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Evaluate a line integral

Short answer

Both poles z=1 and z=2 lie inside |z|=3. Residues are -e and e^2, sum e^2 - e. By the residue theorem, the integral equals 2πi (e^2 - e) = 2πi e(e-1).

Evaluate ∮_C e^z/((z − 1)(z − 2)) dz over the circle C: |z| = 3.

Let f(z) = e^z/((z − 1)(z − 2)). The poles z = 1 and z = 2 are both inside C: |z| = 3. By Cauchy's residue theorem, ∮_C f dz = 2πi × (sum of residues).

Res at z = 1:  e^z/(z − 2) at z = 1 = e/(−1) = −e
Res at z = 2:  e^z/(z − 1) at z = 2 = e²/1 = e²
Sum = e² − e

Answer: ∮_C f(z) dz = 2πi (e² − e) = 2πi e (e − 1).

✓ Answer checked by Claude Opus 5.5 (high effort)

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