Residues & Contour Integration Practice Quiz with a Step-by-Step Interactive Lesson
Use the question set below to practice residues and contour integration: reading the coefficient of \((z-a)^{-1}\), computing residues at simple and higher-order poles, deciding which poles lie inside a contour, applying \(\oint_\Gamma f(z)\,dz=2\pi i\sum\operatorname{Res}(f,a)\), handling cancellations, recognizing removable and essential singularities, and noticing when a pole on the contour blocks the basic theorem. If you need a refresher, open the lesson for mentally followable examples and quick checks.
Answer the question set and review your mistakes at the end.
How this residues and contour integration practice works
1. Take the practice set: answer questions about residues, poles, contour integrals, and theorem hypotheses.
2. Open the lesson: review the residue theorem, residue shortcuts, contour orientation, and inside-versus-outside pole decisions.
3. Retry: return to the question set and first list the singularities enclosed by the contour.
What you will learn in the residues and contour integration lesson
Residues and poles
Residue: the coefficient of \((z-a)^{-1}\) in the Laurent expansion at \(a\).
Simple pole: for \(g(z)/(z-a)\), the residue is \(g(a)\).
Zero shortcut: if \(q(a)=0\) and \(q'(a)≠0\), then \(\operatorname{Res}(p/q,a)=p(a)/q'(a)\).
Contour theorem
Residue theorem: integrate by summing enclosed residues and multiplying by \(2\pi i\).
Inside only: poles outside the contour do not contribute.
Orientation: reversing orientation changes the sign of the integral.
Series and singularities
Series shortcut: expand only far enough to find the \(1/(z-a)\) coefficient.
Higher-order pole: use the derivative formula or a short Taylor expansion.
Classification: no negative Laurent powers means removable; infinitely many negative powers means essential.
Common traps
On-contour pole: the basic residue theorem is not directly applicable.
Residue is not pole order: \(1/(z-a)^2\) has residue \(0\).
Cancellations: several enclosed residues can sum to \(0\).
Purpose: Build a reliable contour-integration routine: locate singularities, classify poles, compute residues from Laurent coefficients or shortcuts, apply the residue theorem with the correct orientation, recognize when residues cancel, and know when the basic theorem does not apply.
Success criteria
Define the residue as the coefficient of \((z-a)^{-1}\) in a Laurent expansion.
Compute residues at simple poles using \(\lim_{z\to a}(z-a)f(z)\).
Use \(\operatorname{Res}(p/q,a)=p(a)/q'(a)\) when \(q\) has a simple zero at \(a\).
Compute residues at higher-order poles with a short Taylor expansion or derivative formula.
Apply the residue theorem only when the contour avoids the singularities.
Ignore exterior poles and include all interior poles, even when their residues cancel.
Track positive orientation and know that reversing a contour changes the sign.
Classify removable singularities, poles of a given order, and essential singularities from the Laurent expansion.
Spot traps involving second-order poles with zero residue, holomorphic integrands, and poles on the contour.
Key vocabulary
Laurent expansion: a series \(\sum_{n=-\infty}^{\infty}c_n(z-a)^n\) valid in an annulus around \(a\).
Residue: \(c_{-1}\), the coefficient of \((z-a)^{-1}\).
Simple pole: a pole whose principal part starts and ends with \((z-a)^{-1}\).
Removable singularity: no negative Laurent powers after simplification.
Contour: an oriented closed path; positive orientation is counterclockwise for a simple closed contour.
Residue theorem: a closed contour integral equals \(2\pi i\) times the sum of enclosed residues.
Quick pre-check
Pre-check: If \(g\) is holomorphic at \(a\), what is \(\operatorname{Res}(g(z)/(z-a),a)\)?
Hint: Around \(a\), the numerator is \(g(a)+g'(a)(z-a)+\cdots\). Only the coefficient multiplying \((z-a)^{-1}\) matters.
Residues are the \(1/(z-a)\) coefficients
Learning goal: Compute simple residues without expanding more than necessary.
