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Residue theorem on |z|=4 with (z+1)(z+3)

Short answer

Assuming numerator e^z, f(z)=e^z/(( z+1)(z+3)) has simple poles at z=−1 and z=−3, both inside |z|=4. Residues are e^{−1}/2 and −e^{−3}/2. Sum = (e^{−1}−e^{−3})/2. Integral = πi(e^{−1}−e^{−3}).

Evaluate the integral ∮_C [?]/((z + 1)(z + 3)) dz where C: |z| = 4, using residue theorem. [numerator unreadable in scan]

Reading of the question: the numerator is unreadable; the usual form of this problem is ∮_C eᶻ / ((z + 1)(z + 3)) dz with C: |z| = 4. I solve it with numerator eᶻ.

f(z) = eᶻ/((z + 1)(z + 3)). Poles are z = −1 and z = −3, both simple. Since |−1| = 1 < 4 and |−3| = 3 < 4, both lie inside C.

Residue at z = −1:

Res = eᶻ/(z + 3) at z = −1 = e⁻¹/2

Residue at z = −3:

Res = eᶻ/(z + 1) at z = −3 = e⁻³/(−2) = −e⁻³/2

Sum of residues = (e⁻¹ − e⁻³)/2.

By Cauchy's residue theorem:

∮_C f dz = 2πi · (e⁻¹ − e⁻³)/2

Answer: πi (e⁻¹ − e⁻³) ≈ 0.9993 i

(Numerically: π(0.36788 − 0.04979) = 0.9993.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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