Why the plane needs a path
In ordinary calculus, the integral from a to b runs along the only road there is — the segment of the real line between two numbers. The plane is roomier. To get from a point a to a point b you can take a straight line, a wide arc, a loop that circles the origin first, a zigzag — infinitely many genuinely different journeys. So an integral in the plane cannot just name two endpoints; it must name a whole path, and that path is part of the question. This is the first and most important mental adjustment of the rung.
The object we are after is the contour integral, usually written as the integral of f(z) dz along a curve C. You should read it as: walk along C, and as you go, accumulate the values of f, weighted by the tiny complex displacements dz you make at each step. Because both f(z) and dz are complex numbers, every little contribution f(z) dz is a complex number too, complete with a direction — and the integral is the running complex sum of all of them. Already this hints that direction and orientation will matter in a way they never did on the real line.
From a curve to an ordinary integral: parametrize
How do you actually compute such a thing? You reduce it to a calculus integral you already know how to do, by laying down a clock. A parametrized path is a function z(t) that, as the real parameter t runs from a to b, traces out the curve C — z(a) is the start, z(b) is the finish, and z(t) is where you are at time t. This is exactly the curves and arcs viewpoint from the previous rung, now put to work. The displacement dz becomes z'(t) dt, the velocity of the tracing point times a sliver of time.
integral over C of f(z) dz
= integral from t=a to t=b of f( z(t) ) * z'(t) dt
z(t) : parametrized path, z(a)=start, z(b)=finish
z'(t) : velocity of the tracing point (a complex number)On the right you now have an ordinary integral over the real interval [a, b], of a complex-valued function of one real variable. That you do componentwise: integrate the real part and the imaginary part separately, exactly as you learned at the start of this whole journey. So a contour integral is never mysterious to evaluate — it is always one substitution away from being plain calculus. The art is choosing a convenient parametrization, not in any new kind of integration.
A worked loop around the origin
Let's do the single most important example in the subject, and do it concretely. Take C to be the unit circle traced once counterclockwise, and integrate f(z) = 1/z around it. Parametrize the circle the natural way with Euler's formula: z(t) = e^(i t) for t from 0 to 2 pi. Then the velocity is z'(t) = i e^(i t), and on the circle f(z(t)) = 1 / e^(i t) = e^(-i t). Multiply these together and watch the exponentials collapse.
- Form the integrand: f(z(t)) * z'(t) = e^(-i t) * i e^(i t) = i, because e^(-i t) e^(i t) = e^0 = 1. The integrand is the constant i.
- Integrate over t from 0 to 2 pi: integral of i dt = i * (2 pi - 0) = 2 pi i.
- Conclusion: the integral of (1/z) dz around the unit circle equals 2 pi i — emphatically NOT zero.
That clean answer 2 pi i is worth sitting with. It does not depend on the radius — redo it with z(t) = R e^(i t) and the R's cancel, leaving 2 pi i again. It is the seed of everything to come in this rung and the next: the residue idea, the count of how many times a loop winds, the recovery of a function from its boundary values. Keep it in your pocket. We are not yet equipped to say why such integrals so often come out to exactly zero for other functions — that is the punchline of guide 4 — but you can already feel that 1/z is special precisely because it is not nice at the origin sitting inside the loop.
The properties that make it behave
A small set of basic properties lets you manipulate contour integrals with the same ease you handle ordinary ones — and all of them follow instantly from the parametrized definition, because once you are on the right-hand side it is just calculus. First, linearity: the integral of (a f + b g) along C is a times the integral of f plus b times the integral of g, for complex constants a and b. You can split sums apart and pull constants out exactly as always.
Second, reversal of orientation: if you walk the same curve backwards, from b to a, every displacement dz flips sign, so the integral over the reversed path -C is exactly minus the integral over C. This is the precise statement of why orientation matters in the plane — the unit circle traced clockwise gives -2 pi i for 1/z. Third, additivity over pieces: if C is C_1 followed by C_2 (the end of the first is the start of the second), then the integral over C is the integral over C_1 plus the integral over C_2. That is what lets you build any polygonal or piecewise-smooth contour out of simple arcs and add up the parts.
A first handle on size, and what comes next
Often you do not need the exact value of an integral — you need to know it is small, or that it tends to zero. For that there is a simple but enormously useful bound. If |f(z)| never exceeds some number M everywhere along C, and C has total length L, then the size of the whole integral cannot exceed M times L. This is the ML inequality, and the intuition is honest book-keeping: you are summing displacements totalling length L, each carrying a value of modulus at most M, so the modulus of the sum is at most M L. The triangle inequality, generalized to integrals, does the rest.
You now have the three things a working complex analyst needs from this guide: a clear meaning for the integral along a path, a reliable recipe to compute it by parametrizing, and the linearity, reversal, and additivity rules — plus the ML bound for sizing. Carry forward two facts in particular. The integral of 1/z once around the origin is 2 pi i, and in general the path is part of the data. From here the road forks toward something wonderful: in guide 3 we ask when an integrand has a primitive and the path stops mattering — path independence — and in guide 4 we meet the Cauchy integral theorem, the statement that for a holomorphic function a closed loop integrates to exactly zero. The example you just computed is precisely the exception that reveals where that miracle can and cannot reach.