Complex analysis: a contour is part of the hypothesis
A contour integral depends on where the singularities sit relative to the path. A correct answer for one circle cannot simply be reused for a larger circle: the analytic domain needed by the theorem may have changed.
Course topic: Complex Analysis
Study route and exercise: University Reading List editorial project. How we curate · Sources checked:
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Scope and responsibility
A first route through holomorphic functions, Cauchy formulas and isolated singularities. Beck and collaborators provide a free first-course spine; Ahlfors is a more compressed classical treatment; Needham adds geometric explanation. This is not a complete course in topology or advanced complex function theory.
Choose your starting point
I need a free first course
Start with Beck and collaborators, building integration and the residue theorem in sequence.
I want a formal classical treatment
Choose Ahlfors after proof-based calculus; expect to supply intermediate arguments.
I need a geometric reason for the formulas
Use Needham alongside an analytic spine, drawing domains and deformations.
Compare the three books
A First Course in Complex Analysis
Matthias Beck; Gerald Marchesi; Dennis Pixton; Lucas Sabalka · Edition: 1.6 · 2025
Free first-course spine
- Choose it when
- You can differentiate and integrate real functions and write short proofs.
- Use another route when
- You need only a quick table of integration tricks.
- Selected reading
- Complex integration, Cauchy formulas, singularities and residues in version 1.6.
- Limits and trade-offs
- The sequence develops needed ideas rather than assuming a substantial real-analysis course, but proofs still require work.
Try before choosing
For e^z/[z²(z−3)], mark all poles and the circle |z|=1 before choosing a theorem.
Check your work: Only 0 lies inside; e^z/(z−3) is analytic on and inside that circle.
Complex Analysis: An Introduction to the Theory of Analytic Functions of One Complex Variable
Lars V. Ahlfors · Edition: 3 · 2021
Compressed analytic reference
- Choose it when
- You are comfortable with limits, series and proof and want precise hypotheses.
- Use another route when
- A first encounter with proofs requires many small scaffolded steps.
- Selected reading
- Analytic functions, complex integration and local behavior at singularities in the selected third-edition reprint.
- Limits and trade-offs
- Concise arguments repay rereading; a reprint date does not make it a newly rewritten course.
Try before choosing
Differentiate e^z/(z−3) at zero and keep both product-rule terms.
Check your work: The derivative is −1/3−1/9=−4/9, not just −1/3.
Visual Complex Analysis
Tristan Needham · Edition: 25th anniversary · 2023
Geometric companion
- Choose it when
- Pictures of winding and deformation help you understand an integral.
- Use another route when
- You require a short computational reference or a replacement for checking analytic domains.
- Selected reading
- Complex Integration: Cauchy’s Theorem and Cauchy’s Formula, with the earlier geometric foundations.
- Limits and trade-offs
- Visual explanation is substantial, but the drawing must still identify every singularity crossed.
Try before choosing
Draw a deformation from radius 1 to radius 4 and mark the obstruction.
Check your work: The pole at 3 lies between the circles, so deformation invariance cannot be invoked across it.
Check your preparation
Can I differentiate a quotient of complex analytic functions?
Away from z=3, F(z)=e^z/(z−3) has F′(z)=e^z/(z−3)−e^z/(z−3)².
What does “analytic inside” include?
The whole enclosed domain needed by the selected theorem, not just points on the contour.
Can I distinguish an orientation from a radius?
Both cases below are counterclockwise. Reversing orientation would separately negate each integral.
A six-step study plan
Steps completed: 0 / 6
These are your own study marks, not an assessment. They are not sent to this site.
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Map the domain
Selected reading: Complex functions and isolated singularities.
What to do: Draw both circles and classify the two poles.
Check your work: Zero is double and 3 is simple; neither lies on either path.
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State the usable theorem
Selected reading: Cauchy derivative formula.
What to do: Write f=F/z² and specify where F is analytic.
Check your work: The hypothesis holds on the small disk and fails on the large disk.
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Compute the local coefficient
Selected reading: Local series or derivative formula.
What to do: Find the coefficient of z^−1 at zero using two methods.
Check your work: Both give −4/9.
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Solve the small contour
Selected reading: Cauchy formula or residues.
What to do: Give the oriented integral and explain which singularity contributes.
Check your work: The answer is −8πi/9.
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Change the domain
Selected reading: Residue theorem and the follow-up.
What to do: Add the newly enclosed simple-pole residue.
Check your work: The additional contribution is 2πi e³/9.
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Audit a deformation claim
Selected reading: Your diagrams and both solutions.
What to do: Write one valid deformation within the punctured domain and one invalid crossing.
Check your work: A theorem failure is explained by a specific pole, not by the circle being larger.
Worked exercise
A double pole inside a small circle
Compute the counterclockwise integral of f(z)=e^z/[z²(z−3)] around |z|=1, giving a domain justification.
Assumptions
Exact complex arithmetic; e^z is entire; each circle is traversed once counterclockwise. There are no singularities on the path.
- The singularities are a double pole at 0 and a simple pole at 3. Only 0 is inside |z|=1. Put F(z)=e^z/(z−3), analytic on a neighborhood of the closed unit disk.
- Cauchy’s derivative formula gives the integral 2πi F′(0). Product differentiation gives F′(0)=−1/3−1/9=−4/9.
- Equivalently, e^z=1+z+… and 1/(z−3)=−1/3−z/9−… near zero, so the z coefficient of F is −4/9 and the residue of f is −4/9.
Answer and interpretation
The integral is −8πi/9.
Test the result
The derivative and local-series calculations agree. The pole at 3 contributes nothing to this contour because it lies outside, not because its residue is zero.
Change an assumption
A new enclosed singularity changes the theorem route
Keep f and orientation, but integrate around |z|=4. Explain why applying the earlier single Cauchy-derivative formula over this whole disk is invalid.
- Now both poles are enclosed. The residue at zero remains −4/9; local behavior does not depend on the chosen enclosing circle.
- The residue at the simple pole 3 is lim(z→3)(z−3)f(z)=e³/9. The residue theorem therefore sums −4/9+e³/9.
- F=e^z/(z−3) is not analytic on the larger disk. Shrinking that circle directly to the small one crosses 3, so the previous analytic-domain hypothesis fails.
Answer and interpretation
The larger integral is 2πi(e³−4)/9; its difference from the small-circle integral is 2πi e³/9.
Test the result
A deformation that avoids both poles preserves the integral. One that crosses 3 changes the winding around that pole, accounting for exactly the computed difference.
Common mistakes
Only checking the path
A function can be finite on a circle and still have a pole inside it.
Dropping a derivative term
Both the exponential and reciprocal factor contribute to F′.
Treating deformation as an unrestricted drawing
The homotopy must stay in a domain where the integrand is analytic.
Before moving on
- State the domain assumptions of a Cauchy formula.
- Compute a double-pole residue by derivative and series.
- Explain a changed contour using the new enclosed pole.
Sources and edition checks
Sources checked:
- Authors: first course, version1.6
The authors’ version 1.6 develops a first rigorous route to residues and is freely available under its stated licence.
- CiNii: exact AMS third-edition reprint
The selected AMS item is a third-edition reprint, not a newly authored 2021 edition.
- Oxford:25th-anniversary geometric approach
The anniversary work provides geometric treatments of complex integration and Cauchy formulas.
These are proposed study activities, not official course assignments or a claim that every book has been read cover to cover.