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Complex Integration | Mathematics

A contour (or piecewise smooth path) in C\mathbb{C} is a continuous function γ:[a,b]C\gamma : [a, b] \to \mathbb{C} that is differentiable except at finitely many points, with a Continuous derivative everywhere it exists.

A simple closed contour is a contour with γ(a)=γ(b)\gamma(a) = \gamma(b) and no other Self-intersections.

Definition. For a contour γ\gamma and a continuous function ff on γ\gamma:

γf(z)dz=abf(γ(t))γ"(t)dt\int_{\gamma} f(z)\, dz = \int_a^b f(\gamma(t))\gamma"(t)\, dt

Proposition 4.1. The complex integral is linear:

γ(af+bg)dz=aγfdz+bγgdz\int_\gamma (af + bg)\, dz = a\int_\gamma f\, dz + b\int_\gamma g\, dz

Proposition 4.2. Reversing orientation changes the sign:

γfdz=γfdz\int_{-\gamma} f\, dz = -\int_\gamma f\, dz

Proposition 4.3. Additivity over contours:

γ1+γ2fdz=γ1fdz+γ2fdz\int_{\gamma_1 + \gamma_2} f\, dz = \int_{\gamma_1} f\, dz + \int_{\gamma_2} f\, dz

Proposition 4.4 (ML Inequality). If f(z)M|f(z)| \leq M for all zz on a contour γ\gamma of length LLThen

γf(z)dzML\left|\int_\gamma f(z)\, dz\right| \leq ML

Proof. abf(γ(t))γ(t)dtabf(γ(t))γ(t)dtMabγ(t)dt=ML\left|\int_a^b f(\gamma(t))\gamma'(t)\, dt\right| \leq \int_a^b |f(\gamma(t))||\gamma'(t)|\, dt \leq M \int_a^b |\gamma'(t)|\, dt = ML. \blacksquare

Solution

Problem. Evaluate γz2dz\int_\gamma z^2\, dz where γ\gamma is the line segment from 00 to 1+i1 + i.

Solution. Parameterize γ(t)=t(1+i)\gamma(t) = t(1 + i) for 0t10 \leq t \leq 1. Then γ(t)=1+i\gamma'(t) = 1 + i.

γz2dz=01(t(1+i))2(1+i)dt=(1+i)301t2dt=(1+i)313\int_\gamma z^2\, dz = \int_0^1 (t(1+i))^2 (1+i)\, dt = (1+i)^3 \int_0^1 t^2\, dt = (1+i)^3 \cdot \frac{1}{3}

(1+i)3=(1+i)(1+i)2=(1+i)(2i)=2i+2i2=2i2=2+2i(1+i)^3 = (1+i)(1+i)^2 = (1+i)(2i) = 2i + 2i^2 = 2i - 2 = -2 + 2i.

γz2dz=2+2i3\int_\gamma z^2\, dz = \frac{-2 + 2i}{3}. \blacksquare

Problem. Evaluate γzˉdz\int_\gamma \bar{z}\, dz where γ\gamma is the unit circle traversed once Counterclockwise.

γ(t)=eit\gamma(t) = e^{it}, 0t2π0 \leq t \leq 2\pi, γ(t)=ieit\gamma'(t) = ie^{it}. zˉ=eit\bar{z} = e^{-it} on γ\gamma.

γzˉdz=02πeitieitdt=02πidt=2πi\int_\gamma \bar{z}\, dz = \int_0^{2\pi} e^{-it} \cdot ie^{it}\, dt = \int_0^{2\pi} i\, dt = 2\pi i.

Note: Since zˉ\bar{z} is not analytic, this result is non-zero, as expected.

Problem. Evaluate γdzz\int_\gamma \frac{dz}{z} where γ\gamma is the upper semicircle z=eiθz = e^{i\theta}, 0θπ0 \leq \theta \leq \pi.

0πieiθeiθdθ=0πidθ=iπ\int_0^\pi \frac{ie^{i\theta}}{e^{i\theta}}\, d\theta = \int_0^\pi i\, d\theta = i\pi.

Problem. Evaluate γzdz\int_\gamma z\, dz where γ\gamma consists of the line segment from 00 to 11 followed by the line segment from 11 to 1+i1 + i.

γ1(t)=t\gamma_1(t) = t, 0t10 \leq t \leq 1: 01t1dt=12\int_0^1 t \cdot 1\, dt = \frac{1}{2}.

γ2(t)=1+it\gamma_2(t) = 1 + it, 0t10 \leq t \leq 1: 01(1+it)idt=01(it)dt=i12\int_0^1 (1 + it) \cdot i\, dt = \int_0^1 (i - t)\, dt = i - \frac{1}{2}.

Total: 12+i12=i\frac{1}{2} + i - \frac{1}{2} = i.

Check: Since zz is entire, the integral from 00 to 1+i1 + i is 12(1+i)2=i\frac{1}{2}(1+i)^2 = i. Consistent. \blacksquare

Solution

Problem. Use the ML inequality to show that limRCReizzdz=0\lim_{R \to \infty} \int_{C_R} \frac{e^{iz}}{z}\, dz = 0 Where CRC_R is the upper semicircle z=R|z| = R, Im(z)0\mathrm{Im}(z) \geq 0.

On CRC_R: z=Reiθz = Re^{i\theta}, 0θπ0 \leq \theta \leq \pi. eiz=eiR(cosθ+isinθ)=eRsinθ|e^{iz}| = |e^{iR(\cos\theta + i\sin\theta)}| = e^{-R\sin\theta}.

