Skip to content
GRE Math overview

Worked example

Free to read

A Contour Integral at a Third-Order Pole

A public complex analysis question that evaluates a contour integral with a pole of order three, where the factorial in Cauchy's formula for derivatives decides the answer.

Question

Evaluate ∮∣z∣=1ezz3 dz\displaystyle \oint_{|z| = 1} \frac{e^{z}}{z^{3}}\,dz, where the circle is traversed once counterclockwise.

  1. a.00
  2. b.πi\pi i
  3. c.2πi2\pi i
  4. d.4πi4\pi i

Correct answer

B. πi\pi i

Full reasoning

  1. 1The integrand ez/z3e^{z}/z^{3} is analytic except at z=0z = 0, which lies inside ∣z∣=1|z| = 1, so Cauchy's theorem does not apply and the singularity must be handled.
  2. 2Match the shape f(z)(z−a)n+1\dfrac{f(z)}{(z-a)^{n+1}} with f(z)=ezf(z) = e^{z}, a=0a = 0, and n+1=3n + 1 = 3, so n=2n = 2. The function ff is entire, so the formula applies.
  3. 3Cauchy's integral formula for derivatives gives ∮∣z∣=1ezz3 dz=2πi2! f′′(0)\displaystyle \oint_{|z|=1} \frac{e^{z}}{z^{3}}\,dz = \frac{2\pi i}{2!}\,f''(0).
  4. 4Since f(z)=ezf(z) = e^{z} satisfies f′′(z)=ezf''(z) = e^{z}, we get f′′(0)=1f''(0) = 1, so the integral is 2πi2=πi\dfrac{2\pi i}{2} = \pi i.
  5. 5Residue cross-check: ezz3=1z3+1z2+12z+16+⋯\dfrac{e^{z}}{z^{3}} = \dfrac{1}{z^{3}} + \dfrac{1}{z^{2}} + \dfrac{1}{2z} + \dfrac{1}{6} + \cdots, so the residue — the coefficient of z−1z^{-1} — is 12\tfrac{1}{2}, and 2πi⋅12=πi2\pi i \cdot \tfrac{1}{2} = \pi i.

Why each choice is right or wrong

Choice A

This applies Cauchy's theorem as though the integrand were analytic inside the contour. It is not: z3z^{3} vanishes at the enclosed point z=0z = 0, so the integrand has a pole there and the theorem does not apply.

Choice B

Correct. With f(z)=ezf(z) = e^{z} and n=2n = 2, the formula gives 2πi2!f′′(0)=2πi2⋅1=πi\dfrac{2\pi i}{2!}f''(0) = \dfrac{2\pi i}{2} \cdot 1 = \pi i; equivalently the residue is the coefficient 12\tfrac{1}{2} of z2z^{2} in eze^{z}.

Choice C

This drops the 1n!\tfrac{1}{n!} factor and computes 2πi⋅f′′(0)=2πi2\pi i \cdot f''(0) = 2\pi i. It is also what you get by treating z=0z = 0 as a simple pole with residue e0=1e^{0} = 1, ignoring the exponent 33.

Choice D

This multiplies by n!=2n! = 2 instead of dividing by it. The factorial sits in the denominator of Cauchy's formula for derivatives, so the correct scaling halves 2πi2\pi i rather than doubling it.

Related formula

Cauchy's integral formula for derivatives

∮Cf(z)(z−a)n+1 dz=2πin! f(n)(a)\oint_{C} \frac{f(z)}{(z-a)^{n+1}}\,dz = \frac{2\pi i}{n!}\, f^{(n)}(a)

Assumptions: C is a simple closed contour traversed once counterclockwise with a inside it. The factorial divides; n = 0 recovers the ordinary Cauchy integral formula.

Related topic and practice

Analytic Functions and ResiduesCreate a free account to practice

This worked example is free to read. A free account unlocks practice questions for this exam.

Sources

Authored for this public explainer registry. It has no database question ID and is not copied from a protected or official exam bank.

  1. Complex Variables with Applications (Orloff), Section 5.2: Cauchy's Integral Formula for Derivatives — LibreTexts Mathematics. Accessed 2026-08-15. Orloff's Complex Variables with Applications on LibreTexts is CC BY-NC-SA 4.0; attribute the author and platform and avoid verbatim reuse beyond short cited references.
  2. Complex Variables with Applications (Orloff), Section 9.5: Cauchy Residue Theorem — LibreTexts Mathematics. Accessed 2026-08-15. Orloff's Complex Variables with Applications on LibreTexts is CC BY-NC-SA 4.0; attribute the author and platform and avoid verbatim reuse beyond short cited references.
Share this worked solution

Sources and corrections

Sources last checked 2026-08-15

Every source cited on this page was checked on the date shown, and we update the page when a source changes. If something looks wrong, tell us and we'll recheck it.

Report an issue with this page