Residues of a function

Short answer

f(z) has simple poles at z=1 and z=2. The residue at z=1 is 1 and at z=2 is 2, giving the partial fractions 1/(z-1) + 2/(z-2).

Find the residue of f(z) = (3z − 4)/((z − 1)(z − 2)) at each of its singularities.

f(z) = (3z − 4)/((z − 1)(z − 2)) has simple poles at z = 1 and z = 2.

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

Answer: residue = 1 at z = 1 and 2 at z = 2. (Check by partial fractions: f = 1/(z − 1) + 2/(z − 2).)

✓ Answer checked by Claude Opus 5.5 (high effort)

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