Key idea
If \(f\) has an isolated singularity at \(a\), write its Laurent expansion as \(f(z)=\cdots+c_{-2}(z-a)^{-2}+c_{-1}(z-a)^{-1}+c_0+\cdots\). The residue is \(c_{-1}\). For a simple pole, \[\operatorname{Res}(f,a)=\lim_{z\to a}(z-a)f(z).\]
Examples
\(\operatorname{Res}(1/(z-a),a)=1\).
\(\operatorname{Res}(2/(z-3),3)=2\).
\(\operatorname{Res}(1/(z-a)^2,a)=0\) because there is no \((z-a)^{-1}\) term.
If \(f(z)=p(z)/q(z)\), \(q(a)=0\), and \(q'(a)≠0\), then \(\operatorname{Res}(f,a)=p(a)/q'(a)\).
Simple-pole shortcut
A useful simple-pole shortcut is to cancel the factor causing the pole, then evaluate what remains. For \(1/((z-1)(z-2))\) at \(z=1\), remove \(z-1\) and evaluate \(1/(z-2)\) at \(1\).
Worked example
Example: Find the residue of \(1/(z^2+1)\) at \(z=i\).
Factor \(z^2+1=(z-i)(z+i)\). Removing the simple factor \(z-i\) leaves \(1/(z+i)\). At \(z=i\), this is \(1/(2i)\).
Try it
Try it: What is \(\operatorname{Res}(1/((z-1)(z-2)),1)\)?
Hint: Remove the factor \(z-1\), then evaluate the remaining factor at \(z=1\).
A closed contour integral is a residue sum
Learning goal: Turn contour integrals into a finite sum over enclosed singularities.
Key idea
Let \(\Gamma\) be a positively oriented closed contour, and suppose \(f\) is holomorphic on and inside \(\Gamma\) except at finitely many isolated singularities \(a_1,\dots,a_n\) inside. Then \[\oint_\Gamma f(z)\,dz=2\pi i\sum_{k=1}^n\operatorname{Res}(f,a_k).\] No pole may lie on the contour for this standard form.
Computation routine
List all singularities of the integrand.
Decide which ones are inside the contour.
Compute the residue at each enclosed singularity.
Add the enclosed residues.
Multiply by \(2\pi i\), with a sign change for negative orientation.
Both poles are inside. The residues are \(-1\) at \(1\) and \(1\) at \(2\). Their sum is \(0\), so the contour integral is \(2\pi i\cdot0=0\).
Try it
Try it: What is \(\displaystyle\oint_{|z|=2}\frac{3\,dz}{z-1}\)?
Hint: The pole \(1\) is inside the circle and the residue is \(3\).
For higher-order poles, extract the coefficient carefully
Learning goal: Compute residues when the pole is not simple, using a derivative formula or a short series.
Key idea
If \(f(z)=h(z)/(z-a)^m\) with \(h\) holomorphic at \(a\), then \[\operatorname{Res}(f,a)=\frac{h^{(m-1)}(a)}{(m-1)!}.\] For \(m=2\), the residue is \(h'(a)\). Equivalently, expand \(h\) just far enough to find the coefficient of \((z-a)^{m-1}\).
Formulas to remember
A second-order pole can have residue \(0\).
\(\operatorname{Res}(h(z)/(z-a)^2,a)=h'(a)\).
\(\operatorname{Res}(h(z)/(z-a)^3,a)=h''(a)/2\).
Do not confuse residue with the coefficient of the highest negative power.
If a numerator has a zero, simplify first; the pole order may drop or disappear.
Worked example
Example: Find \(\operatorname{Res}(e^z/z^3,0)\).
Use \(e^z=1+z+z^2/2+\cdots\). Then \(e^z/z^3=z^{-3}+z^{-2}+\frac{1}{2}z^{-1}+\cdots\). The residue is \(\frac{1}{2}\).
Try it
Try it: What is \(\operatorname{Res}(e^z/z^2,0)\)?
