Math · 11
Attempt these before you open a book or a video. Write the argument in the work box and the conclusion in the answer box. Check counts the problem only when both pass, and it does not open the solution. The solution stays locked for twenty minutes after you start the timer.
Attack 1
Write the Cauchy–Riemann equations for \(f=u+iv\), and check them for \(f(z)=z^2\).
Timer not started. Twenty minutes of real work before a solution.
\(u_x=v_y\) and \(u_y=-v_x\). For \(z^2=(x+iy)^2=x^2-y^2+2ixy\), one has \(u_x=2x=v_y\) and \(u_y=-2y=-v_x\).
Attack 2
Compute \(\int_{|z|=1} dz/z\), counterclockwise.
Timer not started. Twenty minutes of real work before a solution.
Parametrize \(z=e^{i\theta}\), \(dz=ie^{i\theta}d\theta\), \(\theta\) from 0 to \(2\pi\). The integral is \(\int_0^{2\pi} i\,d\theta=2\pi i\).
Attack 3
Why does Cauchy–Riemann for \(f(z)=\bar z\) fail?
Timer not started. Twenty minutes of real work before a solution.
\(u=x\), \(v=-y\). Then \(u_x=1\) and \(v_y=-1\), so \(u_x\ne v_y\). Conjugation is not holomorphic anywhere.
Complex analysis is calculus for functions \(\mathbb{C}\to\mathbb{C}\) that are complex-differentiable on an open set. That single hypothesis is absurdly strong. It implies the function is analytic, that its integral over a triangle vanishes, that Cauchy's integral formula recovers the function from its values on a circle, and that a bounded entire function is constant. Residues then turn real integrals into contour bookkeeping. You are not learning a new computational trick. You are learning why holomorphicity is a rigidity condition.
MIT 18.04 is a full course with notes and problem sets, and it is the working text because a standard book such as Ahlfors or Conway is paid and no free legal scan was verified. Stein and Shakarchi, and Needham’s Visual Complex Analysis, are paid and unnamed by ISBN here for the same reason. Use 18.04’s problem sets as the book. The proofs you must own are Cauchy's theorem for a disk, the integral formula, and the residue theorem at the level of a simple pole.
A formal power series you cannot connect to an integral is not yet complex analysis.
Statements only. Write the argument before you reveal it. A revealed proof you did not attempt is a paragraph you will not be able to rebuild at 2am.
Holomorphic. \(f\) is holomorphic at \(z_0\) when \(\lim_{h\to 0}(f(z_0+h)-f(z_0))/h\) exists in \(\mathbb{C}\). On an open set this is equivalent, given continuous partials, to the Cauchy–Riemann equations.
Approach along real and imaginary directions to get Cauchy–Riemann. The converse, with continuous partials, writes the real differential and matches it to multiplication by a complex number.
Cauchy. If \(f\) is holomorphic inside and on a triangle, the integral over the boundary is 0. Cauchy's integral formula: \(f(a)=\frac{1}{2\pi i}\int_\gamma f(z)/(z-a)\,dz\) for \(a\) inside a suitable closed curve.
Goursat’s proof for a triangle subdivides and uses a uniform estimate on the difference quotient. The integral formula reduces to the integral of \(1/(z-a)\) by writing \(f(z)=f(a)+(f(z)-f(a))\) and showing the second term integrates to 0.
Consequences. Holomorphic functions are analytic. Liouville: a bounded entire function is constant. The identity theorem: if two holomorphic functions on a connected open set agree on a set with a limit point, they agree everywhere.
Expand via Cauchy's formula as a power series. Liouville: Cauchy's estimates force every Taylor coefficient past the constant term to vanish when the function is bounded. The identity theorem uses isolation of zeros of a non identically zero holomorphic function.
Residues. The residue at an isolated singularity is the coefficient of \(1/(z-a)\) in the Laurent series, and the integral over a small circle is \(2\pi i\) times that coefficient. A simple pole of \(p/q\) at a simple zero of \(q\) has residue \(p(a)/q'(a)\).
Laurent coefficients are Cauchy integrals. Only the \(1/(z-a)\) term survives integration over a circle. The simple-pole formula is the limit \((z-a)p(z)/q(z)\) as \(z\to a\), which is \(p(a)/q'(a)\) by differentiability of \(q\).
Real integrals. A standard real integral of a rational function of sine and cosine, or a rational function with enough decay, can be read off from residues of a related complex function, provided you justify the vanishing of the integral on the closing arc.
The justification is an estimate: length of the arc times the maximum of the modulus. Write one such estimate, for a semicircle and a function with degree of the denominator at least two more than the numerator, before you use a table.
Four rungs, in order. Do not skip a rung because the next one looks more serious.
Watch or read once. Do not study from this.
MIT 18.04 course page, the overview only
Read once, do not study from this. Read the overview so you see the destination is contour integrals and residues. The same site is the course; the preview is one pass over the description, then you start the problem sets as Tier 1.
The thing you work through.
MIT 18.04 notes and problem sets
This is the main text. Ahlfors and Conway are the paid classics; editions were not verified, so they are not cited with fake ISBNs.
The problem book.
18.04 exams and the harder problem sets
Do every set. The hardness is justifying arcs and branch cuts, not the residue arithmetic.
The hardest questions that exist at this level.
Putnam problems that are complex analysis in disguise, via Kedlaya
Qualifying exams in complex analysis are the real ceiling. Until you have a public quals PDF you have personally verified, the Putnam archive plus a complete 18.04 exam is the honest public ceiling.
Names and links. The problem text stays on the source site. Copyrighted contest problems are not copied here. What follows is where they are and why they hurt.
Pass: a timed 18.04 exam, plus written proofs of Cauchy's integral formula for a circle and of Liouville’s theorem. Fail if a residue computation has no sentence about the contour.
Five sets. Finish a set before you touch the next. Full sentences, not a circled answer.
Ten problems to meet again. Logging a writeup schedules them at 1 day, 7 days, and 30 days in this browser.
State Cauchy–Riemann, compute the unit-circle integral of \(1/z\), and state Liouville. No notes.