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Cauchy theorem vs formula; evaluate integral

Short answer

Cauchy integral theorem says ∮_ C f(z)dz=0 when f is analytic inside and on C; integral formula gives ∮_ C f(z)/(z-a)dz = 2πi f(a) for a inside C. For ∮_ C e^z/(( z+1)(z-2)) dz with C:|z-1|=3, both poles z=−1,2 lie inside. Write as (1/3)[ e^z/(z-2) − e^z/(z+1) ]; applying formula gives (2πi/3)(e^2 − e^{-1}).

Distinguish between Cauchy integral Theorem and Cauchy integral formula. Using Cauchy integral formula evaluate ∮_C e^z/((z + 1)(z - 2)) dz where C is the circle |z - 1| = 3.

Cauchy integral theorem vs Cauchy integral formula

PointCauchy integral theoremCauchy integral formula
Statement∮_C f(z) dz = 0f(a) = (1/2πi) ∮_C f(z)/(z − a) dz
Conditionf analytic inside and on Cf analytic inside and on C, a inside C
Integrandf(z) itselff(z)/(z − a), has a singularity at a
ResultAlways zeroGives the value of f at the interior point a
UseShow integrals vanish, path independenceEvaluate integrals with poles; get derivatives

Solution

C: |z − 1| = 3, a circle with centre 1 and radius 3.

  • z = −1: |−1 − 1| = 2 < 3, inside
  • z = 2: |2 − 1| = 1 < 3, inside

Both poles are inside C. Partial fractions:

1/((z + 1)(z − 2)) = (1/3) [ 1/(z − 2) − 1/(z + 1) ]

Apply CIF with f(z) = e^z (analytic everywhere):

∮ e^z/(z − 2) dz = 2πi f(2)  = 2πi e²
∮ e^z/(z + 1) dz = 2πi f(−1) = 2πi e⁻¹

I = (1/3) [ 2πi e² − 2πi e⁻¹ ] = (2πi/3)(e² − e⁻¹)

Answer: (2πi/3)(e² − 1/e) ≈ 14.705 i

✓ Answer checked by Claude Opus 5.5 (high effort)

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