Hint: \(e^z=1+z+\cdots\), and division by \(z^2\) turns the \(z\)-term into \(1/z\).
Expand only as much as the residue needs
Learning goal: Use familiar Taylor series to read residues quickly.
Key idea
Many residues near \(0\) are coefficient questions. For example, \(\sin z=z-z^3/6+\cdots\), so \(\sin z/z^2=1/z-z/6+\cdots\). The residue is \(1\). You do not need the whole Laurent series; stop once the \(1/z\) coefficient is clear.
Theorem checklist
Move the singularity to \(0\) if a substitution makes the expansion clearer.
Use standard Taylor series for \(e^z\), \(\sin z\), \(\cos z\), and geometric series.
After division by \((z-a)^m\), look only for the term that becomes \((z-a)^{-1}\).
If the negative powers vanish after simplification, the singularity is removable.
If infinitely many negative powers remain, the singularity is essential, but the residue is still the \(1/(z-a)\) coefficient.
Worked example
Example: Find \(\operatorname{Res}(\cos z/z,0)\).
Since \(\cos z=1-z^2/2+\cdots\), dividing by \(z\) gives \(1/z-z/2+\cdots\). The residue is \(1\).
Try it
Try it: What is \(\operatorname{Res}(\sin z/z^2,0)\)?
Hint: The first term of \(\sin z\) is \(z\).
Count enclosed residues and keep the sign
Learning goal: Decide which singularities contribute and how orientation changes the contour integral.
Key idea
For a positively oriented simple closed contour, \(\oint_\Gamma f(z)\,dz=2\pi i\sum\operatorname{Res}(f,a)\), where the sum is over singularities inside \(\Gamma\). A clockwise traversal reverses the sign, and poles outside the contour contribute nothing.
What the contour proves
Inside poles contribute their residues.
Outside poles do not contribute.
If the integrand is holomorphic inside and on the contour, the integral is \(0\).
Clockwise orientation changes \(2\pi i\) into \(-2\pi i\) for a single residue \(1\).
A pole on the contour blocks the basic residue theorem.
The only pole inside is \(0\), with residue \(1\). Clockwise orientation reverses the positive result, so the integral is \(-2\pi i\).
Try it
Try it: What is \(\displaystyle\oint_{|z|=1}^{\mathrm{clockwise}}\frac{dz}{z}\)?
Hint: The residue is \(1\), but the contour orientation is clockwise.
Some residues are ignored, and some cancel
Learning goal: Combine several residue facts in one contour integral.
Key idea
A contour integral may contain several singularities, but only enclosed singularities enter the sum. Split simple sums when helpful, factor denominators to find poles, and remember that two nonzero enclosed residues can still add to \(0\).
Theorem checklist
Split terms such as \(1/z+1/(z-3)\) into separate residue checks.
Factor denominators such as \(z^2+1=(z-i)(z+i)\).
Ignore poles outside the contour.
Add all enclosed residues before multiplying by \(2\pi i\).
Watch for cancellation, zero residues, and removable singularities.
Worked example
Example: Why is \(\displaystyle\oint_{|z|=2}\frac{dz}{z^2+1}=0\)?
Both poles \(i\) and \(-i\) lie inside. Their residues are \(1/(2i)\) and \(-1/(2i)\), which cancel. The sum of enclosed residues is \(0\), so the integral is \(0\).
Try it
Try it: What is \(\displaystyle\oint_{|z|=2}\left(\frac1z+\frac1{z-3}\right)\,dz\)?
Hint: The pole at \(0\) is inside \(|z|=2\), while the pole at \(3\) is outside.
Most mistakes come from counting the wrong poles
Learning goal: Finish with the checks that prevent common residue-theorem errors.
Common traps
Outside poles: they do not contribute to a contour integral.
On-contour poles: the basic residue theorem is not directly applicable.
Orientation: clockwise orientation gives the negative of the usual result.
Residue zero: a function can have a pole but residue \(0\).
Cancellation: enclosed residues can add to \(0\), making the integral \(0\).
