✓ Answer checked by Claude Opus 5.5 (high effort)
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