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Complex Integration

4.1 Contours

Section titled “4.1 Contours”

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.

4.2 The Complex Integral

Section titled “4.2 The Complex Integral”

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

4.3 Basic Properties

Section titled “4.3 Basic Properties”

Proposition 4.1. The complex integral is linear:

∫γ(af+bg) dz=a∫γf dz+b∫γg dz\int_\gamma (af + bg)\, dz = a\int_\gamma f\, dz + b\int_\gamma g\, dz

Proposition 4.2. Reversing orientation changes the sign:

∫−γf dz=−∫γf dz\int_{-\gamma} f\, dz = -\int_\gamma f\, dz

Proposition 4.3. Additivity over contours:

∫γ1+γ2f dz=∫γ1f dz+∫γ2f dz\int_{\gamma_1 + \gamma_2} f\, dz = \int_{\gamma_1} f\, dz + \int_{\gamma_2} f\, dz

4.4 ML Inequality

Section titled “4.4 ML Inequality”

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

∣∫γf(z) dz∣≤ML\left|\int_\gamma f(z)\, dz\right| \leq ML

Proof. ∣∫abf(γ(t))γ′(t) dt∣≤∫ab∣f(γ(t))∣∣γ′(t)∣ dt≤M∫ab∣γ′(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

4.5 Worked Examples: Contour Integrals

Section titled “4.5 Worked Examples: Contour Integrals”
Solution

Problem. Evaluate ∫γz2 dz\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 0≤t≤10 \leq t \leq 1. Then γ′(t)=1+i\gamma'(t) = 1 + i.

∫γz2 dz=∫01(t(1+i))2(1+i) dt=(1+i)3∫01t2 dt=(1+i)3⋅13\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=2i−2=−2+2i(1+i)^3 = (1+i)(1+i)^2 = (1+i)(2i) = 2i + 2i^2 = 2i - 2 = -2 + 2i.

∫γz2 dz=−2+2i3\int_\gamma z^2\, dz = \frac{-2 + 2i}{3}. ■\blacksquare

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

γ(t)=eit\gamma(t) = e^{it}, 0≤t≤2π0 \leq t \leq 2\pi, γ′(t)=ieit\gamma'(t) = ie^{it}. zˉ=e−it\bar{z} = e^{-it} on γ\gamma.

∫γzˉ dz=∫02πe−it⋅ieit dt=∫02πi dt=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}, 0≤θ≤π0 \leq \theta \leq \pi.

∫0πieiθeiθ dθ=∫0πi dθ=iπ\int_0^\pi \frac{ie^{i\theta}}{e^{i\theta}}\, d\theta = \int_0^\pi i\, d\theta = i\pi.

Problem. Evaluate ∫γz dz\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, 0≤t≤10 \leq t \leq 1: ∫01t⋅1 dt=12\int_0^1 t \cdot 1\, dt = \frac{1}{2}.

γ2(t)=1+it\gamma_2(t) = 1 + it, 0≤t≤10 \leq t \leq 1: ∫01(1+it)⋅i dt=∫01(i−t) dt=i−12\int_0^1 (1 + it) \cdot i\, dt = \int_0^1 (i - t)\, dt = i - \frac{1}{2}.

Total: 12+i−12=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

4.6 ML Inequality Applications

Section titled “4.6 ML Inequality Applications”
Solution

Problem. Use the ML inequality to show that lim⁡R→∞∫CReizz dz=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}, 0≤θ≤π0 \leq \theta \leq \pi. ∣eiz∣=∣eiR(cos⁡θ+isin⁡θ)∣=e−Rsin⁡θ|e^{iz}| = |e^{iR(\cos\theta + i\sin\theta)}| = e^{-R\sin\theta}.

∣∫CReizz dz∣≤∫0πe−Rsin⁡θR⋅R dθ=∫0πe−Rsin⁡θ dθ\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 sin⁡θ≥2θπ\sin\theta \geq \frac{2\theta}{\pi} for θ∈[0,π/2]\theta \in [0, \pi/2]:

≤2∫0π/2e−2Rθ/π dθ=πR(1−e−R)→0\leq 2\int_0^{\pi/2} e^{-2R\theta/\pi}\, d\theta = \frac{\pi}{R}(1 - e^{-R}) \to 0 as R→∞R \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+4∣≥∣z∣2−4=9−4=5|z^2 + 4| \geq |z|^2 - 4 = 9 - 4 = 5 (reverse triangle inequality). So ∣1z2+4∣≤15\left|\frac{1}{z^2 + 4}\right| \leq \frac{1}{5}.

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

∣∫γdzz2+4∣≤15⋅6π=6π5\left|\int_\gamma \frac{dz}{z^2 + 4}\right| \leq \frac{1}{5} \cdot 6\pi = \frac{6\pi}{5}.

4.7 Antiderivative Method

Section titled “4.7 Antiderivative Method”

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 ∫γcos⁡z dz\int_\gamma \cos z\, dz where γ\gamma is any path from 00 to π+i\pi + i.

Since cos⁡z\cos z is entire with antiderivative sin⁡z\sin z:

∫γcos⁡z dz=sin⁡(π+i)−sin⁡(0)=sin⁡(π+i)\int_\gamma \cos z\, dz = \sin(\pi + i) - \sin(0) = \sin(\pi + i).

sin⁡(π+i)=sin⁡πcosh⁡1+icos⁡πsinh⁡1=−isinh⁡1\sin(\pi + i) = \sin\pi\cosh 1 + i\cos\pi\sinh 1 = -i\sinh 1.

So the integral equals −isinh⁡1-i\sinh 1.

Problem. Evaluate ∫γe2z dz\int_\gamma e^{2z}\, dz where γ\gamma is any path from 11 to ii.

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

∫γe2z dz=12(e2i−e2)\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 simply 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.

4.8 Common Mistakes

Section titled “4.8 Common Mistakes”

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 ∫−γf dz=−∫γf dz\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) dz∣≤ML|\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.

Cross-References

Section titled “Cross-References”
Section titled “Related in this track”