Removable singularities: simplify before classifying.
Holomorphic case: no singularities inside and on the contour gives integral \(0\).
Singularity class: a simple pole has lowest Laurent power \(-1\), while an essential singularity has infinitely many negative powers.
Worked example
Example: Why is \(\oint_{|z|=2}\frac{dz}{z^2+1}=0\)?
Both poles \(i\) and \(-i\) lie inside. Their residues are \(1/(2i)\) and \(-1/(2i)\), which cancel. The sum of enclosed residues is \(0\), so the integral is \(0\).
Try it
Try it: If a pole lies exactly on the contour, what should you say about the basic residue theorem?
Hint: The standard theorem assumes the function is holomorphic on the contour itself.
Final recap
The residue is the coefficient of \((z-a)^{-1}\).
For a simple pole, multiply by \(z-a\) and take the limit.
For \(p/q\) with a simple zero \(q(a)=0\), use \(p(a)/q'(a)\).
The residue theorem sums residues inside the contour only.
Positive orientation gives \(2\pi i\) times the residue sum; negative orientation changes the sign.
A second-order pole may have residue \(0\).
Series expansions only need the coefficient that becomes \(1/(z-a)\).
No negative Laurent powers means removable; infinitely many negative powers means essential.
Poles on the contour require extra care beyond the basic theorem.
Residues can cancel, so an integral with enclosed poles can still be \(0\).
Next step: Close this lesson and try the quiz again. For each problem, identify the singularities, keep only the enclosed ones, compute the \(1/(z-a)\) coefficients, then check orientation and theorem hypotheses.
Practice set
Residues & Contour Integration practice questions with instant score
Answer all 10 questions below, then get your final score and a mistake review at the end so you know exactly what to improve.
0/10answered
Question 1Not answered
What is the residue of \(1/(z-a)\) at \(z=a\)?
Correct answer: B. \(1\)
Explanation: The coefficient of \((z-a)^{-1}\) is \(1\).
Question 2Not answered
What is the residue of \(2/(z-3)\) at \(z=3\)?
Correct answer: A. \(2\)
Explanation: The coefficient of \((z-3)^{-1}\) is \(2\).
Question 3Not answered
What is \(\displaystyle\oint_{|z|=1}\frac{dz}{z}\)?
Correct answer: A. \(2\pi i\)
Explanation: The residue at \(0\) is \(1\), so the integral is \(2\pi i\).
Question 4Not answered
What is \(\displaystyle\oint_{|z|=1}\frac{dz}{z-2}\)?
Correct answer: C. \(0\)
Explanation: The pole \(z=2\) is outside the unit circle, so the integral is \(0\).
Question 5Not answered
If \(f\) is holomorphic inside and on a closed contour, what is \(\oint f(z)\,dz\)?
Correct answer: D. \(0\)
Explanation: Cauchy's theorem gives zero when there are no singularities inside.
Question 6Not answered
For holomorphic \(g\), the residue of \(g(z)/(z-a)\) at \(a\) is:
Correct answer: C. \(g(a)\)
Explanation: The coefficient of \((z-a)^{-1}\) is \(g(a)\).
Question 7Not answered
What is the residue of \(1/z^2\) at \(0\)?
Correct answer: A. \(0\)
Explanation: The residue is the coefficient of \(1/z\), and here that coefficient is \(0\).
Question 8Not answered
A simple pole means the Laurent expansion has lowest power:
Correct answer: C. \(-1\)
Explanation: A simple pole has one negative power, namely \((z-a)^{-1}\).
Question 9Not answered
The residue theorem says a contour integral equals \(2\pi i\) times:
Correct answer: D. The sum of residues inside
Explanation: It is \(2\pi i\) times the sum of residues inside the contour.
Question 10Not answered
What is \(\displaystyle\oint_{|z|=2}\frac{3\,dz}{z-1}\)?
Correct answer: C. \(6\pi i\)
Explanation: The pole \(1\) is inside and the residue is \(3\), so the integral is \(6\pi i\).