CReizzdz0πeRsinθRRdθ=0πeRsinθdθ\left|\int_{C_R} \frac{e^{iz}}{z}\, dz\right| \leq \int_0^\pi \frac{e^{-R\sin\theta}}{R} \cdot R\, d\theta = \int_0^\pi e^{-R\sin\theta}\, d\theta.

By Jordan’s inequality sinθ2θπ\sin\theta \geq \frac{2\theta}{\pi} for θ[0,π/2]\theta \in [0, \pi/2]:

20π/2e2Rθ/πdθ=πR(1eR)0\leq 2\int_0^{\pi/2} e^{-2R\theta/\pi}\, d\theta = \frac{\pi}{R}(1 - e^{-R}) \to 0 as RR \to \infty. \blacksquare

Problem. Bound γdzz2+4\left|\int_\gamma \frac{dz}{z^2 + 4}\right| where γ\gamma is z=3|z| = 3.

On γ\gamma: z2+4z24=94=5|z^2 + 4| \geq |z|^2 - 4 = 9 - 4 = 5 (reverse triangle inequality). So 1z2+415\left|\frac{1}{z^2 + 4}\right| \leq \frac{1}{5}.

Length of γ\gamma: L=2π3=6πL = 2\pi \cdot 3 = 6\pi.

γdzz2+4156π=6π5\left|\int_\gamma \frac{dz}{z^2 + 4}\right| \leq \frac{1}{5} \cdot 6\pi = \frac{6\pi}{5}.

When ff is analytic on a connected domain and has a known antiderivative FF with F=fF' = f:

γf(z)dz=F(γ(b))F(γ(a))\int_\gamma f(z)\, dz = F(\gamma(b)) - F(\gamma(a))

This follows from the fundamental theorem of calculus applied to F(γ(t))F(\gamma(t)).

Solution

Problem. Evaluate γcoszdz\int_\gamma \cos z\, dz where γ\gamma is any path from 00 to π+i\pi + i.

Since cosz\cos z is entire with antiderivative sinz\sin z:

γcoszdz=sin(π+i)sin(0)=sin(π+i)\int_\gamma \cos z\, dz = \sin(\pi + i) - \sin(0) = \sin(\pi + i).

sin(π+i)=sinπcosh1+icosπsinh1=isinh1\sin(\pi + i) = \sin\pi\cosh 1 + i\cos\pi\sinh 1 = -i\sinh 1.

So the integral equals isinh1-i\sinh 1.

Problem. Evaluate γe2zdz\int_\gamma e^{2z}\, dz where γ\gamma is any path from 11 to ii.

Antiderivative: 12e2z\frac{1}{2}e^{2z}.

γe2zdz=12(e2ie2)\int_\gamma e^{2z}\, dz = \frac{1}{2}(e^{2i} - e^{2}).

4.7 Intuition: What Makes Complex Integration Special?

Section titled “4.7 Intuition: What Makes Complex Integration Special?”

Complex integration generalises the real line integral to paths in the complex plane. The integral γf(z)dz\int_\gamma f(z)\,dz sums the values of ff along a contour, weighted by the direction of the contour. For analytic functions (those that are complex-differentiable), the remarkable property is path-independence: if ff is analytic on a directly connected domain, the integral depends only on the endpoints, not on the path taken. This is the complex analogue of the fundamental theorem of calculus.

The Cauchy integral formula is the cornerstone: it says that the value of an analytic function at any interior point of a contour is completely determined by its values on the boundary. This is far stronger than anything in real analysis, where a function’s values on an interval tell you nothing about its values elsewhere. The residue theorem, which computes integrals by summing contributions from singularities inside the contour, is a consequence of the Cauchy integral formula. These tools make it possible to evaluate many real integrals that are difficult or impossible by real methods alone.

Mistake 1: Confusing the contour integral with the real line integral. The complex integral γf(z)dz\int_\gamma f(z)\, dz is defined as abf(γ(t))γ(t)dt\int_a^b f(\gamma(t))\gamma'(t)\, dt, which is a real integral of a complex-valued function. Do not confuse this with the real line integral γf(x)dx\int_\gamma f(x)\, dx. The complex integral depends on the path γ\gamma, not just the endpoints.

Mistake 2: Forgetting that reversing orientation changes the sign. If γ\gamma is a contour from aa to bb, then γ-\gamma is the same contour traversed from bb to aa. We have γfdz=γfdz\int_{-\gamma} f\, dz = -\int_\gamma f\, dz. Forgetting this can lead to sign errors in calculations.

Mistake 3: Assuming that the integral of an analytic function over a closed contour is always zero. The Cauchy-Goursat theorem states that the integral of an analytic function over a closed contour is zero only if the function is analytic inside the contour. If the function has singularities inside the contour, the integral may be non-zero. Use the residue theorem to compute such integrals.

Mistake 4: Misapplying the ML inequality. The ML inequality states that γf(z)dzML|\int_\gamma f(z)\, dz| \leq ML, where MM is an upper bound for f(z)|f(z)| on γ\gamma and LL is the length of γ\gamma. Forgetting to find a tight upper bound MM or miscalculating the length LL can lead to incorrect estimates.

Mistake 5: Confusing the parameterization of a contour. The value of the contour integral does not depend on the parameterization of the contour, only on the path traced out. However, different parameterizations can lead to different expressions for the integral. Always ensure that the parameterization is correct and that γ(t)\gamma'(t) is computed properly.

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A[4_Complex Integration] --> B[Key Concepts]
A --> C[Core Principles]
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