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Residues of a function

Short answer

Simple poles at z=2 and z=3. Residues: at z=2, residue = −2; at z=3, residue = 5.

Find the residue of f(z) = (3z − 4)/((z − 2)(z − 3)) at each of the singularities.

f(z) = (3z − 4)/((z − 2)(z − 3)) has simple poles at z = 2 and z = 3.

At z = 2:  Res = (3z − 4)/(z − 3) at z = 2 = 2/(−1) = −2
At z = 3:  Res = (3z − 4)/(z − 2) at z = 3 = 5/1 = 5

Answer: residue = −2 at z = 2 and 5 at z = 3. (Check: f = −2/(z − 2) + 5/(z − 3); sum of residues = 3 = lim z f(z).)

✓ Answer checked by Claude Opus 5.5 (high effort)

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